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Brilliant Summation Contest Season-1 (Part 1)

Hi Brilliant! I've decided to start the first ever summation contest here.

The aims of the Summation contest are to improve skills in the computation of sums, to learn from each other as much as possible, and of course to have fun. Anyone here may participate in this contest.

The rules are as follows:

  • I will start by posting the first problem. If there is a user solves it, then they must post a new one.

  • You may only post a solution of the problem below the thread of problem and post your proposed problem in a new thread. Put them separately.

  • Only make substantial comment that will contribute to the discussion.

  • Make sure you know how to solve your own problem before posting it in case there is no one can answer it within 48 hours, then you must post the solution and you have a right to post another problem.

  • If the one who solves the last problem does not post his/her own problem after solving it within a day, then the one who has a right to post a problem is the last solver before him/her.

  • You are NOT allowed to post a multiple summation problem.

  • Problems must be purely of summation. They shouldn't have integrals and products in them. Solutions can follow methods that use integrals and products.

  • It is NOT compulsory to post original problems. But make sure it has not been posted on brilliant.

  • There is no restriction in the standard of summations.

Please post your solution and your proposed problem in a single new thread.

Format your post is as follows:

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**SOLUTION OF PROBLEM xxx (number of problem) :**

**[Post your solution here]**

**PROBLEM xxx (number of problem) :**

**[Post your problem here]**

The comments will be easiest to follow if you sort by "Newest":

See Part-2.

Note by Aditya Kumar
5 months, 1 week ago

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Problem 2: Prove that \[\displaystyle \sum_{n\geq 1} \frac{(-1)^n}{n^2}\left(\gamma+\psi\left(3n+\frac12\right) \right)=\frac{63}{2}\zeta(3)+\zeta(2)\log 2 -14\pi G\]

\(G\) is Catalan's constant.

This problem has been solved by Ishan Singh.

Proposition:

\[\displaystyle \sum_{n\geq 1} \frac{(-1)^n}{n^2}\left(\gamma+\psi\left(kn+\frac12\right) \right)=\frac72k^2\zeta(3)+2\pi\sum_{j=1}^{2k-1}j\text{Cl}_2(\pi j/k+\frac{\pi}{2k}) - \pi\sum_{j=1}^{k-1}j\text{Cl}_2(2\pi j/k + \pi/k) + \zeta(2)\log 2\]

Aman Rajput · 5 months, 1 week ago

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@Aman Rajput Solution Of Problem 2 :

Proposition : \[ \sum_{n=1}^{\infty} \dfrac{H_{n} x^n}{n^2} = \operatorname{Li_{3}} (x) - \operatorname{Li}_{3} (1-x) + \ln(1-x) \operatorname{Li}_{2}(1-x) + \dfrac{1}{2} \ln x \ln^2 (1-x) + \zeta(3) \]

Proof : It is easy to see that

\[ \displaystyle \sum_{n=1}^{\infty} \dfrac{H_{n} x^n}{n} = \operatorname{Li}_{2}(x) + \dfrac{1}{2} \ln^2(1-x) \]

Dividing by \(x\) and integrating, we have,

\[ \displaystyle \sum_{n=1}^{\infty} \dfrac{H_{n} x^n}{n^2} = \int \dfrac{\operatorname{Li}_{2}(x)}{x}\mathrm{d}x + \dfrac{1}{2} \int \dfrac{\ln^2(1-x)}{x}\mathrm{d}x \]

\[ = \operatorname{Li}_{3}(x) + \dfrac{1}{2} [\ln^2(1-x) \ln x] + \int \dfrac{\ln x \ln (1-x)}{1-x}\mathrm{d}x \]

Let \(x=1-t\)

\[ \implies \displaystyle \sum_{n=1}^{\infty} \dfrac{H_{n} x^n}{n^2} = \operatorname{Li}_{3}(x) + \dfrac{1}{2} [\ln^2(1-x) \ln x] - \int \dfrac{\ln (1-t) \ln t}{t} \mathrm{d}t \]

Now,

\[\int \dfrac{\ln (1-t) \ln t}{t} \mathrm{d}t = \ln t \operatorname{Li}_{2}(t) - \int \dfrac{ \operatorname{Li}_{2}(t) }{t}\mathrm{d}t \]

\[ = \ln t \operatorname{Li}_{2}(t) - \operatorname{Li}_{3}(t) \]

Substituting back, we have,

\[ \displaystyle \sum_{n=1}^{\infty} \dfrac{H_{n} x^n}{n^2} = \operatorname{Li_{3}} (x) - \operatorname{Li}_{3} (1-x) + \ln(1-x) \operatorname{Li}_{2}(1-x) + \dfrac{1}{2} \ln x \ln^2 (1-x) + C\]

Putting \(x=0\), we get \(C=\zeta(3)\), thus,

\[ \sum_{n=1}^{\infty} \dfrac{H_{n} x^n}{n^2} = \operatorname{Li_{3}} (x) - \operatorname{Li}_{3} (1-x) + \ln(1-x) \operatorname{Li}_{2}(1-x) + \dfrac{1}{2} \ln x \ln^2 (1-x) + \zeta(3) \quad \square \]

Now,

\[\displaystyle \text{S} = \sum_{n = 1}^{\infty} \frac{(-1)^n}{n^2}\left(\gamma+\psi\left(3n+\frac{1}{2} \right) \right) = -2\ln 2 \sum_{n = 1}^{\infty} \dfrac{(-1)^n}{n^2} + \sum_{n = 1}^{\infty} \sum_{k=1}^{3n} \dfrac{2}{2k-1} \]

Also,

\[\sum_{k=1}^{3n} \dfrac{1}{2k-1} = \sum_{k=1}^{6n} \dfrac{1}{k} - \sum_{k=1}^{3n}\dfrac{1}{2k}\]

\[ \sum_{k=1}^{3n} \dfrac{1}{2k-1} = H_{6n} - \dfrac{1}{2} H_{3n} \]

\[ \implies \text{S} = \zeta(2) \ln 2 + \sum_{n = 1}^{\infty} \dfrac{(-1)^n}{n^2} [2 H_{6n} - H_{3n}] \]

Using cube roots and sixth roots of unity on the proposition, we have,

\[ \sum_{n = 1}^{\infty} \dfrac{(-1)^n}{n^2} [2 H_{6n} - H_{3n}] = \dfrac{63}{2} \zeta(3) -14 \pi G \]

\[\implies \text{S} = \boxed{\zeta(2) \ln 2 + \dfrac{63}{2} \zeta(3) -14 \pi G}\] Ishan Singh · 5 months ago

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PROBLEM 13:

Show that \[ \sum_{n=1}^\infty \frac{\big(H_n^{(2)}\big)^2}{2^n} \; = \; \tfrac{1}{360}\pi^4 - \tfrac16\pi^2\ln^22 + \tfrac16\ln^42 + 2\mathrm{Li}_4(\tfrac12) + \zeta(3)\ln2 \]

This problem has been solved by Aditya Kumar Mark Hennings · 5 months ago

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@Mark Hennings Solution to Problem 13:

Just to see how nasty the last integral in this problem can get, here are the associated indefinite integrals!

