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Brilliant Summation Contest Season-1 (Part 2)

Hi Brilliant! This is the sequel of the first part of Brilliant Summation Contest Season-1.

Update: This contest has been ended. Thanks for everyone's participation.

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":

Note by Aditya Kumar
1 year, 4 months ago

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Problem 27:

Prove:

\[\sum _{ n=-\infty }^{ \infty }{ { \left( -e^{-\pi} \right) }^{ { n }^{ 2 } } } =\frac { { \left( \frac { \pi }{ 2 } \right) }^{ \frac { 1 }{ 4 } } }{ \Gamma \left( \frac { 3 }{ 4 } \right) } \]

Aditya Kumar - 1 year, 4 months ago

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We have a proof on brilliant

\[\sum _{ n=-\infty }^{ \infty }{ { \left( e^{-\pi} \right) }^{ { n }^{ 2 } } } =\frac { { \left( \pi \right) }^{ \frac { 1 }{ 4 } } }{ \Gamma \left( \frac { 3 }{ 4 } \right) } \]

Now use this formula \[\phi(-e^{-2\pi/n})=\frac { { \left( \pi \right) }^{ \frac { 1 }{ 4 } } }{ \Gamma \left( \frac { 3 }{ 4 } \right) } \frac{n^{1/4}r_{2,2/n^2}}{2^{1/4}r_{2,2}}\]

\[r_{k,n} \] is a function of DedeKind Eta function. Which have some formulaes to solve.

\[r_{2,2}=2^{1/8} \text{and} r_{2,1/2}=2^{-1/8}\]

Thus , you get \[\frac{(\frac{\pi}{2})^{1/4}}{\Gamma(3/4)}\]

Aman Rajput - 1 year, 4 months ago

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This is evaluated in ramanujan's notebook which I have but there's no proof for the formula used :/ . So I was trying to prove the above q-product

Aditya Narayan Sharma - 1 year, 4 months ago

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@Aditya Narayan Sharma share your q product .... if i have time i can solve it easily

Aman Rajput - 1 year, 4 months ago

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@Aman Rajput No, I m not talking bout the q-product actually. I m talking about the above formula , he used q-products and generalisations to come to this. So share if you have a proof of this , coz I didnt find

Aditya Narayan Sharma - 1 year, 4 months ago

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At least could you elaborate on step 2?

Aditya Kumar - 1 year, 4 months ago

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@Aditya Kumar I used this identity from a ramanujan paper

Aman Rajput - 1 year, 4 months ago

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Problem 18:

Prove that:

\[\sum_{k=0}^n \binom{n+k}{n-k}\frac{(-1)^{n+k}}{2k+1}=-\frac{2}{2n+1}\,\cos\left(\frac{2(n-1)\pi}{3}\right)\;\text{?}\]

This Problem has been solved by Ishan Singh.

Aditya Kumar - 1 year, 4 months ago

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Solution Of Problem 18 :

Proposition : \[ \text{S}_{n} = \sum_{r=0}^{n} (-1)^r \dbinom{2n+1-r}{r} = \dfrac{2}{\sqrt{3}} \sin \left( \dfrac{2 (n+1) \pi}{3} \right) \]

Proof : Since \( \dbinom{n}{r} = 0 \) for \( r > n \), we can rewrite the sum as

\[ \text{S}_{n} = \sum_{r=0}^{\infty} (-1)^r \dbinom{2n+1-r}{r} \]

From the Binomial Theorem, we see that the sum is the coefficient of \(x^n\) in

\[f(x) = x^n (1-x)^{2n+1} + x^{n-1} (1-x)^{2n} + \ldots \]

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

Note that,

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

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

Putting \(a = \dfrac{1}{2}\), we see that,

\[ f(x) = (1-x)^{n+1} \sum_{n=0}^{\infty}{{U}_{r} \left( \dfrac{1}{2} \right) {x}^{r}} \quad (*) \]

To calculate the coefficient of \(x^n\) in \( (*) \), we see that coefficient of \(x^{n-r}\) for fixed \(r\) in \((1-x)^{n+1}\) is \( (-1)^{n-r} \dbinom{n+1}{n-r} \), so coefficient of \(x^n\) is,

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

Substituting \( r \mapsto r-1 \), we have,

\[ \text{S}_{n} = \sum_{r=1}^{n+1} (-1)^{n+1-r} \dbinom{n+1}{n+1-r} U_{r-1} \left(\dfrac{1}{2}\right) \]

\[ = (-1)^{n+1} \sum_{r=1}^{n+1} (-1)^{r} \dbinom{n+1}{r} U_{r-1} \left(\dfrac{1}{2}\right) \quad \quad \left( \because \dbinom{n}{n-r} = \dbinom{n}{r} \right) \]

\[ = (-1)^{n+1} \left( \dfrac{2}{\sqrt{3}} \right) \sum_{r=1}^{n+1} (-1)^{r} \dbinom{n+1}{r} \sin \left( \dfrac{r \pi}{3} \right) \quad \quad \left( \because U_{r-1} \left(\dfrac{1}{2}\right) = \sin \left( \dfrac{r \pi}{3} \right) \right) \]

\[ (-1)^{n+1} \left( \dfrac{2}{\sqrt{3}} \right) \sum_{r=0}^{n+1} (-1)^{r} \dbinom{n+1}{r} \sin \left( \dfrac{r \pi}{3} \right) \quad \quad \left( \because \sin 0 = 0 \right) \]

Now, \[\displaystyle \sum_{r=0}^{n+1} (-1)^{r} \dbinom{n+1}{r} \sin \left( \dfrac{r \pi}{3} \right) = \Im \left( \sum_{r=0}^{n+1} (-1)^{r} \dbinom{n+1}{r} e^{\frac{i r \pi}{3}} \right) = (-1)^{n+1} \sin \left( \dfrac{2 (n+1) \pi}{3} \right) \]

\[ \implies \text{S}_{n} = \dfrac{2}{\sqrt{3}} \sin \left( \dfrac{2 (n+1) \pi}{3} \right) \quad \square \]

