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Series Contest (SC2)

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Post Thu May 16, 2013 2:36 pm
Shobhit Site Admin
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Welcome! I announce the Beginning of another epic Series Contest (SC2)! The aim of the Series Contest is to improve skills in the computation of infinite series.

You can check out the previous SC and take some ideas.

I will mention the rules briefly:

  • I will start by posting the first problem. If anybody solves it, he is given the ability to post a new one.
  • You may post only one problem at a time.
  • The Scope of questions is only computation of infinite series.
  • The final answer can contain special functions of any kinds.
  • Every question has a lifespan of 24 hours, then the OP has to post the full solution.
  • If someone solves a problem and he has no another to post then he should indicate that.
  • This time we do not have any points for an answer.

Post Mon May 20, 2013 4:33 am
Shobhit Site Admin
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The first problem of this Contest is going to be a Euler Sum.

  • Problem 1
Prove that \(\displaystyle \sum_{n=0}^\infty \frac{H_{2n+1}}{(2n+1)^2}=\frac{21}{16}\zeta(3)\)

Post Mon May 20, 2013 6:06 pm
galactus User avatar
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Hey E:

You always find the fun sums.

No doubt, sos will have a cool way when he comes around. He is good at evaluating Euler sums.

Anyway, I have some work to do on this one yet, but I'll be back.


First, note the identity \(\displaystyle \psi(2n+2)+\gamma=H_{2n+1}\)

This gives us:

\(\displaystyle \sum_{n=0}^{\infty}\frac{\psi(2n+2)+\gamma}{(2n+1)^{2}}=\sum_{n=0}^{\infty}\frac{\psi(2n+2)}{(2n+1)^{2}}+\gamma\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{2}}\)

The rightmost sum is rather famous, and evaluates to \(\displaystyle \frac{\gamma{\pi}^{2}}{8}\)

The left sum with the digamma term evaluates to......get this..............

\(\displaystyle \frac{21}{16}\zeta(3)-\frac{\gamma{\pi}^{2}}{8}\)

adding them results in our required solution.

I hate to leave it like this. Too banal. I will be back and try writing up an evaluation for the digamma sum.

There are thousands of identities and approaches one could take with this one.

\(\displaystyle -\int_{0}^{1}t^{2k}log(t)dt=\frac{1}{(2k+1)^{2}}\)

\(\displaystyle \frac{1}{(2n+1)^{2}}=\int_{0}^{\infty}ue^{-(2n+1)u}du\)

\(\displaystyle \sum_{n=0}^{\infty}\frac{1}{(2n+1)^{2}}\int_{0}^{1}\frac{1-t^{2n+1}}{1-t}dt\)

are possible starts for this one.

Post Tue May 21, 2013 3:44 am
Shobhit Site Admin
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Great! Here is my approach.

Let \(\displaystyle S_1=\sum_{n=1}^\infty \frac{H_n}{n^2}\) and \(\displaystyle S_2 = \sum_{n=1}^\infty(-1)^{n+1}\frac{H_n}{n^2}\).

Then, our sum can be written as

\(\displaystyle \sum_{n=0}^\infty \frac{H_{2n+1}}{(2n+1)^2} = \frac{S_1+S_2}{2}\)

We need to find \(\displaystyle S_1\) and \(\displaystyle S_2\).

1. Calculation of \(\displaystyle S_1\)

Note that

\(\displaystyle \frac{1}{k^2}= \int_0^1\int_0^1 (xy)^{k-1}dx \ dy\)

\(\displaystyle \frac{1}{n}= \int_0^1 z^{n-1} dz\)

With the help of these, \(\displaystyle S_1\) can be calculated.

