# 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.

3 years, 3 months 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:

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

- 3 years, 3 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,

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

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

- 3 years, 3 months ago

Special Problem :

Evaluate

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

- 3 years, 3 months ago

Solution to Special Problem : $\underline{\text{Method 1}}$

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

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

$=\displaystyle\sum_{r=0}^{n}(-1)^r \dfrac{r!(n-r)!\color{#3D99F6}{(n-r+1+r+1)}}{n!}\color{#D61F06}{\times\dfrac{1}{(n+2)}}$

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

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

$\color{#3D99F6}{=\displaystyle \dfrac{1}{n!(n+2)} \sum_{r=0}^{n} \left( T_{r} - T_{r+1} \right) }\tag{*}$

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,

$\color{#D61F06}{\therefore\text{S}=\boxed{2\dfrac{n+1}{n+2}}}$

$\underline{\text{Method 2}}$

Consider the following Lemma.

Lemma : $\color{#624F41}{\displaystyle \sum_{r=0}^k\dfrac{(-1)^r}{\dbinom{k}{r}}=0}$

where $k$ is an odd positive integer.

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

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

$\displaystyle\color{#3D99F6}{\left(\because \sum_{r=0}^{n} f(r) = \sum_{r=0}^{n} f(n-r) \right)}$

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

$\implies \text{S}=-\text{S}$

$\implies \text{S}=0$

This completes the proof of the Lemma.

Now, since $n$ is even, $n+1$ is odd.

Using the Lemma, we have,

$\color{#3D99F6}{\displaystyle \sum_{r=0}^{n+1}\dfrac{(-1)^r}{\dbinom{n+1}{r}}=0} \tag{1}$

Let,

$\color{#D61F06}{\text{J}=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}}}\tag{2}$

Operating $(1)+(2)$,

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

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

$=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}} + \displaystyle \sum_{r=0}^{n}\dfrac{(-1)^{n-r}}{\dbinom{n+1}{n-r}} -1 \quad \displaystyle\color{#3D99F6}{\left(\because \sum_{r=0}^{n} f(r)=\sum_{r=0}^{n} f(n-r)\right)}$

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

$=\displaystyle \sum_{r=0}^n(-1)^r\left\{\dfrac{1}{\dbinom{n}{r}} + \dfrac{r+1}{n+1} \cdot \dfrac{1}{\dbinom{n}{r}}\right\} -1 \quad \displaystyle\color{#D61F06}{\left[\because \dbinom{n}{r}=\dfrac{n}{r}\dbinom{n-1}{r-1} \ \text{\&} \ \dbinom{n}{r}=\dbinom{n}{n-r} \right]}$

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

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

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

Now,

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

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

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

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

$\implies \text{G}=n\cdot \text{J}-\text{G}$

$\implies\color{#D61F06}{\text{G}=\dfrac{\text{J}n}{2}} \tag{4}$

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

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

$\color{#20A900}{\therefore\text{J}=\boxed{2\dfrac{n+1}{n+2}}} \quad \square$

- 3 years, 3 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{aligned} \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{aligned}

- 3 years, 3 months ago

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

- 3 years, 3 months ago

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.

- 3 years, 3 months ago

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

\begin{aligned} \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{aligned}

where G is Catalan's constant.

- 3 years, 3 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!

- 3 years, 3 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.

- 3 years, 3 months ago

SOLUTION OF PROBLEM 1 :

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

- 3 years, 3 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.

- 3 years, 3 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. :)

- 3 years, 3 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, $S=2^{n-1} \sin nx$

- 3 years, 3 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.

- 3 years, 3 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:

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

- 3 years, 3 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,

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

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

$= H_{n} \quad \square$

- 3 years, 3 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.

- 3 years, 3 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.

- 3 years, 3 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,

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

- 3 years, 3 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

- 3 years, 3 months ago

Start with $\begin{array}{rcl} B(x) & = & \displaystyle \sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2}x^{n+1} \; = \; \int_0^x \left(\int_0^u \sum_{n=1}^\infty H_n^{(2)}v^n\,dv\right)\,\frac{du}{u} \\ & = & \displaystyle \int_0^x \left(\int_0^u \frac{\mathrm{Li}_2(v)}{1-v}\,dv\right)\,\frac{du}{u} \; = \; \int_0^x \left(\int_v^x \frac{\mathrm{Li}_2(v)}{u(1-v)}\,du\right)\,dv \\ & = & \displaystyle \int_0^x \ln\big(\tfrac{x}{v}\big) \frac{\mathrm{Li}_2(v)}{1-v}\,dv \end{array}$