Corollary : \[\text{S}_{n} = \sum_{k=0}^n (-1)^{n+k} \dbinom{n+1+k}{2k+1} = \dfrac{2}{\sqrt{3}} \sin \left( \dfrac{2 (n+1) \pi}{3} \right)\]

Proof : \[\sum_{k=0}^n(-1)^{n+k} \dbinom{n+1+k}{2k+1} = \sum_{k=0}^n(-1)^{n+k} \dbinom{n+1+k}{n-k} = \sum_{k=0}^n(-1)^k \dbinom{2n+1-k}{k} = \text{S}_{n} \]

\[\left( \because \dbinom{n}{r} = \dbinom{n}{n-r} \ \text{&} \ \sum_{r=0}^{n} f(k) = \sum_{r=0}^{n} f(n-k) \right) \quad \square \]

Now,

\[ \lambda = \sum_{k=0}^n\dfrac{(-1)^{n+k}}{2k+1} \dbinom{n+k}{n-k} \]

\[ = \sum_{k=0}^n \dfrac{(-1)^{n+k}(n+k)!}{(2k+1)!(n-k)!} \]

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

\[ =\dfrac{1}{2n+1}\sum_{k=0}^n\left((-1)^{n+k} \dbinom{n+1+k}{2k+1} - (-1)^{n-1+k} \dbinom{n+k}{2k+1} \right) \]

\[ =\dfrac{1}{2n+1} \left(\text{S}_{n} - \text{S}_{n-1}\right) \]

Using the Proposition and simplifying, we have,

\[ \lambda = -\dfrac{2}{2n+1} \, \cos\left(\frac{2(n-1)\pi}{3}\right) \quad \square \]


Note : Coefficient of \(x^n\) in the Proposition can also be found by splitting \(\dfrac{1}{x^2-x+1}\) as partial fraction and expanding using Infinite GP.

Ishan Singh - 1 year, 4 months ago

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Problem 17 :

Prove That

\[ \displaystyle \sum_{n=0}^{\infty} \dfrac{2^{n-1} n!}{\prod_{r=0}^{n} (4r+1)} = \dfrac{\pi + 2\log(1+\sqrt{2})}{4 \sqrt{2}} \]

This problem has been solved by Aditya Kumar.

Ishan Singh - 1 year, 4 months ago

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Solution to Problem 17:

\[S=\sum _{ n=0 }^{ \infty } \frac { 2^{ n-1 }n! }{ \prod _{ r=0 }^{ n } (4r+1) } \]

\[S=\sum _{ n=0 }^{ \infty } \left[ \frac { { 2 }^{ -n-1 }\Gamma \left( n+1 \right) \Gamma \left( \frac { 5 }{ 4 } \right) }{ \Gamma \left( n+\frac { 5 }{ 4 } \right) } \right] \]

Using beta function, we get:

\[S=\sum _{ n=0 }^{ \infty } \left[ { 2 }^{ -n-1 }B\left( n+1,\frac { 5 }{ 4 } \right) .\left( n+\frac { 5 }{ 4 } \right) \right] \]

\[S=\frac { 1 }{ 2 } \sum _{ n=0 }^{ \infty } \left[ { 2 }^{ -n }B\left( n+1,\frac { 5 }{ 4 } \right) .n \right] +\frac { 5 }{ 8 } \sum _{ n=0 }^{ \infty } \left[ { 2 }^{ -n }B\left( n+1,\frac { 5 }{ 4 } \right) \right] \]

\[S=\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \left\{ \sum _{ n=0 }^{ \infty }{ \left[ { 2 }^{ -n }.n.{ t }^{ n }{ \left( 1-t \right) }^{ \frac { 1 }{ 4 } } \right] } dt \right\} } +\frac { 5 }{ 8 } \int _{ 0 }^{ 1 }{ \left\{ \sum _{ n=0 }^{ \infty }{ \left[ { 2 }^{ -n }.{ t }^{ n }{ \left( 1-t \right) }^{ \frac { 1 }{ 4 } } \right] } dt \right\} } \]

Now I'll use the following geometric progression summations:

\[A=\sum _{ n=0 }^{ \infty }{ \left[ { \left( xt \right) }^{ n }{ \left( 1-t \right) }^{ \frac { 1 }{ 4 } } \right] } =\frac { { \left( 1-t \right) }^{ \frac { 1 }{ 4 } } }{ 1-xt } ,\quad B=\sum _{ n=0 }^{ \infty }{ \left[ { \left( xt \right) }^{ n }{ .n.\left( 1-t \right) }^{ \frac { 1 }{ 4 } } \right] } =\frac { { xt\left( 1-t \right) }^{ \frac { 1 }{ 4 } } }{ { \left( 1-xt \right) }^{ 2 } } \]

\[S=\frac { 5 }{ 4 } \int _{ 0 }^{ 1 }{ \frac { { \left( 1-t \right) }^{ \frac { 1 }{ 4 } } }{ 2-t } dt } +\int _{ 0 }^{ 1 }{ \frac { x{ \left( 1-x \right) }^{ \frac { 1 }{ 4 } } }{ { \left( 2-x \right) }^{ 2 } } dx } \]

The first integral can be easily evaluated. The second one is also easily evaluated. I was screwed up and I couldn't evaluate it. So I asked for it here.