\(\displaystyle \begin{aligned}S_1 &= \sum_{k=1}^\infty \frac{1}{k^2}\sum_{n=1}^k \frac{1}{n}\\ &=\sum_{k=1}^\infty \sum_{n=1}^k \int_0^1\int_0^1 (xy)^{k-1}dx \ dy \int_0^1 z^{n-1} dz \\ &= \sum_{n=1}^\infty \sum_{k=n}^\infty \int_0^1\int_0^1 (xy)^{k-1}dx \ dy \int_0^1 z^{n-1} dz \\ &= \int_0^1 \int_0^1 \int_0^1 \left( \sum_{n=1}^\infty \frac{(xy z)^{n-1}}{1-xy}\right)dx \ dy \ dz \\ &= \int_0^1 \int_0^1 \int_0^1 \frac{1}{(1-xy)(1-xyz)}dx \ dy \ dz \\&= \int_0^1 \int_0^1 \frac{\log(1-xy)}{xy(xy-1)}dx \ dy \\ &=-\int_0^1 \int_0^1 \frac{\log(1-xy)}{xy}dx dy -\int_0^1 \int_0^1 \frac{\log(1-xy)}{1-xy}dx \ dy \\ &=-\int_0^1 \int_0^1 \frac{\log(1-xy)}{xy}dx dy+\int_0^1 \frac{\log^2(1-y)}{2y}dy\end{aligned}\)

We have

\(\displaystyle \int_0^1 \int_0^1 \frac{\log(1-xy)}{1-xy}dx \ dy =- \sum_{n=1}^\infty \frac{1}{n} \int_0^1 \int_0^1 (xy)^{n-1}dx \ dy \\ = -\sum_{n=1}^\infty \frac{1}{n^3}=-\zeta(3)\)

and

\(\displaystyle \int_0^1 \frac{\log^2(1-y)}{2y}dy = \zeta(3)\)

So \(\displaystyle S_1 = \boxed{2\zeta(3)}\)

2. Calculation of \(\displaystyle S_2\)

This can be done in the same way as the previous one. We will use

\(\displaystyle \dfrac{(-1)^{k-1}}{k^2} = \int_0^1 (-x)^{k-1} dx \int_0^1 z^{k-1} dz = (-1)^{k-1} \int_0^1 \int_0^1 (xz)^{k-1} dx dz\)

Proceeding like the previous one we have

\(\displaystyle \begin{aligned}S_2 &=\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} \sum_{n=1}^k \dfrac{1}{n} \\ &=\sum_{k=1}^{\infty} \sum_{n=1}^k \int_0^1\int_0^1 (-1)^{k-1} (xz)^{k-1}dxdz \int_0^1 y^{n-1} dy\\& = \int_0^1 \int_0^1 \int_0^1 \sum_{n=1}^{\infty} \dfrac{(-xyz)^{n-1}}{1+xz} dx dy dz\\& = \int_0^1 \int_0^1 \int_0^1 \dfrac1{(1+xz)(1+xyz)} dx dy dz\\& = \int_0^1 \int_0^1 \dfrac{\log(1+xz)}{xz(1+xz)} dx dz\\& = \int_0^1 \int_0^1 \dfrac{\log(1+xz)}{xz} dx dz - \int_0^1 \int_0^1 \dfrac{\log(1+xz)}{1+xz} dx dz\\& = \int_0^1 \int_0^1 \dfrac{\log(1+xz)}{xz} dx dz- \int_0^1 \dfrac{\log^2(1+z)}{2z} dz\end{aligned}\)

Here

\(\displaystyle \int_0^1 \int_0^1 \dfrac{\log(1+xz)}{xz} dx dz = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \int_0^1 \int_0^1 (x z)^{n-1} dx \ dz \\ = \sum_{n=1}^\infty \frac{(-1)^n}{n^3}=\frac{3}{4}\zeta(3)\)

and

\(\displaystyle \int_0^1 \frac{\log^2(1+z)}{2z}dz = \frac{\zeta(3)}{8}\)

So \(\displaystyle S_2 = \boxed{\frac{5}{8}\zeta(3)}\)

3. Final answer

\(\displaystyle \begin{aligned}\sum_{n=0}^\infty \frac{H_{2n+1}}{(2n+1)^2} &= \frac{S_1+S_2}{2} \\ &= \frac{2\zeta(3)+\frac{5}{8}\zeta(3)}{2} \\ &= \boxed{\frac{21}{16}\zeta(3)}\end{aligned}\)

Post Tue May 21, 2013 8:38 am
galactus User avatar
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I decided to go another route with this problem.