Start with \[ \begin{array}{rcl} B(x) & = & \displaystyle \sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2}x^{n+1} \; = \; \int_0^x \left(\int_0^u \sum_{n=1}^\infty H_n^{(2)}v^n\,dv\right)\,\frac{du}{u} \\ & = & \displaystyle \int_0^x \left(\int_0^u \frac{\mathrm{Li}_2(v)}{1-v}\,dv\right)\,\frac{du}{u} \; = \; \int_0^x \left(\int_v^x \frac{\mathrm{Li}_2(v)}{u(1-v)}\,du\right)\,dv \\ & = & \displaystyle \int_0^x \ln\big(\tfrac{x}{v}\big) \frac{\mathrm{Li}_2(v)}{1-v}\,dv \end{array} \] Now (differentiate back to check): \[ \begin{array}{rcl} \displaystyle \int_0^x \frac{\mathrm{Li}_2(v)}{1-v}\,dv & = & \displaystyle -\tfrac13\pi^2 \ln(1 - x) + \ln^2(1 - x) \ln x + \ln(1 - x)\mathrm{Li}_2(x) + 2\mathrm{Li}_3(1 - x) - 2 \zeta(3) \\ \displaystyle \int_0^x \frac{\mathrm{Li}_2(v)}{1-v}\ln v\,dv & = & \displaystyle \tfrac{1}{45}\pi^4 - \tfrac16 \pi^2 \ln^2(1 - x) + \tfrac14 \ln^4(1 - x) + \tfrac12\ln^2(1 - x) \ln^2x \\ & & \displaystyle + (\ln^2(1 - x) - \ln(1 - x) \ln x + \ln^2x) \mathrm{Li}_2(x) - \tfrac12 \mathrm{Li}_2(x)^2 \\ & & \displaystyle + \ln^2\left(\frac{x}{1-x}\right) \mathrm{Li}_2\left(-\frac{x}{1-x}\right) + 2 \ln(1 - x) \mathrm{Li}_3(1 - x) - 2 \ln x \mathrm{Li}_3(x) \\ & & \displaystyle + 2 \ln(1 - x) \mathrm{Li}_3\left(-\frac{x}{1-x}\right) - 2 \ln x \mathrm{Li}_3\left(-\frac{x}{1 - x}\right) - 2 \mathrm{Li}_4(1 - x) \\ & & \displaystyle + 2 \mathrm{Li}_4(x) + 2 \mathrm{Li}_4\left(-\frac{x}{1 - x}\right) \end{array} \] and hence \[ B(\tfrac12) \; = \; \tfrac{1}{1440}\pi^4 - \tfrac{1}{24}\pi^2 \ln^22 + \tfrac{1}{24}\ln^42 + \tfrac14\zeta(3)\ln 2 \] Now consider \[ \begin{array}{rcl} \displaystyle A(x) & = & \displaystyle \sum_{n=1}^\infty \big(H_n^{(2)}\big)^2x^n \; =\; \sum_{n=1}^\infty \left(H_{n-1}^{(2)} + \tfrac{1}{n^2}\right)^2 x^n \\ & = & \displaystyle\sum_{n=1}^\infty \sum_{n=1}^\infty \big(H_n^{(2)}\big)^2 x^{n+1} + 2\sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2} x^{n+1} + \mathrm{Li}_4(x) \\ & =& \displaystyle xA(x) + 2 B(x) + \mathrm{Li}_4(x) \end{array} \] so that \[ \begin{array}{rcl} \displaystyle \sum_{n=1}^\infty \frac{\big(H_n^{(2)}\big)^2}{2^n} & = & \displaystyle A(\tfrac12) \; =\; 4B(\tfrac12) + 2\mathrm{Li}_4(\tfrac12) \\ & = & \displaystyle \tfrac{1}{360}\pi^4 - \tfrac16\pi^2\ln^22 + \tfrac16\ln^42 + 2\mathrm{Li}_4(\tfrac12) + \zeta(3)\ln2 \end{array} \] as required. Mark Hennings · 5 months ago

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@Mark Hennings Can you please help me with this query which is related to the above integrals? Ishan Singh · 5 months ago

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@Mark Hennings The main integral to evaluate here is \( \displaystyle \int \dfrac{\ln^2(1-x) \ln x}{x} \mathrm{d}x\). Rest of the integrals follow from this using Euler's Reflection Formula and/or IBP. The integral \( \displaystyle \int \dfrac{\operatorname{Li}_{2}(x)}{1-x} \mathrm{d}x\) is evaluated in my solution of Problem 2. Ishan Singh · 5 months ago

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@Mark Hennings I was wondering whether we can extend the result to other weights too? Ishan Singh · 5 months ago

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@Mark Hennings I have solved the indefinite integrals by using Landen's Identity. Ishan Singh · 5 months ago

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@Mark Hennings Solution to Problem 13:

Consider the function: \(\displaystyle f\left( x \right) =\sum _{ n=1 }^{ \infty }{ { \left( { H }_{ n }^{ (2) } \right) }^{ 2 }{ x }^{ n } } \).

\[f\left( x \right) =\sum _{ n=1 }^{ \infty }{ { \left( { H }_{ n-1 }^{ (2) }+\frac { 1 }{ { n }^{ 2 } } \right) }^{ 2 }{ x }^{ n } } \]

\[f\left( x \right) =\sum _{ n=1 }^{ \infty }{ { \left( { H }_{ n-1 }^{ (2) } \right) }^{ 2 }{ x }^{ n } } +{ \text{Li} }_{ 4 }(x)+2\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n-1 }^{ (2) }{ x }^{ n } }{ { n }^{ 2 } } } \]

On changing the summation index from \(n\) to \(n+1\), we get:

\[f\left( x \right) =xf\left( x \right) +{ \text{Li} }_{ 4 }(x)+2\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) }{ x }^{ n+1 } }{ { \left( n+1 \right) }^{ 2 } } } \quad \quad (...1)\]

We have to find \(\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) }{ x }^{ n+1 } }{ { \left( n+1 \right) }^{ 2 } } } \)

\[\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) }{ x }^{ n+1 } }{ { \left( n+1 \right) }^{ 2 } } } =\int _{ 0 }^{ x }{ \left[ \sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) }{ x }^{ n } }{ { \left( n+1 \right) } } } \right] dx } \quad \quad (...2)\]

Now we have to find \(\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) }{ x }^{ n } }{ { \left( n+1 \right) } } } \)

I'll use the identity: \(\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) }{ x }^{ n } }{ n } } =\int _{ 0 }^{ x }{ \frac { { \text{Li} }_{ 2 }\left( t \right) }{ t\left( 1-t \right) } dt } \).

Now, we differentiate w.r.t. x and get:

\[\sum _{ n=1 }^{ \infty }{ { H }_{ n }^{ (2) }{ x }^{ n-1 } } =\frac { { \text{Li} }_{ 2 }\left( x \right) }{ x\left( 1-x \right) } \]

We multiply both sides by x:

\[\sum _{ n=1 }^{ \infty }{ { H }_{ n }^{ (2) }{ x }^{ n } } =\frac { {\text{Li} }_{ 2 }\left( x \right) }{ \left( 1-x \right) } \]

Now, we integrate both sides w.r.t. \(x\) from 0 to x:

\[\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) }{ x }^{ n+1 } }{ n+1 } } =2{ \text{Li} }_{ 3 }\left( 1-x \right) -2{ \text{Li} }_{ 2 }\left( 1-x \right) \ln { \left( 1-x \right) } -{ \text{Li} }_{ 2 }\left( x \right) \ln { \left( 1-x \right) } -\ln { \left( x \right) } { \left( \ln { \left( 1-x \right) } \right) }^{ 2 }-2\zeta \left( 3 \right) \]

On dividing both sides by \(x\):

\[\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) }{ x }^{ n } }{ n+1 } } =\frac { 2{ \text{Li} }_{ 3 }\left( 1-x \right) }{ x } -\frac { 2{ \text{Li} }_{ 2 }\left( 1-x \right) \ln { \left( 1-x \right) } }{ x } -\frac { {\text{Li} }_{ 2 }\left( x \right) \ln { \left( 1-x \right) } }{ x } -\frac { \ln { \left( x \right) } { \left( \ln { \left( 1-x \right) } \right) }^{ 2 } }{ x } -\frac { 2\zeta \left( 3 \right) }{ x } \]

Now plugging this in eqn \(2\), we get:

\[\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) }{ x }^{ n+1 } }{ { \left( n+1 \right) }^{ 2 } } } =\int _{ 0 }^{ x }{ \left[ \frac { 2{ \text{Li} }_{ 3 }\left( 1-x \right) }{ x } -\frac { 2{ \text{Li} }_{ 2 }\left( 1-x \right) \ln { \left( 1-x \right) } }{ x } -\frac { { \text{Li} }_{ 2 }\left( x \right) \ln { \left( 1-x \right) } }{ x } -\frac { \ln { \left( x \right) } { \left( \ln { \left( 1-x \right) } \right) }^{ 2 } }{ x } -\frac { 2\zeta \left( 3 \right) }{ x } \right] dx } \]

Now plugging this in eqn \(1\), we get:

\[f\left( x \right) =\frac { { \text{Li} }_{ 4 }\left( x \right) }{ 1-x } +\frac { 2 }{ 1-x } \int _{ 0 }^{ x }{ \left[ \frac { 2{ \text{Li} }_{ 3 }\left( 1-x \right) }{ x } -\frac { 2{\text{Li} }_{ 2 }\left( 1-x \right) \ln { \left( 1-x \right) } }{ x } -\frac { { \text{Li} }_{ 2 }\left( x \right) \ln { \left( 1-x \right) } }{ x } -\frac { \ln { \left( x \right) } { \left( \ln { \left( 1-x \right) } \right) }^{ 2 } }{ x } -\frac { 2\zeta \left( 3 \right) }{ x } \right] dx } \]

Now we take \(x=\frac { 1 }{ 2 } \) and compute the integral. Then we finally get:

\[\boxed{\displaystyle \sum_{n=1}^\infty \frac{\big(H_n^{(2)}\big)^2}{2^n} \; = \; \tfrac{1}{360}\pi^4 - \tfrac16\pi^2\ln^22 + \tfrac16\ln^42 + 2\mathrm{Li}_4(\tfrac12) + \zeta(3)\ln2}\]

The final integral is left to the reader's exercise. I didn't post it as it would've made this solution too long. Aditya Kumar · 5 months ago

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@Aditya Kumar I'm impressed. The last integral is a brute. Mark Hennings · 5 months ago

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@Mark Hennings In fact we can even generalize the result by finding the indefinite integral of the last one and in turn the generating function \( \displaystyle \sum_{n=1}^{\infty} (H_{n} ^{(2)})^2 x^n \) . The solution gets too long though. Ishan Singh · 5 months ago