Hence the final answer on doing some simplification is:

\[\boxed{\displaystyle\sum_{n=0}^{\infty} \dfrac{2^{n-1} n!}{\prod_{r=0}^{n} (4r+1)} = \dfrac{\pi + 2\log(1+\sqrt{2})}{4 \sqrt{2}}}\]

Aditya Kumar - 1 year, 4 months ago

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Problem 24:

Prove that \[\displaystyle \sum_{k \geq 1} \frac{F_{2k}H_{k-1}^{(2)}}{k^2\binom{2k}{k}}=\frac{2\pi^4}{375\sqrt5}\]

\[F_n - \text{ Fibonacci number }\]

This problem has been solved by Aditya Kumar.

Aman Rajput - 1 year, 4 months ago

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Solution of Problem 24:

Lemma 1: \[{ x }^{ 2 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \left[ \frac { { 2 }^{ 2n } }{ { n }^{ 2 } \dbinom{2n}{n} } \sin ^{ 2n }{ x } \right] } \]

Proof:

Consider: \({ f }_{ n }\left( x \right) =\frac { \sin ^{ 2n }{ x } }{ \left( 2n \right) ! } \).

Then we get: \(\\ { f" }_{ n }\left( x \right) ={ f }_{ n-1 }\left( x \right) -{ \left( 2n \right) }^{ 2 }{ f }_{ n }\left( x \right) \)

We can write \(\displaystyle { x }^{ 2 }=\sum _{ n=0 }^{ \infty }{ { a }_{ n }{ f }_{ n }\left( x \right) } \) for some function \(a_n\).

On differentiating both sides twice:

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

From this relation \(a_1=2\) and \(a_{n+1}=a_n{(2n)}^{2}\)

Thus, \(\displaystyle \frac { { a }_{ n } }{ 2 } =\prod _{ k=1 }^{ n-1 }{ \frac { { a }_{ k+1 } }{ { a }_{ k } } } ={ 2 }^{ 2n-2 }{ \left( n-1 \right) ! }^{ 2 }\).

Hence, we get : \(\displaystyle { x }^{ 2 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \left[ \frac { { 2 }^{ 2n } }{ { n }^{ 2 } \dbinom{2n}{n} } \sin ^{ 2n }{ x } \right] } \)

Lemma 2: \[{ x }^{ 4 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \left[ \frac { { 2 }^{ 2n }\sin ^{ 2n }{ x } { H }_{ n-1 }^{ (2) } }{ { n }^{ 2 }\dbinom{2n}{n} } \right] } \]

Proof:

From Lemma 1, we can write: \(\displaystyle { x }^{ 2 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \left[ \frac { { 2 }^{ 2n } }{ { n }^{ 2 } \dbinom{2n}{n} } \sin ^{ 2n }{ x } \right] }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ { b }_{ n }^{ 2 }{ f }_{ n }\left( x \right) } \)

Here, \(b_n={ 2 }^{ n }(n-1)!\) and \(b_{n+1}=2nb_n\).

We can write: \(\displaystyle { x }^{ 4 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ { a }_{ n }{ b }_{ n }^{ 2 }{ f }_{ n }\left( x \right) } \) for some function \(a_n\).

On differentiating both sides twice:

\[12{ x }^{ 2 }=\frac { 1 }{ 2 } \sum _{ n=0 }^{ \infty }{ { a }_{ n+1 }{ b }_{ n+1 }^{ 2 }{ f }_{ n }\left( x \right) } -\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ { a }_{ n }{ b }_{ n }^{ 2 }{ \left( 2n \right) }^{ 2 }{ f }_{ n }\left( x \right) } \]

Thus, we get: \(({ a }_{ n+1 }-{ a }_{ n }){ b }_{ n }^{ 2 }{ \left( 2n \right) }^{ 2 }=12{ b }_{ n }^{ 2 }\)

From here we get: \(({ a }_{ n+1 }-{ a }_{ n })=\frac { 12 }{ { \left( 2n \right) }^{ 2 } } \).

Hence, \({ a }_{ n }={ H }_{ n-1 }^{ (2) }\).

Therefore, we get: \(\displaystyle { x }^{ 4 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \left[ \frac { { 2 }^{ 2n }\sin ^{ 2n }{ x } { H }_{ n-1 }^{ (2) } }{ { n }^{ 2 }\dbinom{2n}{n} } \right] }\).

From lemma 2, we get: \[\displaystyle (\sin^{-1} x)^4 = \frac32 \sum_{n=1}^\infty \frac{2^{2n} H_{n-1}^{(2)}}{n^2 \binom{2n}{n}} \,x^{2n} \ \quad ...(A)\]

Now, in the problem, we use: \({ F }_{ 2k }=\frac { { \varphi }^{ 2k }-\frac { 1 }{ { \varphi }^{ 2k } } }{ \sqrt { 5 } } \)

On substituting and using \(A\), we get: \[\boxed{\displaystyle \sum_{k \geq 1} \frac{F_{2k}H_{k-1}^{(2)}}{k^2\binom{2k}{k}}=\frac{2\pi^4}{375\sqrt5}}\]

Aditya Kumar - 1 year, 4 months ago

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(+1) Nicely explained! I solved using Beta Functions and Integration.

Ishan Singh - 1 year, 4 months ago

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Problem 22:

Prove the identity:

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

This problem has been solved by Aditya Sharma, Mark Hennings and Ishan Singh.

