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Series Contest (SC4)

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Post Tue Jul 08, 2014 10:59 am
galactus User avatar
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Yes, I thought of that but was not sure exactly how to deal with it. Upon finding the residues, one manages to get the series in question among them. Then, solving for it would do the trick. But, of course, it does not work due to the branch. It sure looks like it should though. I wonder if there is a way to deal with it in this manner?.

i.e.

The series at the integers n:

$$\frac{(-1)\ln(n)}{(z-n)^{2}}+\frac{(-1)^{n}(1/n+H_{n}\ln(n))}{z-n}+.....$$

some interesting series are involved. such as, $$\sum_{n=1}^{\infty}\frac{(-1)^{n}\ln(n)}{n^{2}}=1/2\zeta'(2)+\frac{\pi^{2}}{12}\ln(2)$$

Of course, the negative integers have complex involved due to the negative.

anyway, I just played with it a little. It looks like it should work if handled properly.

Maybe not though.

sorry, gotta go to work now. Be back later.

Post Tue Jul 08, 2014 11:18 am
Shobhit Site Admin
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Anyway, here's another:

Problem 20

\(\displaystyle \sum_{k=1}^\infty \frac{(-1)^{k-1}k}{\sinh(\pi k)}=\frac{1}{4\pi}\)

Post Tue Jul 08, 2014 12:26 pm
Random Variable Integration Guru
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The Mellin transform of $ \displaystyle \frac{x}{\sinh \pi x}$ is $$\begin{align} \Big\{\mathcal{M}\ \frac{x}{\sinh \pi x} \Big\} (s) &= 2 \pi^{-s-1} \Gamma(s+1) \chi_{s+1}(1) \\ &= \pi^{-s-1} \Gamma(s+1) \Big(\zeta(s+1) + 2^{-s} \zeta(s+1) \Big) . \end{align}$$

Then using the inverse Mellin transform, $$ \frac{x}{\sinh \pi x} = \frac{1}{2 \pi i } \int_{c- i \infty}^{c+ i \infty} \pi^{-s-1}\Gamma(s+1) \Big(\zeta(s+1) + 2^{-s} \zeta(s+1) \Big) x^{-s} \ ds $$ where $c$ is any value greater than $0$.


Replacing $x$ with $-n$ and summing both sides, $$ \begin{align} \sum_{k=1}^{\infty} \frac{(-1)^{k-1} k}{\sinh \pi k} &= \frac{1}{2 \pi i }\int_{c- i \infty}^{c+ i \infty} \pi^{-s-1} \Gamma(s+1) \Big(\zeta(s+1) + 2^{-s} \zeta(s+1) \Big) (1-2^{1-s}) \zeta(s) \ ds \\ &= \frac{1}{2 \pi i } \int_{c- i \infty}^{c+ i \infty} f(s) \ ds . \end{align}$$

The integrand has a simple pole at the origin with residue $ \displaystyle \frac{1}{2 \pi}$.

The integrand is also odd along the imaginary axis. I used Wolfram Alpha to confirm this.

So we can shift the contour to the imaginary axis.

Then $$ \sum_{k=1}^{\infty} \frac{(-1)^{k-1} k}{\sinh \pi k} = \frac{1}{2 \pi i } \pi i \ \text{Res}[f(s),0] = \frac{1}{4 \pi} .$$

Post Tue Jul 08, 2014 1:14 pm
Shobhit Site Admin
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Very nice RV! :D Mellin Transform is nice way to solve hyperbolic sums.

Post Tue Jul 08, 2014 10:02 pm
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Sometimes that method works. But often you can't find anywhere to shift the contour so that the integral can be evaluated exactly.

Post Fri Jul 11, 2014 2:22 pm
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The Mellin transform is a cool way to go G.

As I am sure you all know, this one can also be done by the standard residue sum method by considering

$$\oint_{C}\frac{z\pi \csc(\pi z)}{\sinh(\pi z)}$$

The poles lie at $z=n$ and $z=ni$

The residue at 0 is:

$$\frac{\pi z}{\sin(\pi z)\sinh(\pi z)}=\frac{\pi z}{\left(\pi z-\frac{(\pi z)^{3}}{3!}+\frac{(\pi z)^{5}}{5!}-\cdot\cdot\cdot \right)\left(\pi z+\frac{(\pi z)^{3}}{3!}+\frac{(\pi z)^{5}}{5!}+\cdot\cdot\cdot \right)}$$

$$=\frac{1}{\pi z}+\frac{\pi^{3}}{90}z^{3}+\cdot\cdot\cdot $$

The coefficient of the 1/z term being $\frac{1}{\pi}$

By L'Hopital, the sum of residues at $z=n, \;\ n=\pm 1, \pm 2, \pm 3,....$:

$$\sum_{n=1}^{\infty}\frac{(-1)^{n}n}{\sinh(\pi n)}$$

Likewise, sum of residues at $z=ni$:

$$\sum_{n=1}^{\infty}\frac{(-1)^{n}n}{\sinh(\pi n)}$$

By the residue theorem, the negative sum of the residues, $\displaystyle \sum_{-\infty}^{\infty}f(k)=-S$

$$\underbrace{\oint \frac{\pi csc(\pi z)z}{\sinh(\pi z)}dz}_{\text{0}}=-\frac{1}{\pi}+4\sum_{n=1}^{N}\frac{(-1)^{n}n}{\sinh(\pi n)}$$

By the usual argument, the integral on the left tends to 0 and solving for the required sum results in

$$\frac{1}{4\pi}$$

Post Fri Jul 11, 2014 3:54 pm
Random Variable Integration Guru
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I didn't even consider that approach because I didn't think the integral would vanish.

But the $\csc (\pi z)$ (as opposed to $\cot (\pi z)$) should cause the integral to vanish along the top and bottom of the square. And the $\text{csch} (\pi z)$ should cause the integral to vanish along the left and right sides of the square.

Post Fri Jul 11, 2014 4:45 pm
galactus User avatar
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Yes, G, it happened to work out. But, that Mellin thing looks pretty cool and can be mighty powerful.

You know something from the past I saw you do that I liked was the use of that nth root of unity thing for solving partial sums. Remember what I mean?.

Post Fri Jul 18, 2014 10:49 am
galactus User avatar
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Here's another sum to try. It may be a pretty challenging Euler sum. I am trying to revive the contest because it's been stagnant lately. S, you're pretty dang good at these. where ya' been lately?. :)


$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}H_{n}}{n}\left(1-\frac{1}{3}+\frac{1}{5}-\cdot\cdot\cdot +\frac{(-1)^{n+1}}{2n-1}\right)$$

Post Fri Jul 18, 2014 7:13 pm
Random Variable Integration Guru
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What does that sum evaluate to?


Something I came across fairly recently is $$ \sum_{n=0}^{\infty} \left(H_{n}^{-} - \log 2\right) ^{2} = \log 2$$

where $\displaystyle H_{n}^{-} = \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k}$ (i.e., the alternating harmonic numbers) and where $H_{0}^{-}$ is defined to be zero.

The sum converges absolutely because as $n \to \infty$, $H_{n}^{-} \sim \log(2) + \frac{(-1)^{n-1}}{2n}$.

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