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Problem 12:

Prove the following for \(|x| < \dfrac{4}{27}\).

\[\sum_{n=0}^\infty \binom{3n}{n}x^n = \frac{2\cos(\frac{1}{3} \sin^{-1}(\frac{3\sqrt{3x}}{2}))}{\sqrt{4-27x}}\]

This problem has been solved by Mark Hennings and Ishan Singh. Jake Lai · 5 months ago

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@Jake Lai Solution To Problem 12 :

Proposition : \[ f(a,z) = \int_{0}^{\infty} \dfrac{x^z}{x^2 +2ax +1} \mathrm{d}x = \dfrac{\pi}{\sin \pi z} \dfrac{\sin((1-z) \cos^{-1}(a))}{\sin(\cos^{-1}(a))}\]

Proof : Note that,

\[ \sum_{n=0}^{\infty}{{U}_{n}(a) {(-x)}^{n}} = \dfrac{1}{x^2 +2ax+1} \]

where \( U_{n} (x) \) is the Chebyshev Polynomial of the second kind.

\[ \implies \sum_{n=0}^{\infty}{{U}_{n}(a) \Gamma(n+1) \frac{{(-x)}^{n}}{n!}} = \frac{1}{x^2 +2ax +1} \]

Using Ramanujan Master Theorem, we have,

\[ f(a,z) = \dfrac{\pi}{\sin \pi z} U_{-s} (a) \]

\[ = \dfrac{\pi}{\sin \pi z} \dfrac{\sin((1-z) \cos^{-1}(a))}{\sin(\cos^{-1}(a))} \quad \square \]

Now, using Gamma Triplication Formula,

\[\text{S} = \sum_{n=0}^{\infty} \dbinom{3n}{n} x^n = \dfrac{\sqrt{3}}{2 \pi} \int_0^\infty \frac{x^{\frac{1}{3}}(1 + x^2)}{(x^2 + 2ax + 1)(x^2 - 2ax + 1)} \mathrm{d}x \]

where \(a = \dfrac{3}{2} \sqrt{3x} \)

Using Partial Fraction and the Proposition, we have,

\[\text{S} = \dfrac{\sqrt{3}}{2\pi} \left[ \dfrac{1}{2} f \left( \dfrac{1}{3} , -a \right) + \dfrac{1}{2} f\left( \dfrac{7}{3} , -a \right) - \dfrac{1}{4a} f\left( \dfrac{4}{3} , -a \right) - \dfrac{1}{4a} f\left( \dfrac{10}{3} , -a \right) + \dfrac{1}{2} f \left( \dfrac{1}{3} , a \right) + \dfrac{1}{2} f\left( \dfrac{7}{3} , a \right) + \dfrac{1}{4a} f\left( \dfrac{4}{3} , a \right) + \dfrac{1}{4a} f\left( \dfrac{10}{3} , a \right)\right] \]

After simplification, we have,

\[ \text{S} = \dfrac{2\cos(\frac{1}{3} \sin^{-1}(\frac{3\sqrt{3x}}{2}))}{\sqrt{4-27x}} \quad \square \] Ishan Singh · 5 months ago

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@Jake Lai Solution to Problem 12:

By the multiplication formula for the Gamma function, \[ (3n)! \; = \; \Gamma(3n+1) \; = \; \tfrac{1}{2\pi} 3^{2n+\frac12} \Gamma(n+\tfrac13)\Gamma(n+\tfrac23)\Gamma(n+1) \] and hence \[ {3n \choose n} \; = \; \frac{3^{3n+\frac12}}{2\pi} \times \frac{\Gamma(n+\frac13)\Gamma(n+\frac23)}{\Gamma(2n+1)} \; = \;\frac{3^{2n+\frac12}}{2\pi}B(n+\tfrac23,n+\tfrac13) \] Thus \[ \begin{array}{rcl}\displaystyle S \; =\; \sum_{n=0}^\infty {3n \choose n}x^n & = & \displaystyle\frac{\sqrt{3}}{2\pi}\sum_{n=0}^\infty 3^{2n} x^n \int_0^1 u^{n-\frac13}(1-u)^{n-\frac23}\,du \\ & = & \displaystyle \frac{\sqrt{3}}{2\pi} \int_0^1 \frac{u^{-\frac13}(1-u)^{-\frac23}}{1 - 27xu(1-u)}\,du \; =\; \frac{\sqrt{3}}{2\pi}\int_0^1 \frac{u^{-\frac13}(1-u)^{-\frac23}}{1 - 4\sin^23\alpha\, u(1-u)}\,du \end{array} \] where \(0 < \alpha < \tfrac16\pi\) is such that \(\sin3\alpha = \tfrac12\sqrt{27x}\). The substitutions \(u = \sin^2\theta\) and then \(x = \tan\theta\) yield \[ \begin{array}{rcl} S & = & \displaystyle \frac{\sqrt{3}}{2\pi} \int_0^{\frac12\pi} \frac{\sin^{-\frac23}\theta \cos^{-\frac43}\theta \times 2\sin\theta \cos\theta}{1 - \sin^23\alpha\, \sin^22\theta}\,d\theta \; = \; \displaystyle \frac{\sqrt{3}}{\pi}\int_0^{\frac12\pi} \frac{\tan^{\frac13}\theta}{1 - \sin^23\alpha\, \sin^22\theta}\,d\theta \\ & = & \displaystyle\frac{\sqrt{3}}{\pi} \int_0^\infty \frac{x^{\frac13}(1 + x^2)}{(1 + x^2)^2 - 4\sin^23\alpha x^2}\,dx \; = \; \displaystyle\frac{\sqrt{3}}{\pi} \int_0^\infty \frac{x^{\frac13}(1 + x^2)}{(x^2 + 2x \sin3\alpha + 1)(x^2 - 2x \sin3\alpha + 1)}\,dx \\ & = & \displaystyle \frac{\sqrt{3}}{\pi} \int_0^\infty x^{\frac13}f(x)\,dx \; = \; \frac{\sqrt{3}}{\pi}I \end{array} \] where the meromorphic function \[ f(z) \; =\; \frac{z^2 + 1}{(z^2 + 2z \sin3\alpha + 1)(z^2 - 2z \sin3\alpha + 1)} \] has simple poles at each of the points in \(\mathcal{R} \,=\, \{ \pm ie^{3i\alpha}, \pm i e^{-3i\alpha}\}\).

Cutting the complex plane along the positive real axis, and integrating around the "keyhole contour" \(\gamma_1 + \gamma_2 - \gamma_3 - \gamma_4\) where

  • \(\gamma_1\) is the straight line segment from \(\epsilon\) to \(R\), just above the cut,
  • \(\gamma_2\) is the circular segment \(z = Re^{i\theta}\) for \(0 \le \theta \le 2\pi\),
  • \(\gamma_3\) is the straight line segment from \(\epsilon e^{2\pi i}\) to \(R e^{2 \pi i}\), just below the cut,
  • \(\gamma_4\) is the circular segment \(z = \epsilon e^{i\theta}\) for \(0 \le \theta \le 2\pi\)

for any \(0 < \epsilon < 1 < R\), then \[ \left(\int_{\gamma_1} + \int_{\gamma_2} - \int_{\gamma_3} - \int_{\gamma_4}\right) z^{\frac13} f(z\,dz \; = \; 2\pi i \sum_{u \in \mathcal{R}} \mathrm{Res}_{z=u} z^{\frac13}f(z) \] so that \[ (1 - \omega)\int_\epsilon^R x^{\frac13}f(x)\,dx + \left(\int_{\gamma_2} - \int_{\gamma_4}\right) z^{\frac13}f(z)\,dz \; = \; 2\pi i\sum_{u \in \mathcal{R}} \mathrm{Res}_{z=u} z^{\frac13}f(z) \] Letting \(\epsilon \to 0\) and \(R \to \infty\), we obtain \[ (1 - \omega)I \; = \; 2\pi i\sum_{u \in \mathcal{R}} \mathrm{Res}_{z=u} z^{\frac13}f(z) \] and hence \[ I \; = \; \frac{\pi}{\sqrt{3}(2\cos2\alpha - 1)} \; = \; \frac{\pi}{\sqrt{3}(4\cos^2\alpha - 3)} \; = \; \frac{\pi \cos\alpha}{\sqrt{3}\cos3\alpha} \] so that \[ S \; = \; \frac{\cos\alpha}{\cos3\alpha} \; = \; \frac{2\cos\left(\frac13\sin^{-1}\left(\frac{3\sqrt{3x}}{2}\right)\right)}{\sqrt{4 - 27x}} \] as required. Mark Hennings · 5 months ago

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PROBLEM 11

Evaluate \[ \sum_{n=1}^\infty (-1)^n \frac{H_n}{2n+1} \;.\]

This problem has been solved by Jake Lai and Aditya Kumar. Mark Hennings · 5 months ago

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@Mark Hennings It is well-known that \(\displaystyle -\frac{\ln(1-x)}{1-x} = \sum_{n=1}^\infty H_nx^n\). So

\[\begin{align} \sum_{n=1}^\infty \frac{(-1)^nH_n}{2n+1} &= \int_0^1 \sum_{n=1}^\infty H_n(-x^2)^n \ dx \\ &= \int_0^1 \frac{-\ln(1+x^2)}{1+x^2} \ dx & (\text{Substitute } u = \tan^{-1} x) \\ &= \int_0^{\pi/4} -\ln(1+\tan^2 u) \ du \\ &= \int_0^{\pi/4} -\ln(\sec^2 u) \ du \\ &= \int_0^{\pi/4} 2 \ln \cos u \ du \\ &= G - \frac{\pi}{2} \ln 2 \end{align}\]

where G is Catalan's constant. Jake Lai · 5 months ago

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@Jake Lai Reversing the order of integration and summation to obtain the second line of your argument takes a bit of justification. The generating function series is not monotonic, nor is it uniformly bounded by an integrable function, so the result is not automatic.