Aditya Kumar - 1 year, 4 months ago

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We have,

\[ \text{S} = \sum_{k=1}^{n}\dfrac{H_{k}}{n-k+1} \]

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

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

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

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

\[ = \sum_{r=1}^{n} \sum_{k=r}^{n} \dfrac{1}{(k+1)} \left[\dfrac{1}{(k-r+1)} + \dfrac{1}{r}\right] \]

\[ = \sum_{k=1}^{n} \sum_{r=1}^{k} \dfrac{1}{(k+1)} \left[\dfrac{1}{(k-r+1)} + \dfrac{1}{r}\right] \]

\[ = 2\sum_{k=1}^{n} \dfrac{H_{k}}{k+1} \]

Substitute \( k+1 \mapsto k \)

\[\implies \text{S} = 2\sum_{k=2}^{n+1} \dfrac{H_{k-1}}{k} \]

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

\[ = 2\sum_{k=1}^{n+1} \dfrac{H_{k}}{k} - 2H_{n+1}^{(2)} \]

Since \(\displaystyle \sum_{k=1}^{n+1} \dfrac{H_{k}}{k} = \dfrac{1}{2} (H_{n+1}^2 + H_{n+1}^{(2)}) \) (I have proved it here), we have,

\[\text{S} = (H_{n+1})^2 - H_{n+1}^{(2)} \quad \square \]

Ishan Singh - 1 year, 4 months ago

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Let \(\displaystyle S(r)=\sum_{k=0}^{n} \frac{H_{r-k}}{k} \) for \(r>0\) with \(S(0)=0\).

We have , \(\displaystyle S(n+1) = \sum_{k=1}^{n} \frac{H_k}{n-k+1} = \sum_{k=1}^{n}\frac{H_{n-k+1}}{k} = \sum_{k=1}^{n}\frac{H_{n-k}}{k} + \frac{2H_n}{n+1} = S(n)+\frac{2H_n}{n+1}\)

Telescoping above we have , \(\displaystyle S(n+1) = \sum_{k=0}^{n}\frac{2H_k}{k+1}\)

\(\displaystyle \text{Lemma :} \large \boxed{\sum_{k=0}^{n} [(k+1)^p-k^p]H_{k}^{(m)} = (n+1)^p H_{n}^{(m)}-H_{n}^{(m-p)}}\)

\(\text{Proof :}\) Using summation by parts we have ,

\(\displaystyle \sum_{k=0}^{n} [(k+1)^p-k^p]H_{k}^{(m)} = (n+1)^p H_{n+1}^{(p)} - \sum_{k=0}^{n} (k+1)^{(m-p)} = (n+1)^p H_{n+1}^{(p)} - H_{n+1}^{(m-p)} = (n+1)H_{n}^{(p)} - H_{n}^{(m-p)}\) & thus proved.

We'll use a well known Harmonic sum identity , \(\displaystyle \sum_{k=1}^{n} \frac{H_k}{k} = \frac{1}{2}((H_n)^2+H_{n}^{(2)})\)

Applying the lemma for \(p=-1,m=1\) we have,

\(\displaystyle S(n+1) = 2\sum_{k=0}^{n} \frac{H_k}{k+1} = \sum_{k=1}^{n} \frac{2H_k}{k} + \frac{H_n}{n+1}-H_{n}^{(2)}\)

Simplifying we get , \(\displaystyle \sum_{k=1}^{n} \frac{H_k}{n-k+1} = (H_{n+1})^2-H_{n+1}^{(2)}\)

Aditya Narayan Sharma - 1 year, 4 months ago

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The generating function of the \(S(n)\) is \[ \sum_{n=1}^\infty S(n) x^{n+1} = \sum_{n=1}^\infty H_n x^n \times \sum_{n=1}^\infty \tfrac{1}{n} x^n = \frac{\ln^2(1-x)}{1-x} \] so the result follows using the known formulae for the generating functions for \(H_n^{(2)}\) and for \((H_n)^2\).

Mark Hennings - 1 year, 4 months ago

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Problem 29:

Prove:

\[\displaystyle \sum\limits_{m=1}^{99}{\frac{\sin{\left(\frac{17 m \pi}{100}\right)} \sin{\left(\frac{39 m \pi}{100}\right)}}{1+\cos{\left( \frac{m\pi}{100} \right) }}}=1037 \]

Due to time constraint, I have decided to post the solution.

Jack Lam - 1 year, 4 months ago

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Define \( \displaystyle T(k) = \sum_{m=1}^{n-1} \frac{\cos{\frac{mk\pi}{n}}}{1+\cos{\frac{m\pi}{n}}}\)

Observing that \(\cos{(k+1)\theta} + 2\cos{k\theta} + \cos{(k-1)\theta} = 2\cos{k\theta}(1+\cos{\theta}) \)

The following recurrence relation is obtained:

\(\displaystyle T(k+1) + 2T(k) + T(k-1) = 2 \sum_{m=1}^{n-1} \cos{\frac{mk\pi}{n}} = 1+\cos{k\pi} + \sin{k\pi} \tan{\frac{k\pi}{2n}}\)

Note that an implicit restriction of \( k \notin n \mathbb{Z} \) is forced because the GP formula is invalid when the common ratio is \(\pm 1\).

Evaluate the recurrence relation at \(k+1\) and add together to yield

\( T(k+2) + 3T(k+1) + 3T(k) + T(k-1) = 2 \)

Evaluate this recurrence relation at \(k\) and take the difference to homogenise the recurrence relation:

\( T(k+2) + 2T(k+1) - 2T(k-1) - T(k-2) = 0 \)

The characteristic roots are \(1\) and \(-1\) with multiplicity \(3\).

The general solution to the recurrence relation is thus

\( (C_1 k^2 + C_2 k+C_3)(-1)^k + C_4\)

Returning to the original problem, define \(\displaystyle S(a,b,n) = \sum_{m=1}^{n-1} \frac{\sin{\frac{am\pi}{n}}\sin{\frac{bm\pi}{n}}}{1+\cos{\frac{m\pi}{n}}}\) where \( 0 \le b \le a \le n \)

We find that \(2S(a,b,n) = T(a-b)-T(a+b)\)

Note that \( (-1)^{a+b} \equiv (-1)^{a-b}\)

Substituting \(T(k)\) into the above and expanding, we obtain

\(\displaystyle S(a,b,n) = (2C_1 a + C_2)b(-1)^{a+b+1} = (-1)^{a+b}\hat{C_1}b(\hat{C_2}-a)\)

A short computation yields \(S(1,1,n) = n-1\)

From \( S(n,b,n) = 0 \, \& \, S(1,1,n)>0\), \(\hat{C_2} = n\)

From \(S(1,1,n) = n-1\), \(\hat{C_1} = 1\)

Therefore, we conclude \( S(a,b,n) = (-1)^{a+b} b(n-a) \)

Substituting \(b=17, a=39, n=100\), and the result which was to be shown follows.