I had begun as E had with \(\displaystyle 1/2\sum_{n=1}^{\infty}\frac{H_{n}}{n^{2}}-1/2\sum_{n=1}^{\infty}\frac{H_{n}}{n^{2}}\), but tried the digamma thing instead. It is too tedious.

Begin with \(\displaystyle H_{n}=-n\int_{0}^{1}(1-x)^{n-1}\ln(x)dx=-\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^{k}}{k}\)

The above can be shown by integrating by parts: \(\displaystyle \int_{0}^{1}\frac{1-(1-x)^{n}}{x}dx\)

Now, using the binomial series for \(\displaystyle (1-x)^{n}\):

\(\displaystyle \int_{0}^{1}\frac{1-(1-x)^{n}}{x}dx=\int_{0}^{1}\sum_{k=1}^{n}\binom{n}{k}(-1)^{k+1}x^{k-1}dx\)

\(\displaystyle =\sum_{k=1}^{n}\int_{0}^{1}\binom{n}{k}(-1)^{k+1}x^{k-1}dx\)

\(\displaystyle =\sum_{k=1}^{n}\frac{\binom{m}{k}(-1)^{k+1}}{k}\)

If we use \(\displaystyle \int_{0}^{u}\frac{1-(1-x)^{n}}{x}dx\) and let \(\displaystyle t=1-x\), then

\(\displaystyle \int_{1-u}^{1}\frac{1-t^{n}}{1-t}dt\)

Now, use a partial geometric series:

\(\displaystyle \int_{1-u}^{1}\sum_{k=1}^{n}t^{k-1}dt=\sum_{k=1}^{n}\frac{1-(1-t)^{k}}{k}\)

Now, let \(\displaystyle u=1\) and we get \(\displaystyle \sum_{k=1}^{n}\frac{1}{k}=H_{n}\)

Now, using the dilog:

\(\displaystyle \sum_{n=1}^{\infty}\frac{H_{n}}{n^{2}}=-\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{1}(1-x)^{n-1}\ln(x)dx\)

\(\displaystyle =-\int_{0}^{1}\sum_{n=1}^{\infty}\frac{(1-x)^{n-1}}{n}\ln(x)dx\)

By the definition of the polylog:

\(\displaystyle \sum_{n=1}^{\infty}\frac{(1-x)^{n-1}}{n}=Li_{1}(1-x)\)

\(\displaystyle \sum_{n=1}^{\infty}\frac{H_{n}}{n^{2}}=-\int_{0}^{1}\frac{Li_{1}(1-x)\ln(x)}{1-x}dx\)

\(\displaystyle =1/2\int_{0}^{1}\frac{ln^{2}(x)}{1-x}dx=\zeta(3)\)

The other sum can be done pretty much the same way exceot it's alternating and the result is

\(\displaystyle 1/2\int_{0}^{1}\frac{Li_{1}(x-1)\ln(x)}{1-x}dx=\frac{-5}{16}\zeta(3)\)

Now, subtract them and it is just as E showed.


EDIT: I found what I thought was a cool way to evaluate this one using digamma series.

That is, using the technique I mentioned previously by evaluating \(\displaystyle \sum_{n=0}^{\infty}\frac{\psi(2n+2)}{(2n+1)^{2}}=\frac{21}{16}\zeta(3)-\frac{{\pi}^{2}\gamma}{8}\)

It should work, but for some reason the numbers do not add up. It may merely be a sign issue. If I find the snafu I will post it.