You need to integrate from \(0\) to \(x < 1\), obtaining \[ \sum_{N=1}^\infty \frac{(-1)^n H_n}{2n+1} x^{2n+1} \; =\; 2\int_0^{\tan^{-1}x} \ln \cos u\,du \] and then let \(x\) tend to \(1-\), using Abel's Theorem to get the answer (having used the Alternating Series Test to prove that the series \(\sum_n (-1)^n\frac{H_n}{2n+1}\) is convergent).

Anyway, you are up for the next question! Mark Hennings · 5 months ago

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@Jake Lai Alternate Method:

\[\sum _{ n=1 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n } }{ 2n+1 } { H }_{ n } } =\sum _{ n=1 }^{ \infty }{ \frac { 2{ H }_{ 2n } }{ 4n+1 } } -\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n } }{ 2n+1 } } \]

Then we can use appropriate generating functions for evaluating it. The integrals have to be carefully handled in order to get the solution. Aditya Kumar · 5 months ago

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Problem 6 :

Prove That

\[\sum_{r=1}^n \frac{(-1)^{r-1}}r \dbinom{n}{r} =H_{n} \]

Notation : \(H_{n}\) denotes the Harmonic Number.

This problem has been solved by Aditya Kumar. Ishan Singh · 5 months ago

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@Ishan Singh Consider the binomial expansion:

\[\sum_{r=0}^n (-x)^{r} \dbinom{n}{r} =\left(1-x\right)^{n} \]

We'll write it as:

\[\sum_{r=1}^n (-x)^{r} \dbinom{n}{r} =\left(1-x\right)^{n}-1 \]

On dividing both the sides by \(-x\), we get:

\[\sum_{r=1}^n (-x)^{r-1} \dbinom{n}{r} =\frac{1-\left(1-x\right)^{n}}{x} \]

On integrating both the sides w.r.t. \(x\) from 0 to 1, we get:

\[\sum_{r=1}^n \frac{(-1)^{r-1}}r \dbinom{n}{r} =\int _{ 0 }^{ 1 }{ \frac { 1-{ \left( 1-x \right) }^{ n } }{ x } dx } =\int _{ 0 }^{ 1 }{ \frac { 1-{ x }^{ n } }{ 1-x } dx } ={ H }_{ n }\\ \] Aditya Kumar · 5 months ago

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@Ishan Singh Solution To Problem 6 :

\[ \text{S} = \sum_{r=1}^{n} \dfrac{(-1)^{r-1}}{r} \dbinom{n}{r} \]

Note that, \( \displaystyle \sum_{k=r}^{n} \dbinom{k-1}{r-1} = \dbinom{n}{r} \), it follows,

\[\text{S} = \sum_{r=1}^{n} \dfrac{(-1)^{r-1}}{r} \sum_{k=r}^{n} \dbinom{k-1}{r-1} \]

Since \( \displaystyle \dbinom{k-1}{r-1} = \dfrac{r}{k} \dbinom{k}{r} \), we have,

\[ \text{S} = \sum_{r=1}^{n} \sum_{k=r}^{n} \dfrac{(-1)^{r-1}}{k} \dbinom{k}{r} \]

Interchanging order of summation, we have,

\[\text{S} = \sum_{k=1}^{n} \sum_{r=1}^{k} \dfrac{(-1)^{r-1}}{k} \dbinom{k}{r} \]

\[ = \sum_{k=1}^{n} \dfrac{1}{k} \]

\[ = H_{n} \quad \square \] Ishan Singh · 5 months ago

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Special Problem :

Evaluate

\[ \sum_{r=0}^{n} \dfrac{(-1)^r}{\dbinom{n}{r}} \] Ishan Singh · 5 months ago

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@Ishan Singh Solution to Special Problem : \[\underline{\text{Method 1}}\]

\[\text{S}=\displaystyle\sum_{r=0}^{n}\dfrac{(-1)^r}{\dbinom{n}{r}}\]

\[=\displaystyle\sum_{r=0}^{n}(-1)^r\dfrac{r!(n-r)!}{n!}\]

\[=\displaystyle\sum_{r=0}^{n}(-1)^r \dfrac{r!(n-r)!\color{blue}{(n-r+1+r+1)}}{n!}\color{red}{\times\dfrac{1}{(n+2)}}\]

\[=\displaystyle \dfrac{1}{n!(n+2)} \sum_{r=0}^{n} \left[(-1)^r r!(n-r+1)! + (-1)^r (r+1)!(n-r)! \right]\]

\[=\displaystyle \dfrac{1}{n!(n+2)} \sum_{r=0}^{n} \left[(-1)^r r!(n-r+1)! - (-1)^{r+1} (r+1)!(n-r)! \right]\]

\[\color{blue}{=\displaystyle \dfrac{1}{n!(n+2)} \sum_{r=0}^{n} \left( T_{r} - T_{r+1} \right) }\tag{*}\]

where \(T_{r}= (-1)^r r!(n-r+1)!\)

Clearly, \((*)\) is a telescoping series. Evaluating it, we have,

\[\text{S}=(-1)^nT_{n+1}+T_{0}\]

\[=(-1)^n\dfrac{(n+1)!}{n!(n+2)} + \dfrac{(n+1)!}{n!(n+2)}\]

Since \(n\) is even,

\[\color{red}{\therefore\text{S}=\boxed{2\dfrac{n+1}{n+2}}}\]


\[\underline{\text{Method 2}}\]

Consider the following Lemma.

Lemma : \[\color{brown}{\displaystyle \sum_{r=0}^k\dfrac{(-1)^r}{\dbinom{k}{r}}=0}\]

where \(k\) is an odd positive integer.

Proof : \[\text{S}=\displaystyle \sum_{r=0}^k\dfrac{(-1)^r}{\dbinom{k}{r}}\]

\[=\displaystyle \sum_{r=0}^k\dfrac{(-1)^{k-r}}{\dbinom{k}{k-r}}\]

\[\displaystyle\color{blue}{\left(\because \sum_{r=0}^{n} f(r) = \sum_{r=0}^{n} f(n-r) \right)} \]

\[= -\displaystyle \sum_{r=0}^k\dfrac{(-1)^r}{\dbinom{k}{r}}\]

\[\implies \text{S}=-\text{S}\]

\[\implies \text{S}=0\]

This completes the proof of the Lemma.

Now, since \(n\) is even, \(n+1\) is odd.

Using the Lemma, we have,

\[\color{blue}{\displaystyle \sum_{r=0}^{n+1}\dfrac{(-1)^r}{\dbinom{n+1}{r}}=0} \tag{1}\]

Let,

\[\color{red}{\text{J}=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}}}\tag{2}\]

Operating \((1)+(2)\),

\[\implies \text{J}+0=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}} + \displaystyle \sum_{r=0}^{n+1}\dfrac{(-1)^r}{\dbinom{n+1}{r}}\]

\[=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}} + \displaystyle \sum_{r=0}^{n}\dfrac{(-1)^r}{\dbinom{n+1}{r}} +\dfrac{(-1)}{\dbinom{n+1}{n+1}}\]

\[=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}} + \displaystyle \sum_{r=0}^{n}\dfrac{(-1)^{n-r}}{\dbinom{n+1}{n-r}} -1 \quad \displaystyle\color{blue}{\left(\because \sum_{r=0}^{n} f(r)=\sum_{r=0}^{n} f(n-r)\right)}\]

\[=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}} + \displaystyle \sum_{r=0}^{n}\dfrac{(-1)^{r}}{\dbinom{n+1}{r+1}} -1\]

\[=\displaystyle \sum_{r=0}^n(-1)^r\left\{\dfrac{1}{\dbinom{n}{r}} + \dfrac{r+1}{n+1} \cdot \dfrac{1}{\dbinom{n}{r}}\right\} -1 \quad \displaystyle\color{red}{\left[\because \dbinom{n}{r}=\dfrac{n}{r}\dbinom{n-1}{r-1} \ \text{&} \ \dbinom{n}{r}=\dbinom{n}{n-r} \right]}\]

\[\implies \text{J} = \displaystyle\left(\dfrac{n+2}{n+1}\right) \sum_{r=0}^{n}\dfrac{(-1)^r}{\dbinom{n}{r}}+ \dfrac{1}{n+1} \sum_{r=0}^{n}\dfrac{(-1)^r\cdot r}{\dbinom{n}{r}} -1\]

\[\implies \text{J}+1 = \displaystyle\left(\dfrac{n+2}{n+1}\right) \text{J}+ \dfrac{1}{n+1} \underbrace{\color{#66f}{\sum_{r=0}^{n}\dfrac{(-1)^r\cdot r}{\dbinom{n}{r}}}}_{\huge{= \ \text{G} \ (\text{let})}} \]

\[ \implies \text{J}+1= \displaystyle\left(\dfrac{n+2}{n+1}\right) \text{J}+ \dfrac{\text{G}}{n+1} \tag{3} \]