Jack Lam - 1 year, 4 months ago

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This was the best solution!

Deeparaj Bhat - 1 year, 4 months ago

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Nice solution. I missed the 48 hour time.

Aditya Kumar - 1 year, 4 months ago

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Problem 21:

Prove that: \[\large \displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} 5^{-n} = \sqrt{5}\]

This problem has been solved by Aditya Kumar.

Aditya Narayan Sharma - 1 year, 4 months ago

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Solution to Problem 21:

We use the well known generating function of Central Binomial coefficient:

\[\frac{1}{\sqrt{1 - 4x}} = \sum_{n=0}^\infty \frac{(2n)!}{(n!)^2}x^n\]

Here substitute \(x=\frac{1}{5}\).

Hence, \(\large \boxed{\displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} 5^{-n} = \sqrt{5}}\)

Aditya Kumar - 1 year, 4 months ago

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Proof of that generating function?Thanks.

Harsh Shrivastava - 1 year, 4 months ago

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@Harsh Shrivastava There are many ways but I learnt it here.

Aditya Kumar - 1 year, 4 months ago

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Problem 20 :

Evaluate \[\sum_{i_m=1}^n \sum_{i_{m-1}=1}^{i_m} \cdots \sum_{i_2=1}^{i_3} \sum_{i_1=1}^{i_2} 1 \]

This problem has been solved by Aditya Sharma and Ameya Daigavane.

Deeparaj Bhat - 1 year, 4 months ago

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Although @Aditya Sharma has solved the question - I thought of an approach that might be considered more "illuminating" than induction.

Let's think of the \( i_1, i_2,...i_m \) as numbers on a register - each \( i_k\) is one number from \(1\) to \(n\). Initially, the register reads \( m \) ones, because all the \(i_k\) are \(1\).
For each unique number on the register, we add one to our count.

Now, we turn the first number (\(i_m\)) to \(2\).
The register reads \(2\) followed by \(m - 1\) \(1\)s.
Note that a digit to the right of another digit can never be greater than it, because \(i_j \geq i_k\)if \(j \geq k\).

So, the number on our register is an \(m\) digit number whose digits are non-decreasing when read from left to right.
We can keep turning the register and counting how many unique numbers we get, until we reach \(m\) digits with value \(n\). After this, our register stops working.

It is easy to see that every single number satisfying the non-decreasing digit condition is encountered, and only once.

So our question now becomes:
What is the number of tuples \( i_1, i_2,...i_m \) satisfying,

\[ 1 \leq i_1 \leq i_2 \leq \ldots \leq i_m \leq n \]

Because for each one of these tuples, we have a number on our register.

If we write the numbers in a row, \[ i_1 i_2 \ldots i_m \] and divide them into \(n\) blocks, such that the numbers enclosed in the \(n^{th}\) block get value \( n \), then we have found one such possibility. We need \(n - 1\) bars to separate the numbers into \(n\) blocks.
For example, if \(m = 4, n = 4\), we can divide the numbers as, \[ i_1 | | i_2 i_3 | i_4 \] Here 0 \[ i_1 = 1, i_2 = i_3 = 3, i_4 = 4 \]

Clearly, the number of ways we can separate these numbers, is the number of tuples we need, because numbers with higher indices are always given higher (or equal) values.

So, we can finish with stars-and-bars, as we have \(m\) numbers, and \(n -1 \) separators, so the total number of separations/tuples is: \[ \binom {m + n - 1}{n - 1} \] which is the same answer as above.

Ameya Daigavane - 1 year, 4 months ago

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Comment deleted Jul 12, 2016

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@Ameya Daigavane Ok sorry for that. It is just that brilliant is the only website that doesn't use $. So that made me think that it was from other source.

Aditya Kumar - 1 year, 4 months ago

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@Aditya Kumar I work with LaTeX on other sites as well. I had thought about this question for some time; in fact, I think there was a similar question in the FIITJEE GMP for JEE Advanced. I really liked my solution there, so I thought I'd just extend it here.

Ameya Daigavane - 1 year, 4 months ago

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Let \[S =\sum_{i_m=1}^n \sum_{i_{m-1}=1}^{i_m} \cdots \sum_{i_2=1}^{i_3} \sum_{i_1=1}^{i_2} 1 \]

\(\displaystyle S=\sum_{i_m=1}^n \sum_{i_{m-1}=1}^{i_m} \cdots \sum_{i_2=1}^{i_3} \binom{i_2}{1} \)

\(\displaystyle S = \sum_{i_m=1}^n \sum_{i_{m-1}=1}^{i_m} \cdots \sum_{i_3=1}^{i_4} \binom{i_3+1}{2}\)

\(\displaystyle S = \sum_{i_m=1}^n \sum_{i_{m-1}=1}^{i_m} \cdots \sum_{i_4=1}^{i_5} \binom{i_4+2}{3} \)

It's obvious from induction that \(\displaystyle S=\sum_{i_m=1}^n \binom{i_m+m-2}{m-1} = \frac{n}{m}\binom{m+n-1}{n}\)

Aditya Narayan Sharma - 1 year, 4 months ago

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Problem 26: Prove : \[\displaystyle \sum_{k \geq 1}\frac{H_k^{(4)}-H_{k-\frac12}^{(4)}}{2k+1}=2\zeta(2)\zeta(3)-(15\ln 2-14)\zeta(4)\]

This problem has been solved by Aditya Kumar.