Post Wed May 22, 2013 5:52 am
Shobhit Site Admin
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Very nicely done, galactus! I think its time to post another series.

  • Problem 2

Prove that

\(\displaystyle \sum_{n=1}^\infty \frac{H_n}{2^n n^2}=-\frac{\pi^2}{12}\log(2)+\zeta(3)\)

Post Wed May 22, 2013 9:11 am
galactus User avatar
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It would appear this one can be related to the first integral in the three log integrals you posted. The one with the upper limit of 1/2.

\(\displaystyle \sum_{n=1}^{\infty}\frac{H_{n}}{n^{2}}x^{n}=\int_{0}^{x}\frac{Li_{2}(t)+1/2log^{2}(1-t)}{t}dt\).....[1]

By letting \(\displaystyle x=1/2\) and integrating we get:

\(\displaystyle \int_{0}^{x}\frac{Li_{2}(t)+1/2log^{2}(1-t)}{t}dt=\left -Li_{3}(1-t)+Li_{3}(t)+Li_{2}(1-t)log(1-t)+1/2log(t)log^{2}(1-t)\right |_{0}^{1/2}\)

Note that \(\displaystyle Li_{3}(1/2)=\frac{7}{8}\zeta(3)+\frac{1}{6}\log^{3}(2)-\frac{{\pi}^{2}}{12}log(2)\)

and \(\displaystyle Li_{2}(1/2)=\frac{{\pi}^{2}}{12}-\frac{log^{2}(2)}{2}\)

So, we arrive at:

\(\displaystyle =\boxed{\zeta(3)-\frac{{\pi}^{2}}{12}log(2)}\)


The integral in [1] can be derived by beginning with:

\(\displaystyle \sum_{n=1}^{\infty}\frac{H_{n}}{n}x^{n}=\int_{0}^{x}\sum_{n=1}^{\infty}H_{n}t^{n-1}dt\)

\(\displaystyle =-\int_{0}^{x}\frac{log(1-t)}{t}dt-\int_{0}^{x}\frac{log(1-t)}{1-t}dt\)

Dividing by t and integrating gives our series in question.

Of course, let x = 1/2.



problem #3:

I reckon I should post one.

Here is a good one. It is a little more challenging in that the exponent is even.

\(\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{4}}\)

Post Wed May 22, 2013 6:40 pm
Shobhit Site Admin
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galactus wrote:
problem #3:

I reckon I should post one.

Here is a good one. It is a little more challenging in that the exponent is even.

\(\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{4}}\)



galactus, I can express it in terms of Hurwitz Zeta Function but I don't know how to simplify it further. :cry:

Post Wed May 22, 2013 7:03 pm
zaidalyafey Global Moderator
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ETYUCAN wrote:


galactus, I can express it in terms of Hurwitz Zeta Function but I don't know how to simplify it further. :cry:


I think solving \(\displaystyle \zeta(2,a)\) will solve a family of series .
Wanna learn what we discuss , see Book

Post Wed May 22, 2013 7:05 pm
galactus User avatar
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Break it up into its odd and even components and use the series for digamma derivatives.

The solution should be

\(\displaystyle \frac{{\pi}^{4}}{96}-\frac{1}{768}\psi^{(3)}(3/4)\) or some other equivalent form.


The odd and even components thing makes it \(\displaystyle \sum_{n=0}^{\infty}\frac{1}{(4n+1)^{4}}-\sum_{n=0}^{\infty}\frac{1}{(4n+3)^{4}}\)

Now, with a wee bit of algebra they can be put into a form that is the third derivative of digamma series.

Does that help any?. That's how I approached it. Your way may be better using the zeta thing. As Z said, a general form.

But, If I am not mistaken, there is no general form known when the exponent is even. If it is odd, then yes there is a general form involving the Euler numbers.

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