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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.

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]** 

6 months, 3 weeks ago

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Problem 27:

Prove:

$\sum _{ n=-\infty }^{ \infty }{ { \left( -e^{-\pi} \right) }^{ { n }^{ 2 } } } =\frac { { \left( \frac { \pi }{ 2 } \right) }^{ \frac { 1 }{ 4 } } }{ \Gamma \left( \frac { 3 }{ 4 } \right) }$ · 6 months, 1 week 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 $\frac{(\frac{\pi}{2})^{1/4}}{\Gamma(3/4)}$ · 6 months, 1 week 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 · 6 months, 1 week ago

share your q product .... if i have time i can solve it easily · 6 months, 1 week 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 · 6 months, 1 week ago

At least could you elaborate on step 2? · 6 months, 1 week ago

I used this identity from a ramanujan paper · 6 months, 1 week 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. · 6 months, 2 weeks 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.

· 6 months, 2 weeks 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. · 6 months, 3 weeks 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:

$\boxed{\displaystyle\sum_{n=0}^{\infty} \dfrac{2^{n-1} n!}{\prod_{r=0}^{n} (4r+1)} = \dfrac{\pi + 2\log(1+\sqrt{2})}{4 \sqrt{2}}}$ · 6 months, 2 weeks 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. · 6 months, 2 weeks 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: $\boxed{\displaystyle \sum_{k \geq 1} \frac{F_{2k}H_{k-1}^{(2)}}{k^2\binom{2k}{k}}=\frac{2\pi^4}{375\sqrt5}}$ · 6 months, 1 week ago

(+1) Nicely explained! I solved using Beta Functions and Integration. · 6 months, 1 week 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. · 6 months, 2 weeks 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,

$\text{S} = (H_{n+1})^2 - H_{n+1}^{(2)} \quad \square$ · 6 months, 2 weeks 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 , $$\displaystyle \sum_{k=1}^{n} \frac{H_k}{n-k+1} = (H_{n+1})^2-H_{n+1}^{(2)}$$ · 6 months, 2 weeks 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 $$(H_n)^2$$. · 6 months, 2 weeks 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. · 6 months, 1 week 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. · 6 months, 1 week ago

This was the best solution! · 6 months ago

Nice solution. I missed the 48 hour time. · 6 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. · 6 months, 2 weeks 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. · 6 months, 1 week ago

Comment deleted 6 months ago

Comment deleted 6 months ago

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. · 6 months, 1 week ago

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. · 6 months, 1 week 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 $$\displaystyle S=\sum_{i_m=1}^n \binom{i_m+m-2}{m-1} = \frac{n}{m}\binom{m+n-1}{n}$$ · 6 months, 2 weeks ago

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. · 6 months, 1 week ago

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. · 6 months, 1 week ago

justify your last double summation. partial fraction is justified. @Aditya Kumar · 6 months, 1 week ago

See formula 2 for that. · 6 months, 1 week ago

Problem 25:

Evaluate:

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

This problem has been solved by Aman Rajput. · 6 months, 1 week ago

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

$=\displaystyle (n+1)H_n^2-nH_n-nH_{n-1}+n-1+n-H_n$ · 6 months, 1 week 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. · 6 months, 2 weeks 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 ,

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

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. · 6 months, 2 weeks ago

firstly , i thought it can be solved using laurent series... but i thought we dont go to the complex analysis . :) nice · 6 months, 2 weeks ago

Its nice to have both real and complex solutions of a problem for enlightenment as we now have for this one · 6 months, 2 weeks 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 · 6 months, 2 weeks ago

Yup its ok. Nice solution . · 6 months, 2 weeks 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. · 6 months, 2 weeks 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, $$\large \boxed{\displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} 5^{-n} = \sqrt{5}}$$ · 6 months, 2 weeks ago

Proof of that generating function?Thanks. · 6 months, 2 weeks ago

There are many ways but I learnt it here. · 6 months, 2 weeks ago

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. · 6 months, 2 weeks ago

\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} · 6 months, 2 weeks ago

Problem 30 :

Evaluate:

$\displaystyle \sum_{n=1}^{\infty} \frac{3^n}{\binom{2n}{n}}$ · 6 months, 1 week ago

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 $$x=3$$

$$= 3+ \dfrac{4 \pi}{\sqrt{3}}$$ · 6 months ago

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

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

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

$\boxed{\frac13(9+4\sqrt3\pi)}$ · 6 months ago

Problem 28: Prove:

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

· 6 months, 1 week ago

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. · 6 months, 1 week ago

This is called solution . Very nice sir ! Jusified clearly · 6 months, 1 week ago

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.

$$\frac{1}{n^2 \binom{n+4}{4}} \equiv \frac{1}{n^2} - \frac{25}{12n} + \frac{4}{n+1} - \frac{3}{n+2} + \frac{4}{3(n+3)} - \frac{1}{4(n+4)}$$ · 6 months, 1 week ago

You can proceed here by using the generating function. · 6 months, 1 week ago

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. · 6 months, 1 week ago

its better if you paste screenshots of wolframalpha.. that would help here · 6 months, 1 week ago

· 6 months, 1 week ago

Hahaha ... 😂😂😂😂😂😂😂 · 6 months, 1 week ago

You asked for it, I presented it :) · 6 months, 1 week ago

i was not asking these .. i was asking the solution to the problem. · 6 months, 1 week ago

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. · 6 months, 1 week ago

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 · 6 months, 1 week ago

i also can add a solution to your problem using wolframalpha and put a screenshot there. But that wont be good. · 6 months, 1 week ago

you should post it . how we can believe your solution is correct. · 6 months, 1 week ago

yes. That was one method. Posting all the steps is indeed tedious. · 6 months, 1 week ago

I look forward to an elegant solution :) · 6 months, 1 week ago

Bonus Problem: Evaluate :

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

None has solved this question till 5 days · 6 months ago

...knock knock, is this comp over? · 6 months ago

Comment deleted 6 months ago

Apart from GPs and simple algebraic telescoping, I can't do summations · 6 months ago

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

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

· 6 months ago