Aman Rajput - 1 year, 4 months ago

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Partial fractions used here: \[\frac { { 32n }^{ 3 }-{ 24n }^{ 2 }+8n-1 }{ \left( 2n+1 \right) { \left( n \right) }^{ 4 }{ \left( 2n-1 \right) }^{ 4 } } =-\frac { 1 }{ { n }^{ 4 } } +\frac { 2 }{ { n }^{ 3 } } -\frac { 4 }{ { n }^{ 2 } } -\frac { 1 }{ { 2n }-1 } -\frac { 15 }{ 2{ n }+1 } +\frac { 2 }{ { \left( 2n-1 \right) }^{ 2 } } -\frac { 4 }{ { \left( 2n-1 \right) }^{ 3 } } +\frac { 8 }{ { \left( 2n-1 \right) }^{ 4 } } +\frac { 8 }{ { n } } \]

Formula (1) used: \[{ H }_{ \frac { q }{ p } }^{ (m) }=\zeta \left( m \right) -{ p }^{ m }\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { \left( q+pn \right) }^{ m } } } \]

Formula (2) used: \[\sum _{ k=1 }^{ p }{ \frac { 1 }{ 2k+1 } } =\frac { 1 }{ 2 } \left( \Psi \left( p+\frac { 3 }{ 2 } \right) -\Psi \left( \frac { 3 }{ 2 } \right) \right) \]

\[\sum _{ n=1 }^{ \infty }{ \left[ \frac { { H }_{ n }^{ (4) }-{ H }_{ \frac { 2n-1 }{ 2 } }^{ (m) } }{ 2n+1 } \right] } =\sum _{ n=1 }^{ \infty }{ \left[ \frac { 16\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { \left( 2n+2k-1 \right) }^{ 4 } } } -\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { \left( n+k \right) }^{ 4 } } } }{ 2n+1 } \right] } \]

\[=\sum _{ n=1 }^{ \infty }{ \sum _{ k=1 }^{ \infty }{ \left[ \frac { 16{ \left( n+k \right) }^{ 4 }-{ \left( 2n+2k-1 \right) }^{ 4 } }{ \left( 2n+1 \right) { \left( n+k \right) }^{ 4 }{ \left( 2n+2k-1 \right) }^{ 4 } } \right] } } \]

\[=\sum _{ n=1 }^{ \infty }{ \sum _{ k=n+1 }^{ \infty }{ \left[ \frac { 16{ \left( k \right) }^{ 4 }-{ \left( 2k-1 \right) }^{ 4 } }{ \left( 2n+1 \right) { \left( k \right) }^{ 4 }{ \left( 2k-1 \right) }^{ 4 } } \right] } } \]

\[=\sum _{ n=1 }^{ \infty }{ \sum _{ k=n+1 }^{ \infty }{ \left[ \frac { 32{ \left( k \right) }^{ 3 }-{ 24\left( k \right) }^{ 2 }+8k-1 }{ \left( 2n+1 \right) { \left( k \right) }^{ 4 }{ \left( 2k-1 \right) }^{ 4 } } \right] } } \]

\[=\sum _{ n=1 }^{ \infty }{ \left[ \sum _{ k=n }^{ \infty }{ \left[ \frac { 32{ \left( k \right) }^{ 3 }-{ 24\left( k \right) }^{ 2 }+8k-1 }{ \left( 2n+1 \right) { \left( k \right) }^{ 4 }{ \left( 2k-1 \right) }^{ 4 } } \right] } -\frac { 32{ \left( n \right) }^{ 3 }-{ 24\left( n \right) }^{ 2 }+8n-1 }{ \left( 2n+1 \right) { \left( n \right) }^{ 4 }{ \left( 2n-1 \right) }^{ 4 } } \right] } \]

Now, we change the order of the double sum:

\[=\sum _{ k=1 }^{ \infty }{ \sum _{ n=1 }^{ k }{ \left[ \frac { 32{ \left( k \right) }^{ 3 }-{ 24\left( k \right) }^{ 2 }+8k-1 }{ \left( 2n+1 \right) { \left( k \right) }^{ 4 }{ \left( 2k-1 \right) }^{ 4 } } \right] } } -\sum _{ n=1 }^{ \infty }{ \left[ \frac { 32{ \left( n \right) }^{ 3 }-{ 24\left( n \right) }^{ 2 }+8n-1 }{ \left( 2n+1 \right) { \left( n \right) }^{ 4 }{ \left( 2n-1 \right) }^{ 4 } } \right] } \]

Now using the 3 formulas I have provided above and evaluating each summation (which is basic wrt this contest), on simplification gives:

\[\displaystyle \sum_{k \geq 1}\frac{H_k^{(4)}-H_{k-\frac12}^{(4)}}{2k+1}=2\zeta(2)\zeta(3)-(15\ln 2-14)\zeta(4)\]

The simplification is left as an exercise to the readers.

Aditya Kumar - 1 year, 4 months ago

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justify your last double summation. partial fraction is justified. @Aditya Kumar

Aman Rajput - 1 year, 4 months ago

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@Aman Rajput See formula 2 for that.

Aditya Kumar - 1 year, 4 months ago

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Problem 25:

Evaluate:

\[\sum _{ r=1 }^{ n }{ { H }_{ r }^{ 2 } } \]

This problem has been solved by Aman Rajput.