Now,

\[\text{G}=\displaystyle\sum_{r=0}^{n}\dfrac{(-1)^r\cdot r}{\dbinom{n}{r}}\]

\[=\displaystyle\sum_{r=0}^{n}\dfrac{(-1)^{n-r}\cdot (n-r)}{\dbinom{n}{n-r}}\]

\[=\displaystyle\sum_{r=0}^{n}\dfrac{(-1)^{r}\cdot (n-r)}{\dbinom{n}{r}}\]

\[=\displaystyle n\sum_{r=0}^{n}\dfrac{(-1)^{r}}{\dbinom{n}{r}} - \sum_{r=0}^{n}\dfrac{(-1)^{r}\cdot r}{\dbinom{n}{r}}\]

\[\implies \text{G}=n\cdot \text{J}-\text{G}\]

\[\implies\color{red}{\text{G}=\dfrac{\text{J}n}{2}} \tag{4}\]

From \((3)\ \text{&} \ (4)\),

\[\implies \text{J}+1= \displaystyle\left(\dfrac{n+2}{n+1}\right) \text{J}+ \dfrac{\text{J}n}{2(n+1)}\]

\[\color{green}{\therefore\text{J}=\boxed{2\dfrac{n+1}{n+2}}} \quad \square\] Ishan Singh · 5 months ago

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@Ishan Singh Start by writing \(\displaystyle \dbinom nr ^{-1} = (n+1) \int_0^1 u^r (1-u)^{n-r} \, du \).

Multiply both sides by \( (-1)^r \) and adding over \(r\) gives:

\[ \begin{eqnarray} \sum_{r=0}^n\dfrac{(-1)^r}{ \binom{n}{r}} &=& (n+1) \int_0^1 \sum_{r=0}^n (-u)^r (1-u)^{n-r} \, du \\ &=& (n+1) \int_0^1 \left [(1-u)^{n+1} + (-1)^n u^{n+1} \right ] \, du \\ &=& \dfrac{n+1}{n+2} ( 1 + (-1)^n ) = \begin{cases} 0 , \qquad \quad n \text{ odd} \\ \dfrac{n+1}{2n+4} , \quad n \text{ even} \end{cases} \end{eqnarray} \] Pi Han Goh · 5 months ago

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@Pi Han Goh This is the same method which Ishan talked about. Thanks for posting! Aditya Kumar · 5 months ago

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@Ishan Singh Solution to Special Problem:

\[ \begin{array}{rcl} \displaystyle \sum_{r=0}^n \frac{(-1)^r}{{n \choose r}} & = & \displaystyle \frac{1}{n!}\sum_{r=0}^n (-1)^r r! (n-r)! \; = \; \frac{1}{n!}\sum_{r=0}^n \int_0^\infty \int_0^\infty x^r y^{n-r} e^{-x-y}\,dx\, dy \\ & = & \displaystyle \frac{1}{n!}\int_0^\infty \int_0^\infty \left(\sum_{r=0}^n (-1)^r x^r y^{n-r}\right) e^{-x-y}\,dx\,dy \\ & = & \displaystyle \frac{1}{n!}\int_0^\infty \int_0^\infty \frac{y^{n+1} + (-1)^n x^{n+1}}{x+y} e^{-x-y}\,dx\,dy \\ & = & \displaystyle \frac{1 + (-1)^n}{n!} \int_0^\infty \int_0^\infty \frac{x^{n+1}}{x+y} e^{-x-y}\,dx\,dy \end{array} \] by symmetry. Note that this sum equals \(0\) when \(n\) is odd. Thus \[ \begin{array}{rcl} \displaystyle \sum_{r=0}^n \frac{(-1)^r}{{n \choose r}} & = & \displaystyle \frac{1 + (-1)^n}{n!} \int_0^\infty \int_0^u \frac{x^{n+1}}{u} e^{-u}\,dx\,du \\ & = & \displaystyle \frac{1 + (-1)^n}{n! (n+2)} \int_0^\infty u^{n+1} e^{-u}\,du \; = \; \frac{(1 + (-1)^n) (n+1)!}{n! (n+2)} \\ & = & \displaystyle (1 + (-1)^n)\,\frac{n+1}{n+2} \end{array} \] Mark Hennings · 5 months ago

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@Mark Hennings (+1) Nice Solution! I've posted \(2\) of my \(3\) methods, the third one used Beta Function (which I haven't posted). Ishan Singh · 5 months ago

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Problem 3:
Evaluate :

\[\sum_{r=0}^n \left[\binom{n}{r}\cdot\sin rx \cdot \cos (n-r)x\right]\]

This problem has been solved by Deeparaj Bhat and Aditya Kumar. Ishan Singh · 5 months ago

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@Ishan Singh Call the given sum \(S\).

Then, by using the definition of trigonometric functions via complex exponentials, we have \[ S = \frac{1}{4i} \sum_{r=0}^{n} {n \choose r} \left( e^{inx}-e^{-inx} + e^{(2r-n)ix} - e^{-(2r-n)ix} \right) \]

Now, using binomial theorem and a bit of simplification, we get that \[ S=2^{n-1} \sin (nx) \]

PS: I don't have good summation problems. So, anyone can post in my behalf. :) Deeparaj Bhat · 5 months ago

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@Deeparaj Bhat Alternate Solution:

We use the identity: \(\displaystyle \sum _{ r=0 }^{ n }{ f\left( r \right) } =\sum _{ r=0 }^{ n }{ f\left( n-r \right) } \)

We take:

\[ S=\sum_{r=0}^n \left[\binom{n}{r}\cdot\sin rx \cdot \cos (n-r)x\right] \quad \quad (1)\]

On applying the identity, we get

\[ S=\sum_{r=0}^n \left[\binom{n}{n-r}\cdot\sin (n-rx) \cdot \cos rx\right] \quad \quad (2)\]

On adding equations 1 and 2, we get:

\[2S=\sum_{r=0}^n \binom{n}{r} \sin nx \]

Therefore, \(S=2^{n-1} \sin nx\) Aditya Kumar · 5 months ago

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SOLUTION OF PROBLEM 1 :

\[ \begin{array}{rcl} \displaystyle \sum_{n \ge 0} (-1)^n \psi^{(3)}(2n+2) & = & \displaystyle 3!\sum_{n \ge 0} (-1)^n\sum_{k \ge 0}\frac{1}{(k+2n+2)^4} \\ & = & \displaystyle 6\sum_{n,k \ge 0} \frac{(-1)^n}{(k+2n+2)^4} \; = \; 6\sum_{K \ge 0} \left(\sum_{n=0}^{\lfloor K/2 \rfloor} (-1)^n\right) \frac{1}{(K+2)^4} \\ & = & \displaystyle 6\sum_{K \ge 0\,,\,K \equiv 0,1 \; (4)} \frac{1}{(K+2)^4} \\ & = & \displaystyle \tfrac38\sum_{K \ge 0} \frac{1}{(2K+1)^4} + 6\sum_{K \ge 0} \frac{1}{(4K+3)^4} \\ & = & \displaystyle \tfrac38\big(1 - \tfrac{1}{16}\big)\tfrac{1}{90}\pi^4 + \tfrac{3}{128}\zeta(4,\tfrac32) \\ & = & \tfrac{1}{256}\pi^4 + \tfrac{1}{256}\psi^{(3)}\big(\tfrac34\big) \end{array} \] Mark Hennings · 5 months ago

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Problem 8: Show that \[ \sum_{n=0}^\infty \frac{1}{n^4 + 4} \; = \; \tfrac18(1 + \pi \mathrm{coth}\,\pi) \]

This problem has been solved by Ishan Singh and Aditya Kumar. Mark Hennings · 5 months ago

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@Mark Hennings Solution To Problem 8 :

Lemma : \[\sum_{n=0}^{\infty} \dfrac{1}{n^2+1} = \dfrac{1}{2} (1 + \pi \coth(\pi)) \]

