# 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, 2 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

- 4 years, 2 months ago

@Aman Rajput Sir, could you please suggest me some good books from where I can learn these type of summations from???

- 1 year, 10 months ago

Such kind of things are not in books. Go for research papers developed by great mathematicians from different universities. Sometimes it requires computers mathematical software like mathematica or maple to solve questions.

- 1 year, 10 months ago

Ohhh......I see........Well, I was already studying some books regarding integrals, and they turned out great!!! So, yeah, now I'll browse through papers too...........!!! Thanks..... :)

- 1 year, 10 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

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

- 4 years, 2 months ago

...knock knock, is this comp over?

- 4 years, 2 months ago

Problem 30 :

Evaluate:

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

- 4 years, 2 months 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 $x=3$

$= 3+ \dfrac{4 \pi}{\sqrt{3}}$

- 4 years, 2 months ago

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

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

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

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

- 4 years, 2 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, 2 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, 2 months ago

This was the best solution!

- 4 years, 2 months ago

Nice solution. I missed the 48 hour time.

- 4 years, 2 months ago

Problem 28: Prove:

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

- 4 years, 2 months 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.

- 4 years, 2 months ago

This is called solution . Very nice sir ! Jusified clearly

- 4 years, 2 months 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.

- 4 years, 2 months ago

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

- 4 years, 2 months ago

- 4 years, 2 months ago

Hahaha ... 😂😂😂😂😂😂😂

- 4 years, 2 months ago

You asked for it, I presented it :)

- 4 years, 2 months ago

i was not asking these .. i was asking the solution to the problem.

- 4 years, 2 months 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.

- 4 years, 2 months 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

- 4 years, 2 months ago

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

- 4 years, 2 months ago

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

- 4 years, 2 months ago

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

- 4 years, 2 months ago

I look forward to an elegant solution :)

- 4 years, 2 months 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.

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

- 4 years, 2 months ago

You can proceed here by using the generating function.

- 4 years, 2 months ago

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, 2 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, 2 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, 2 months ago

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

- 4 years, 2 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, 2 months ago

At least could you elaborate on step 2?

- 4 years, 2 months ago

I used this identity from a ramanujan paper

- 4 years, 2 months 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.

- 4 years, 2 months 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.

- 4 years, 2 months ago

- 4 years, 2 months ago

See formula 2 for that.

- 4 years, 2 months ago

Problem 25:

Evaluate:

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

This problem has been solved by Aman Rajput.

- 4 years, 2 months ago

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

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

- 4 years, 2 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, 2 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, 2 months ago

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

- 4 years, 2 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, 2 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 ,

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

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.

- 4 years, 2 months ago

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

- 4 years, 2 months ago

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

- 4 years, 2 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, 2 months ago

Yup its ok. Nice solution .

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