Aditya Kumar - 1 year, 4 months ago

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Using Summation by parts \[\displaystyle \sum _{ k=1 }^{ n }{ { H }_{ k }^{ 2 } } = H_n((n+1)H_n-n)-\sum_{1\leq k \leq n-1}\frac{(k+1)H_k-k}{k+1}\]

\[=\displaystyle (n+1)H_n^2-nH_n-nH_{n-1}+n-1+n-H_n\]

Aman Rajput - 1 year, 4 months ago

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\(\text{Problem 23 :}\)

Prove that : \(\displaystyle \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{(n+a)^4} = \frac{\pi^4}{6} \frac{\cos(\pi a)}{\sin^2 (\pi a)} (6\csc^2 (\pi a) -1) \) for \(a \in \mathbb{R} \) and \(a\) is not an integer.

This problem has been solved by Aman Rajput.

Aditya Narayan Sharma - 1 year, 4 months ago

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\(\text{Alternative :} \)

Consider the function \(\displaystyle f(z)=\frac{1}{(z+a)^4}\) which analytic and has poles at the point \(z=-a\) of order 4.

Considering the square contour positively oriented having vertices \( \pm (1+i)\) and by residue lemma we derive ,

\( \displaystyle Res(-a) = \frac{1}{6} \lim_{z\to -a} \frac{d^3}{dz^3}(π\csc(πz))\)

Finally using alternate summation theorem ,

\(\displaystyle \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{(n+a)^4} =- \frac{1}{6} \lim_{z\to -a} \frac{d^3}{dz^3}(π\csc(πz)) = \frac{\pi^4}{6} \frac{\cos(\pi a)}{\sin^2 (\pi a)} (6\csc^2 (\pi a) -1) \)

which gives the result as desired.

Aditya Narayan Sharma - 1 year, 4 months ago

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firstly , i thought it can be solved using laurent series... but i thought we dont go to the complex analysis . :) nice

Aman Rajput - 1 year, 4 months ago

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@Aman Rajput Its nice to have both real and complex solutions of a problem for enlightenment as we now have for this one

Aditya Narayan Sharma - 1 year, 4 months ago

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\[\displaystyle \sum_{n\geq -\infty} \frac{(-1)^n}{(n+a)^4}=\frac{(-1)^4}{4^2}\left( \sum_{k \geq 0}\frac{1}{(k+1-\frac{a}2)^4}+\frac{1}{(k+\frac{a}2)^4}-\frac{1}{(k-\frac{1-a}2)^4}-\frac{1}{(k-\frac{1+a}2)^4} \right)\]

=\[\frac{1}{16} (\frac16(\psi^{(3)}(1-\frac{a}2)+\psi^{(3)}(\frac{a}2)-\psi^{(3)}(\frac{1-a}2)-\psi^{(3)}(\frac{1+a}2)))\]

Using reflection formula and On a little bit solving you get

\[\frac{\pi^4}{6}\frac{cos(\pi a)}{\sin^2(\pi a)}(6\csc^2(\pi a)-1)\] Sorry for short solution

Aman Rajput - 1 year, 4 months ago

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Yup its ok. Nice solution .

Aditya Narayan Sharma - 1 year, 4 months ago

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Problem 19 :

Prove That

\[ \sum_{r=1}^{n} H_{r}^{(2)} = (n+1)H_{n}^{(2)} - H_{n} \]

Notation : \(\displaystyle H_{n}^{(m)} = \sum_{k=1}^{n} \dfrac{1}{k^m} \) denotes the Generalized Harmonic Number.

This problem has been solved by Deeparaj Bhat.

Ishan Singh - 1 year, 4 months ago

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\[ \begin{align} \sum_{r=1}^n H_r^{(2)} &= \sum_{r=0}^{n-1} \frac{n-r}{(1+r)^2} \\&= \sum_{r=0}^{n-1} \frac{n+1}{(1+r)^2} - \frac{1}{1+r} \\&= (n+1)H_n^{(2)} - H_n \\& Q. E. D. \end{align} \]

Deeparaj Bhat - 1 year, 4 months ago

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Problem 30 :

Evaluate:

\[\displaystyle \sum_{n=1}^{\infty} \frac{3^n}{\binom{2n}{n}} \]

Jack Lam - 1 year, 4 months ago

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Note that,

\[ f(x) = 2 \arcsin \left( \dfrac{x}{2} \right) = \sum_{n=1}^{\infty} \dfrac{x^n}{n^2 \dbinom{2n}{n}} \]

The series is \( \displaystyle x \dfrac{\text{d}}{\text{d}x} \left(x \dfrac{\text{d}}{{\text{d}x}} f(x) \right) \) at \(x=3\)

\( = 3+ \dfrac{4 \pi}{\sqrt{3}} \)

Ishan Singh - 1 year, 4 months ago

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\[\displaystyle \int\limits_{0}^{1}\sum_{n\geq 1} (2n+1)(3x-3x^2)^n dx\]

\[\displaystyle -9\int\limits_{0}^{1} \frac{x^4-2x^3+2x^2-x}{(3x^2-3x+1)^2} dx\]

\[\frac{-9}{27}(-9-4\sqrt3\pi)\]

\[\boxed{\frac13(9+4\sqrt3\pi)}\]

Aman Rajput - 1 year, 4 months ago

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Problem 28: Prove:

\[\displaystyle \sum_{n \geq 1} \frac{(-1)^{n+1}H_n}{n^2 \binom{n+4}{4}}=\frac58\zeta(3)-\frac{25}{24}\zeta(2)+\frac{16}{3}\ln^2(2)-\frac{28}{3}\ln 2 +\frac{727}{144}\]

