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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.

Format your post is as follows:

 1 2 3 4 5 6 7 **SOLUTION OF PROBLEM xxx (number of problem) :** **[Post your solution here]** **PROBLEM xxx (number of problem) :** **[Post your problem here]** 

See Part-2.

1 year, 4 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.

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

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!

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

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:

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

- 1 year, 4 months ago

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,

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

$\implies \text{S} = \boxed{\zeta(2) \ln 2 + \dfrac{63}{2} \zeta(3) -14 \pi G}$

- 1 year, 4 months ago

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

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

I was wondering whether we can extend the result to other weights too?

- 1 year, 4 months ago

I have solved the indefinite integrals by using Landen's Identity.

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

I'm impressed. The last integral is a brute.

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

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,

$\text{S} = \dfrac{2\cos(\frac{1}{3} \sin^{-1}(\frac{3\sqrt{3x}}{2}))}{\sqrt{4-27x}} \quad \square$

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

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:

$\sum_{r=1}^n \frac{(-1)^{r-1}}r \dbinom{n}{r} =\int _{ 0 }^{ 1 }{ \frac { 1-{ \left( 1-x \right) }^{ n } }{ x } dx } =\int _{ 0 }^{ 1 }{ \frac { 1-{ x }^{ n } }{ 1-x } dx } ={ H }_{ n }\\$

- 1 year, 4 months ago

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,

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

$= \sum_{k=1}^{n} \dfrac{1}{k}$

$= H_{n} \quad \square$

- 1 year, 4 months ago

Special Problem :

Evaluate

$\sum_{r=0}^{n} \dfrac{(-1)^r}{\dbinom{n}{r}}$

- 1 year, 4 months ago

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

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

$\color{green}{\therefore\text{J}=\boxed{2\dfrac{n+1}{n+2}}} \quad \square$

- 1 year, 4 months ago

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:

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

- 1 year, 4 months ago

This is the same method which Ishan talked about. Thanks for posting!

- 1 year, 4 months ago

Comment deleted Jun 30, 2016

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 $\begin{array}{rcl} \displaystyle \sum_{r=0}^n \frac{(-1)^r}{{n \choose r}} & = & \displaystyle \frac{1 + (-1)^n}{n!} \int_0^\infty \int_0^u \frac{x^{n+1}}{u} e^{-u}\,dx\,du \\ & = & \displaystyle \frac{1 + (-1)^n}{n! (n+2)} \int_0^\infty u^{n+1} e^{-u}\,du \; = \; \frac{(1 + (-1)^n) (n+1)!}{n! (n+2)} \\ & = & \displaystyle (1 + (-1)^n)\,\frac{n+1}{n+2} \end{array}$

- 1 year, 4 months ago

(+1) Nice Solution! I've posted $$2$$ of my $$3$$ methods, the third one used Beta Function (which I haven't posted).

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

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. :)

- 1 year, 4 months ago

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, $$S=2^{n-1} \sin nx$$

- 1 year, 4 months ago

SOLUTION OF PROBLEM 1 :

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

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

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,

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

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

$= \dfrac{1}{8} (1 +\pi \coth(\pi)) \quad \square$

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

$S=\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { n }^{ 4 }+4 } }$

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: $\boxed{\displaystyle \sum_{n=0}^\infty \frac{1}{n^4 + 4} \; = \; \tfrac18(1 + \pi\mathrm{coth}\,\pi)}$

- 1 year, 4 months ago

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 $\boxed{\frac{\pi^4}{256}+\frac{\psi^3(3/4)}{256}}$

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

PROBLEM 16: If $$0 < \theta < 2\pi$$, evaluate $\sum_{n=1}^\infty \frac{\cos n\theta}{n}$ Don't forget to justify the convergence!

- 1 year, 4 months ago

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,

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

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

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

$= - \ln\left| 2\sin \left( \dfrac{\theta}{2} \right) \right| \quad \square$

- 1 year, 4 months ago

The contest has shifted to this note.

- 1 year, 4 months ago

The next problem would be posted here.

- 1 year, 4 months ago

Problem 15:

Prove that:

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{(n^{2}+1)^{2}}=\frac{\pi}{4}\left(\coth(\pi)+\pi \text{csch}^{2}(\pi)-\frac{2}{\pi}\right)$

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

Sir I can't think of a good problem and Ishan is down with fever. Can you post the next problem here?

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

This question was essentially the same as the previous question. @Aditya Kumar If you set the next question, please post a new one.

- 1 year, 4 months ago

Can you post the problem? Post it in this note as it will be the 15th problem (don't count spl prob).

- 1 year, 4 months ago

Problem 10:

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

This problem has been solved by Mark Hennings.

- 1 year, 4 months ago

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 $S \; = \; \lim_{N \to \infty}S_N \; =\; \tfrac{13}{12} - 2\ln A - \tfrac12\ln2 - \tfrac13\gamma$

- 1 year, 4 months ago

Awesome solution!

- 1 year, 4 months ago

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,

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

$= \dfrac{13}{12} -2\log A - \dfrac{1}{2} \log 2 - \dfrac{\gamma}{3} \quad \square$

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

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 $$|z| < 1$$.

- 1 year, 4 months ago

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}

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

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: $$\boxed{{ \left( \ln { 2 } \right) }^{ 2 }+\frac { { \pi }^{ 2 } }{ 6 } }$$

- 1 year, 4 months ago

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,

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

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

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

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

$\implies f(x) = \dfrac{\ln^2(1-x) + \operatorname{Li}_{2}(x)}{1-x}$

- 1 year, 4 months ago