# 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.

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]** 

4 years, 8 months ago

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Problem 27:

Prove:

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

- 4 years, 7 months ago

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 $\frac{(\frac{\pi}{2})^{1/4}}{\Gamma(3/4)}$

- 4 years, 7 months ago

At least could you elaborate on step 2?

- 4 years, 7 months ago

I used this identity from a ramanujan paper

- 4 years, 7 months ago

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

- 4 years, 7 months ago

share your q product .... if i have time i can solve it easily

- 4 years, 7 months ago

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

- 4 years, 7 months ago

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.

- 4 years, 8 months ago

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:

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

- 4 years, 8 months ago

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.

- 4 years, 8 months ago

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.

- 4 years, 7 months ago

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.

- 4 years, 7 months ago

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

- 4 years, 7 months ago

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 $(H_n)^2$.

- 4 years, 7 months ago

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,

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

- 4 years, 7 months ago

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.

- 4 years, 7 months ago

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: $\boxed{\displaystyle \sum_{k \geq 1} \frac{F_{2k}H_{k-1}^{(2)}}{k^2\binom{2k}{k}}=\frac{2\pi^4}{375\sqrt5}}$

- 4 years, 7 months ago

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

- 4 years, 7 months ago

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.

- 4 years, 7 months ago

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.

- 4 years, 7 months ago

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 $\displaystyle S=\sum_{i_m=1}^n \binom{i_m+m-2}{m-1} = \frac{n}{m}\binom{m+n-1}{n}$

- 4 years, 7 months ago

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.

- 4 years, 7 months ago

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

- 4 years, 7 months ago

Proof of that generating function?Thanks.

- 4 years, 7 months ago

There are many ways but I learnt it here.

- 4 years, 7 months ago

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.

- 4 years, 7 months ago

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.

- 4 years, 7 months ago

Nice solution. I missed the 48 hour time.

- 4 years, 7 months ago

This was the best solution!

- 4 years, 7 months ago

$\text{Problem 23 :}$

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.

- 4 years, 7 months ago

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

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

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

- 4 years, 7 months ago

Yup its ok. Nice solution .

- 4 years, 7 months ago

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