Proof : Let \[\text{S} = \sum_{n=0}^{\infty} \dfrac{1}{n^2+1}\]

\[\implies \text{S} = \dfrac{1}{2i} \sum_{n=0}^{\infty} \left(\dfrac{1}{n-i} - \dfrac{1}{n+i}\right) \]

Using Digamma Regularization, we have,

\[\text{S} = \dfrac{1}{2i} [\psi(i) - \psi(-i)] \]

Simplifying using Digamma Reflection Formula, we have,

\[\text{S} = \dfrac{1}{2} (1 + i \pi \cot(i\pi)) \]

\[ = \dfrac{1}{2} (1 + \pi \coth(\pi)) \quad \square \]

Now,

\[ \text{J} = \sum _{n=0}^{\infty} \dfrac{1}{n^4+4} \]

\[ = \sum _{n=0}^{\infty} \dfrac{1}{(n^2+2n+2)(n^2-2n+2)} \]

Using Partial Fraction,

\[ \text{J} = \dfrac{1}{4} \sum _{n=0}^{\infty} \dfrac{1}{(n^2-2n+2)} + \dfrac{1}{8} \sum _{n=0}^{\infty} \left( \dfrac{n+2}{n^2+2n +2} - \dfrac{n}{n^2-2n+2} \right) \]

\[ =\dfrac{1}{8} + \dfrac{1}{4} \sum _{n=1}^{\infty} \dfrac{1}{(n-1)^2+1} + \dfrac{1}{8} \sum _{n=0}^{\infty} \left( \dfrac{n+2}{n^2+2n +2} - \dfrac{n}{n^2-2n+2} \right) \]

\[ =\dfrac{1}{8} + \dfrac{1}{4} \sum _{n=0}^{\infty} \dfrac{1}{n^2+1} +\dfrac{1}{8} \sum _{n=0}^{\infty} \left( T_{n+2} - T_{n} \right) \]

where \(T_{n} = \dfrac{n}{n^2-2n+2} \)

Note that the latter sum telescopes. Evaluating it, we have,

\[\sum _{n=0}^{\infty} \left( T_{n+2} - T_{n} \right) = -1\]

\[ \implies \text{J} = \dfrac{1}{4} \sum _{n=0}^{\infty} \dfrac{1}{n^2+1} \]

\[ = \dfrac{1}{8} (1 +\pi \coth(\pi)) \quad \square \] Ishan Singh · 5 months ago

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@Mark Hennings Claim: \( \displaystyle \sum_{a=1}^\infty \frac {8b^4}{a^4+4b^4} = b \pi \coth (b \pi) - 1 \)

Proof: Read Brian Chen's comment here.

The result follows. Pi Han Goh · 5 months ago

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@Mark Hennings Solution to Problem 8:

For any positive integer \(N>2\), let \(C_N\) be the positively oriented square contour with corners at \(\pm(N+\tfrac12) \pm i(N+\tfrac12)\). Let \(R\) be the set \(\{1+i,1-i,-1-i,-1+i\}\). Then, since \(\pi \cot\pi z\) is a meromorphic function, periodic of period \(1\), with a simple pole of residue \(1\) at each integer, we see that \[ \begin{array}{rcl} \displaystyle \frac{1}{2\pi i}\int_{C_N} \frac{\pi \cot \pi z}{z^4 + 4}\,dz & = & \displaystyle \sum_{n=-N}^N \mathrm{Res}_{z=n}\frac{\pi \cot\pi z}{z^4+4} + \sum_{u \in R}\mathrm{Res}_{z=u}\frac{\pi \cot\pi z}{z^4+4} \\ & = & \displaystyle -\tfrac14 + 2\sum_{n=0}^N \frac{1}{n^4 + 4} + \sum_{u \in R} \mathrm{Res}_{z=u} \frac{\pi \cot\pi z}{z^4 + 4} \\ & = & \displaystyle -\tfrac14 + 2\sum_{n=0}^N \frac{1}{n^4 + 4} - \tfrac{1}{16}\pi \sum_{u \in R} u \cot \pi u \end{array} \] Since \[ \cot \pi(\epsilon + \eta i) \; = \; - i \mathrm{coth}\,\pi\eta \] for \(\epsilon,\eta \in \{1,-1\}\), it follows that \[ \frac{1}{2\pi i}\int_{C_N} \frac{\pi \cot \pi z}{z^4 + 4}\,dz \; = \; -\tfrac14 + 2\sum_{n=0}^N \frac{1}{n^4 + 4} - \tfrac14\pi \mathrm{coth}\,\pi \] Since \(\pi \cot \pi z\) is uniformly bounded on \(C_N\) for all integers \(N\), we deduce that \[ \lim_{N \to \infty} \frac{1}{2\pi i}\int_{C_N} \frac{\pi \cot \pi z}{z^4 + 4}\,dz \; = \; 0 \] which gives the result. Mark Hennings · 5 months ago

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@Mark Hennings \[S=\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { n }^{ 4 }+4 } } \]

On splitting it using partial fractions, we get:

\[S=\frac { 1 }{ 4i } \left\{ \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 }-2i } } -\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 }+2i } } \right\} \]

We use the identity: \(\displaystyle\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}\). See the proof here.

On simplifying (exercise to readers), we get: \[\boxed{\displaystyle \sum_{n=0}^\infty \frac{1}{n^4 + 4} \; = \; \tfrac18(1 + \pi\mathrm{coth}\,\pi)} \] Aditya Kumar · 5 months ago

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Solution to Problem 1:

Use the relation \[\displaystyle k^{m+1}\psi^3(kz)=\sum_{0\leq n \leq k-1}\psi^3(z+n/k)\]

Which reduces

\[\displaystyle \sum_{n\geq 0} (-1)^n\psi^3(2n+2)=\sum_{k \geq 1} (-1)^{k-1}\frac{1}{16}[\psi^3(k)+\psi^3(k+\frac{1}{2})\]

\[=\frac{3}{256}[\zeta(4,3/4)-\zeta(4,5/4)-256+3\pi^4]\]

\[=\displaystyle \frac{-\psi^3(1/4)}{512}+\frac{\psi^3(3/4)}{512}+\frac{9\pi^4}{256}\]

and thus , we have \[\boxed{\frac{\pi^4}{256}+\frac{\psi^3(3/4)}{256}}\] Aman Rajput · 5 months ago

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Problem 1:

Prove that: \[\sum_{n=0}^{\infty}(-1)^{n}\psi_{3}(2n+2) = \frac{\pi^4}{256}+\frac{\psi^{(3)}\left(\frac{3}{4} \right)}{256}.\]

This problem has been first solved by Aman Rajput and second by Mark Hennings. Aditya Kumar · 5 months, 1 week ago

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PROBLEM 16: If \(0 < \theta < 2\pi\), evaluate \[ \sum_{n=1}^\infty \frac{\cos n\theta}{n} \] Don't forget to justify the convergence! Mark Hennings · 5 months ago

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@Mark Hennings Solution To Problem 16 :

Let \(a_{n} = \dfrac{1}{n}\) and \(b_{n} = \cos n \theta\). Note that both sequences satisfy the criterion of Dirchlet's Test. Therefore, the sum \( \displaystyle \text{S} = \sum_{n=1}^{\infty} \dfrac{\cos n \theta}{n}\) converges.

Now,

\[ \text{S} = \sum_{n=1}^{\infty} \dfrac{\cos n \theta}{n} \]

\[ = \Re \left( \sum_{n=1}^{\infty} \dfrac{e^{i n \theta}}{n} \right) \]

\[ = -\Re \left(\ln (1-e^{i \theta}) \right) \]

\[ = - \ln\left| 2\sin \left( \dfrac{\theta}{2} \right) \right| \quad \square \] Ishan Singh · 5 months ago

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@Mark Hennings The contest has shifted to this note. Aditya Kumar · 5 months ago

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The next problem would be posted here. Aditya Kumar · 5 months ago

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Problem 15:

Prove that:

\[\displaystyle \sum_{n=1}^{\infty}\frac{1}{(n^{2}+1)^{2}}=\frac{\pi}{4}\left(\coth(\pi)+\pi \text{csch}^{2}(\pi)-\frac{2}{\pi}\right)\] Aditya Kumar · 5 months ago

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@Aditya Kumar Solution to Problem 15:

Refer to my solution to Problem 8 for some of the technical details required about \(C_N\) and \(\pi \cot \pi z\).