Aman Rajput - 1 year, 4 months ago

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Starting with \[ \sum_{n=1}^\infty \frac{H_n x^n}{n} = -\int_0^x \frac{\ln(1-u)}{u(1-u)}\,du = \mathrm{Li}_2(x) + \tfrac{1}{2}\ln^2(1-x)\] we have \[f(x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}H_n x^n}{n} = -\mathrm{Li}_2(-x) -\tfrac12\ln^2(1+x)\] Then, on reordering a multiple integral, \[\begin{array}{rcl} S &=& \displaystyle\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{n^2 \binom{n+4}{4}} \; = \; 24\sum_{ n=1}^{\infty} \frac{(-1)^{n+1}H_n}{n^2(n+1)(n+2)(n+3)(n+4)}\\ &=& \displaystyle24\int_0^1dp \int_0^ p dq\int_0^q dr \int_0^r ds\int_0^s \frac{f(t)}{t}\, dt \\ &=& \displaystyle\int_0^1 \frac{(1-t)^4 f(t)}{t}\,dt \\ &=&\displaystyle -\int_0^1 \frac{(1-t)^4}{t}\big\{\mathrm{Li}_2(-t)+\tfrac12\ln^2(1+t)\big\}\, dt \end{array}\] These integral are easy -even WA can do them -and the answer is a simple calculation.

Let Jack post the next question.

Mark Hennings - 1 year, 4 months ago

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This is called solution . Very nice sir ! Jusified clearly

Aman Rajput - 1 year, 4 months ago

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So far I've managed to arrive at a sum of sums by the method of Partial Fractions.

In the hopes it will help anyone, here is the decomposition.

\( \frac{1}{n^2 \binom{n+4}{4}} \equiv \frac{1}{n^2} - \frac{25}{12n} + \frac{4}{n+1} - \frac{3}{n+2} + \frac{4}{3(n+3)} - \frac{1}{4(n+4)} \)

Jack Lam - 1 year, 4 months ago

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You can proceed here by using the generating function.

Aditya Kumar - 1 year, 4 months ago

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Proceeding from the decomposition of the denominator, we break the sum into six parts.

\( \frac{(-1)^{n+1}H_n}{n^2 \binom{n+4}{4}} \equiv -\frac{(-1)^n H_n}{n^2} + \frac{25}{12} \cdot \frac{(-1)^n H_n}{n} + \frac{4(-1)^{n+1} H_n}{n+1} + \frac{3(-1)^{n+2}H_n}{n+2} + \frac{4}{3} \cdot \frac{(-1)^{n+3} H_n}{n+3} + \frac{(-1)^{n+4}}{4(n+4)} \)

Note the powers have been adjusted to match the denominator.

By using the Generating Function for the Harmonic Numbers and swapping the order of integration and summation, we obtain lots of logarithms, and a few Dilogarithms and Trilogarithms.

Use the Polylogarithm identities found on here: http://mathworld.wolfram.com/Trilogarithm.html and here: http://mathworld.wolfram.com/Dilogarithm.html to simplify everything.

Jack Lam - 1 year, 4 months ago

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its better if you paste screenshots of wolframalpha.. that would help here

Aman Rajput - 1 year, 4 months ago

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@Aman Rajput

Jack Lam - 1 year, 4 months ago

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@Jack Lam Hahaha ... 😂😂😂😂😂😂😂

Aman Rajput - 1 year, 4 months ago

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@Aman Rajput You asked for it, I presented it :)

Jack Lam - 1 year, 4 months ago

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@Jack Lam i was not asking these .. i was asking the solution to the problem.

Aman Rajput - 1 year, 4 months ago

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@Aman Rajput The solution I happen to have come up with is extremely tedious. One must use all of the above to evaluate the sum.

I do not know of a faster way to achieve the result.

Jack Lam - 1 year, 4 months ago

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@Jack Lam i know that the result you have posted and the screenshot does not match at all. I know this cant be solved using that. If you know then prove that you are right and i am wrong . If you cant then delete your question within a day. you have one more day to justify your result

Aman Rajput - 1 year, 4 months ago

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@Jack Lam i also can add a solution to your problem using wolframalpha and put a screenshot there. But that wont be good.

Aman Rajput - 1 year, 4 months ago

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@Jack Lam you should post it . how we can believe your solution is correct.

Aman Rajput - 1 year, 4 months ago

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yes. That was one method. Posting all the steps is indeed tedious.

Aditya Kumar - 1 year, 4 months ago

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@Aditya Kumar I look forward to an elegant solution :)

Jack Lam - 1 year, 4 months ago

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Bonus Problem: Evaluate :

\[\displaystyle \sum_{n=-\infty}^{\infty}(-e^{-12\pi})^{n^2}\]


None has solved this question till 5 days

Aman Rajput - 1 year, 4 months ago

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...knock knock, is this comp over?

Hung Woei Neoh - 1 year, 4 months ago

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Comment deleted Jul 22, 2016

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@Aman Rajput Apart from GPs and simple algebraic telescoping, I can't do summations

Hung Woei Neoh - 1 year, 4 months ago

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Let \[k=\frac{\pi^{1/4}}{\Gamma(3/4)}\]

\[\phi(-e^{-2n\pi})=\frac{kr_{2,2n^2}}{n^{1/4}2^{1/4}r_{n,n}}\]

Put \(n=6\) We have \[r_{6,6}=\frac{3^{1/8}\sqrt{1+\sqrt3}(1+\sqrt3+\sqrt23^{3/4})^{1/3}}{2^{13/24}}\]

\[r_{2,72}=\frac{\eta(6i)}{2^{1/4}\eta(12i)}\]

solving this function, and putting back into equation we get

\[\phi(-e^{-12\pi})=\frac{k2^{-19/48}3^{-3/8}(2-3\sqrt2+3^{5/4}+3^{3/4})^{1/3}}{(\sqrt2-1)^{1/12}(\sqrt3+1)^{1/6}(-1-\sqrt3+\sqrt23^{3/4})^{1/3}}\]

Aman Rajput - 1 year, 4 months ago

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