Integrating \( \frac{\pi \cot \pi z}{(z^2+1)^2} \) around the positively-oriented square contour \(C_N\) with corners at \(\pm(N+\tfrac12) \pm i(N+\tfrac12)\), we deduce that \[\begin{array}{rcl} \displaystyle \frac{1}{2\pi i}\int_{C_N}\frac{\pi \cot \pi z}{(z^2+1)^2}\,dz & = & \displaystyle \left(\sum_{n =-N}^N\mathrm{Res}_{z=n} + \mathrm{Res}_{z=i} + \mathrm{Res}_{z=-i}\right) \frac{\pi \cot\pi z}{(z^2 + 1)^2}\,dz \\ & = & \displaystyle 1 + 2\sum_{n=1}^N \frac{1}{(n^2+1)^2} + \left(\mathrm{Res}_{z=i} + \mathrm{Res}_{z=-i}\right)\frac{\pi \cot\pi z}{(z^2 + 1)^2} \end{array} \] and \[ \mathrm{Res}_{z=\pm i}\frac{\pi \cot \pi z}{(z^2+1)^2} \; = \; \frac{d}{dz} \frac{\pi \cot \pi z}{(z \pm i)^2} \Big|_{z=\pm i} \; = \; -\tfrac14\pi^2 \mathrm{cosech}^2\pi - \tfrac14\pi\mathrm{coth}\pi \] so that \[ \frac{1}{2\pi i}\int_{C_N} \frac{\pi \cot \pi z}{(z^2+1)^2}\,dz \; = \; 1 + 2\sum_{n=1}^N \frac{1}{(n^2+1)^2} - \tfrac12\pi^2 \mathrm{cosech}^2\pi - \tfrac12\pi\mathrm{coth}\pi \] Since the integral around \(C_N\) tends to \(0\) as \(N \to \infty\), the result follows.

As before, I pass on setting the next problem. Mark Hennings · 5 months ago

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@Mark Hennings Alternate Method:

\[\displaystyle \sum_{n=1}^\infty \frac{1}{(n^2+1)^2}=\frac{1}{4}\sum_{n=1}^\infty \left\{ \frac{1}{(-1+in)^2}-\frac{1}{-1+in}+\frac{1}{(1+in)^2}+\frac{1}{1+in}\right\}\]

Here, we can use polygamma function and use their reflection formulas to get to the final result. Aditya Kumar · 5 months ago

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@Mark Hennings Sir I can't think of a good problem and Ishan is down with fever. Can you post the next problem here? Aditya Kumar · 5 months ago

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Problem 14:

Prove that:

\[\sum_{n=1}^\infty\frac{\psi_1(n)}{2^n n^2}=\frac{19\pi^4}{1440}-\frac{\pi^2}{24}\log^2(2)-\frac{\log^4(2)}{24}-\frac{\zeta(3) \log(2)}{4}\]

This problem has been solved by Mark Hennings. Aditya Kumar · 5 months ago

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@Aditya Kumar Solution to Problem 14:

We note that \[ \begin{array}{rcl} \displaystyle \sum_{n=1}^\infty \frac{\psi^{(1)}(n)}{2^n n^2} & = & \displaystyle \tfrac12\zeta(2) - \sum_{n=1}^\infty \frac{1}{2^{n+1}(n+1)^2}\big[H^{(2)}_n - \zeta(2)\big] \\ & = & \displaystyle\zeta(2)\sum_{n=1}^\infty \frac{1}{2^n n^2} - \sum_{n=1}^\infty \frac{H_n^{(2)}}{2^{n+1}(n+1)^2} \\ & = & \displaystyle\zeta(2)\Big[ \tfrac{1}{12}\pi^2 - \tfrac12\ln^22\Big] - \sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2 2^{n+1}} \end{array} \] Going back to the previous problem, we have \[ \begin{array}{rcl} \displaystyle A(x) & = & \displaystyle \sum_{n=1}^\infty \big(H_n^{(2)}\big)^2x^n \; =\; \sum_{n=1}^\infty \left(H_{n-1}^{(2)} + \tfrac{1}{n^2}\right)^2 x^n \\ & = & \displaystyle \sum_{n=1}^\infty \big(H_n^{(2)}\big)^2 x^{n+1} + 2\sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2} x^{n+1} + \mathrm{Li}_4(x) \\ & =& \displaystyle xA(x) + 2\sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2} x^{n+1} + \mathrm{Li}_4(x) \\ (1 - x)A(x) & = & \displaystyle 2\sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2} x^{n+1} + \mathrm{Li}_4(x) \end{array} \] and hence, putting \(x=\tfrac12\), \[ \tfrac12\sum_{n=1}^\infty \frac{\big(H_n^{(2)}\big)^2}{2^n} \; = \; 2\sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2 2^{n+1}} + \mathrm{Li}_4(\tfrac12) \] and so \[ \sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2 2^{n+1}} \; = \; \tfrac{1}{1440}\pi^4 -\tfrac{1}{24}\pi^2\ln^22 + \tfrac{1}{24}\ln^42 + \tfrac14\zeta(3)\ln2 \] so that \[ \sum_{n=1}^\infty \frac{\psi^{(1)}(n)}{2^n n^2} \; = \; \tfrac{19}{1440}\pi^4 - \tfrac{1}{24}\pi^2\ln^22 - \tfrac{1}{24}\ln^42 - \tfrac14\zeta(3)\ln2 \] Someone else can post the next question. Mark Hennings · 5 months ago

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@Mark Hennings This question was essentially the same as the previous question. @Aditya Kumar If you set the next question, please post a new one. Ishan Singh · 5 months ago

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@Ishan Singh Can you post the problem? Post it in this note as it will be the 15th problem (don't count spl prob). Aditya Kumar · 5 months ago

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Problem 10:

Evaluate: \[\displaystyle \sum_{k=1}^\infty\frac{\zeta(2k+1)-1}{2k+3}\]

This problem has been solved by Mark Hennings. Aditya Kumar · 5 months ago

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@Aditya Kumar Solution to Problem 10:

The sum is \[ S \; =\; \sum_{k \ge 1}\frac{\zeta(2k+1)-1}{2k+3} \; =\; \sum_{k \ge 1} \sum_{n \ge 2} \frac{1}{(2k+3)n^{2k+1}} \; = \; \sum_{n \ge 2} F\big(\tfrac{1}{n}\big) \] where \[ F(x) \; = \; \sum_{k \ge 1} \frac{x^{2k+1}}{2k+3} \; = \; \frac{1}{2x^2}\ln\left(\frac{1+x}{1-x}\right) - \tfrac13x - x^{-1} \qquad |x| < 1 \] Thus \[ S \; = \; \sum_{n \ge 2} \left(\tfrac12n^2\ln\left(\frac{n+1}{n-1}\right) - \frac{1}{3n} - n\right) \] After a shed-load of simplification, the partial sum \[ \begin{array}{rcl} S_N & = & \displaystyle \sum_{n = 2}^{N+1} \left(\tfrac12n^2\ln\big(\frac{n+1}{n-1}\big) - \frac{1}{3n} - n\right) \\ & = & \displaystyle \tfrac12(N+1)^2\ln(N+2) + \frac12N^2\ln(N+1) + 2\ln\left(\frac{G(N+1)}{\Gamma(N+1)^N}\right) \\ & & {} \displaystyle- \tfrac12\ln2 + \tfrac43 - \tfrac13H_{N+1} - \tfrac12(N+1)(N+2) \end{array} \] where \(G\) is the Barnes G-function. With the known asymptotics \[ \begin{array}{rcl} \displaystyle \ln G(N+1) & \sim & \displaystyle \tfrac{1}{12} - \ln A + \tfrac12N\ln(2\pi) + \tfrac12(N^2 - \tfrac16)\ln N - \tfrac34N^2 \\ \displaystyle \ln \Gamma(N+1) & \sim & (N+1)\ln(N+1) - (N+1) - \tfrac12\ln(N+1) + \tfrac12\ln(2\pi) + \frac{1}{12(N+1)} \end{array}\] as \(N \to \infty\), where \(A\) is the Glaisher constant, we deduce that \[ \ln\left(\frac{G(N+1)}{\Gamma(N+1)^N}\right) \; \sim \; -\ln A + \tfrac14N(N+4) + \tfrac12(N^2 - \tfrac16)\ln N - \tfrac12N(2N+1)\ln(N+1) \] as \(N \to \infty\) and hence we can deduce (after even more simplification) that \[ S \; = \; \lim_{N \to \infty}S_N \; =\; \tfrac{13}{12} - 2\ln A - \tfrac12\ln2 - \tfrac13\gamma \] Mark Hennings · 5 months ago

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@Mark Hennings Awesome solution! Aditya Kumar · 5 months ago

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@Aditya Kumar Solution To Problem 10 :

From the generating function of Riemann Zeta, we have,

\[ \sum_{n=1}^{\infty} \zeta(n+1) \ x^{n} = - \gamma - \psi(1-x) \]

\[ \implies \sum_{n=1}^{\infty} (1 + (-1)^{n}) \zeta(n+1) \ x^{n} = - 2 \gamma - \psi(1-x) - \psi(1+x) \]

\[ \implies \sum_{n=1}^{\infty} \zeta(2n+1) \ x^{2n+1} = -\gamma x - x \psi(x) - \dfrac{\pi x}{2} \cot (\pi x) - \dfrac{1}{2} \quad \quad (*) \]

Now,

\[ \text{S} = \sum_{k=1}^\infty\frac{\zeta(2k+1)-1}{2k+3} \]

\[ = \int_{0}^{1} x \left[ \sum_{k=1}^{\infty} ( x^{2k+1} \zeta(2k+1) - x^{2k+1} ) \right] \mathrm{d}x \]

Using \((*)\), we have,

\[ \text{S} = - \int_{0}^{1} \left( \gamma x^2 + x^2 \psi (x) + \dfrac{x}{2} + \dfrac{\pi x^2}{2} \cot(\pi x) + \dfrac{x^4}{1-x^2} \right) \mathrm{d}x \]

\[ = \dfrac{13}{12} -2\log A - \dfrac{1}{2} \log 2 - \dfrac{\gamma}{3} \quad \square \] Ishan Singh · 5 months ago

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Problem 7:

Prove that:

\[\sum _{ n=1 }^{ \infty }{ { H }_{ n }^{ \left( r \right) }{ z }^{ n } } =\frac { { \text{Li} }_{ r }\left( z \right) }{ 1-z } \]

Here \({ H }_{ n }^{ \left( r \right) }\) is generalized harmonic number and \({ Li }_{ r }\left( z \right) \) is polylogarithm.

This problem has been solved by Mark Hennings and Deeparaj Bhat. Aditya Kumar · 5 months ago

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@Aditya Kumar Solution to Problem 7:

\[ \begin{array}{rcl} \displaystyle (1-z)\sum_{n=1}^\infty H_n^{(r)}z^n & = & \displaystyle \sum_{n=1}^\infty H^{(r)}_n z^n - \sum_{n=1}^\infty H^{(r)}_n z^{n+1} \\ & = & \displaystyle z + \sum_{n=2}^\infty \big[H^{(r)}_n - H^{(r)}_{n-1}\big]z^n \; = \; z + \sum_{n=2}^\infty \frac{z^n}{n^r} \\ & = & \displaystyle \mathrm{Li}_r(z) \end{array} \] for \(|z| < 1\). Mark Hennings · 5 months ago

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@Aditya Kumar Consider the following summation: \[ \begin{align} & (1-z) \sum_{n=1}^{\infty} H_n^{(r)} z^n \\ &= \sum_{n=1}^{\infty} (H_n^{(r)} - H_{n-1}^{(r)})z^n \quad (let \, H_0^{(r)}=0 )\\&= \text{Li}_r(z) \\ \large Q. E. D. \end{align} \] Deeparaj Bhat · 5 months ago

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Problem 5:

Prove That \[ \displaystyle \sum_{r=2}^{\infty} \dfrac{(-1)^r}{F_{r} F_{r-1}} = \phi - 1 \]

Notation : \(F_{r}\) denotes Fibonacci Number.

This problem has been solved my Mark Hennings. Ishan Singh · 5 months ago

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@Ishan Singh Solution to Problem 5:

Since \[ F_n \; =\; \tfrac{1}{\sqrt{5}}\big[\phi^n - (-\phi^{-1})^n\big] \] we have \[ \begin{array}{rcl} \displaystyle \frac{1}{F_n F_{n-1}} & = & \displaystyle \frac{5}{\big(\phi^n - (-\phi^{-1})^n\big)\big(\phi^{n-1} - (-\phi^{-1})^{n-1}\big)} \; = \; \frac{5\phi^{2n-1}}{\big(\phi^{2n} - (-1)^n\big)\big(\phi^{2n-2} - (-1)^{n-1}\big)} \\ & = & \displaystyle \frac{5}{\phi + \phi^{-1}}\left[ \frac{1}{\phi^{2n} - (-1)^n} + \frac{1}{\phi^{2n-2} - (-1)^{n-1}}\right] \end{array} \] and hence (the series telescopes): \[ \begin{array}{rcl} \displaystyle \sum_{r=2}^\infty \frac{(-1)^n}{F_nF_{n-1}} & = & \displaystyle \frac{5}{\phi+\phi^{-1}}\sum_{n=2}^\infty \left[ \frac{(-1)^n}{\phi^{2n} - (-1)^n} - \frac{(-1)^{n-1}}{\phi^{2n-2} - (-1)^{n-1}}\right] \\ & = & \displaystyle \frac{5}{\phi + \phi^{-1}} \times \frac{1}{\phi^2 + 1} \; =\; \frac{5\phi}{(\phi^2 + 1)^2} \; = \; \frac{5\phi}{(\phi+2)^2} \\ & = & \displaystyle \frac{\phi}{\phi+1} \; = \; \phi - 1 \end{array} \] as required. I am happy to let Ishan set another problem; I am very busy at present. Mark Hennings · 5 months ago

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Problem 4:

Evaluate \[\large \sum_{\substack {0 \leq r \leq n \\ r \equiv 1 \pmod 3 }} {n \choose r} \]

This problem has been solved by Ishan Singh. Deeparaj Bhat · 5 months ago

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@Deeparaj Bhat Solution Of Problem 4 :

Let \[\text{S} = \sum_{\substack {0 \leq r \leq n \\ r \equiv 1 \pmod 3 }} {\dbinom{n}{r}} \]

Since,

\[\sum_{r=1}^{n} \dbinom{n}{r} x^r = (1+x)^n \]

\[ \implies \sum_{r=1}^{n} \dbinom{n}{r} {\omega}^r = (1+\omega)^n \tag{1} \]

\[\sum_{r=1}^{n} \dbinom{n}{r} {\omega}^{2r} = (1 + {\omega}^2)^n \tag{2} \]

\[ \sum_{r=1}^{n} \dbinom{n}{r} = 2^n \tag{3} \]

where \(\omega = e^{i \frac{\pi}{3}}\).

Operating \( {\omega}^2 \cdot (1) + \omega \cdot (2) + (3) \), we have,

\[\text{S} = \dfrac{1}{3} \left[ {\omega}^2(1+\omega)^n + \omega (1 + {\omega}^2)^n + 2^n \right] \]


Similarly, we get,

Generalization : \[ \sum_{\substack {0 \leq r \leq n \\ r \equiv k \pmod m }} {\dbinom{n}{r}} = \sum_{r \geq 0} \dbinom{n}{rm + k} = \dfrac{1}{m} \sum_{r=0}^{m-1} {\omega}^{-rk} (1 + \omega^r)^n \]

where \(\omega\) represents one of the non real \(m^{th}\) roots of unity. Ishan Singh · 5 months ago

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Problem 9 :

Evaluate

\[\sum_{n=1}^{\infty} \dfrac{(H_{n})^2}{2^n}\]

Notation : \(H_{n}\) denotes the Harmonic Number.

This problem has been solved by Aditya Kumar. Ishan Singh · 5 months ago

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@Ishan Singh Solution to Problem 9:

I'll use the generating function: \[\frac{\log^{2}(1-t)+Li_{2}(t)}{(1-t)}=\sum_{n=1}^{\infty}H_{n}^{2}t^{n}\]

Substitute: \(t=\frac{1}{2}\).

Hence the final answer is: \(\boxed{{ \left( \ln { 2 } \right) }^{ 2 }+\frac { { \pi }^{ 2 } }{ 6 } }\) Aditya Kumar · 5 months ago

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@Aditya Kumar Proof For Generating Function :

\[ f(x) = \sum_{n=1}^{\infty} H^2 _{n} x^n \]

Since \(H_n = H_{n-1} +\dfrac{1}{n} \implies H^2 _{n} = H^2 _{n-1} +\dfrac{1}{n^2} + 2\dfrac{H_{n-1}}{n}\) and we have,

\[f(x) = \sum_{n=1}^{\infty} H^2 _{n} x^n = \sum_{n=1}^{\infty} \left(H^2 _{n-1} x^{n} +\dfrac{x^n}{n^2} + 2\dfrac{H_{n-1}}{n} x^n\right) \]

Changing summation index \( n-1 \mapsto n \) in the first and third sum, we have,

\[ f(x) = x\sum_{n=1}^{\infty} H_{n} x^n + \operatorname{Li}_{2}(x) + 2\sum_{n=1}^{\infty} \dfrac{H_{n}}{n+1} x^{n+1} \]

\[\implies f(x) = xf(x) + \operatorname{Li}_{2}(x) + 2\sum_{n=1}^{\infty} \dfrac{H_{n}}{n+1} x^{n+1} \]

Also,

\[ \sum_{n=1}^{\infty} \dfrac{H_{n}}{n+1} x^{n+1} = \int_{0}^{x} \sum_{n=1}^{\infty} H_{n} t^{n} \mathrm{d}t\]

\[ = -\int_{0}^{x}\dfrac{\ln(1-t)}{1-t} \mathrm{d}t \]

\[ = \dfrac{\ln^2(1-x)}{2} \]

\[\implies f(x) = xf(x) + \operatorname{Li}_{2}(x) + \ln^2(1-x) \]

\[ \implies f(x) = \dfrac{\ln^2(1-x) + \operatorname{Li}_{2}(x)}{1-x} \] Ishan Singh · 5 months ago

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