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## Series Contest (SC2)

This is not an ordinary forum. This is the place for the most brilliant ideas and techniques.

Moderators: Random Variable, sos440

### Re: Series Contest (SC2)

Wed Aug 07, 2013 12:33 pm
galactus
Global Moderator

Posts: 902
Nice method, G.

I like that product identity you used at the beginning. I have not seen it before.

### Re: Series Contest (SC2)

Wed Dec 10, 2014 4:45 am
baran

I have a question.
why the integral tend to zero when N tend to infinite?
and what is the pass of the integral?

### Re: Series Contest (SC2)

Wed Dec 10, 2014 4:56 am
baran

galactus wrote:
I think G is into other math pursuits for the time being.

Since the series is alternating, use $\displaystyle \pi\csc(\pi z)$

And we have $\displaystyle \oint\frac{\pi \csc(\pi z)}{(2z+1)\cosh(\frac{\pi}{2}(2z+1))}dz$

The poles are at $\displaystyle z=-1/2, \;\ z=n, \;\ z=\frac{(2n+1)i}{2}-\frac{1}{2}$

$\displaystyle Res(z=-1/2)=\frac{-\pi}{2}$

$\displaystyle Res(z=n)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)\cosh(\frac{\pi}{2}(2n+1))}$

$\displaystyle Res(z=\frac{(2n+1)i}{2}-1/2)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)\cosh(\frac{\pi}{2}(2n+1))}$

Hence:

$\displaystyle \oint\frac{\pi \csc(\pi z)}{(2z+1)\cosh(\frac{\pi}{2}(2z+1))}dz=\frac{-\pi}{2}+4\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)\cosh(\frac{\pi}{2}(2n+1))}$

As $\displaystyle N\to \infty$, then the integral on the left $\displaystyle \to 0$,

So, we have $\displaystyle 0=\frac{-\pi}{2}+4S$

solving for S gives $\displaystyle \frac{\pi}{8}$

I have some question.
why the integeal tend to 0 As $\displaystyle N\to \infty$, and what is N? and Integral Path?

### Re: Series Contest (SC2)

Tue Sep 08, 2015 9:46 pm

Posts: 50
Location: Redmond, WA

sos440 wrote:
Putting together, we obtain
$S = \frac{2}{27}\left\{ \sqrt{3}\pi (2 - \log 3) + \psi_{1}\left( \frac{1}{3} \right) - \psi_{1}\left( \frac{2}{3} \right) \right\}.$

I know this is an old topic, but I would like to add that because of
$$\psi_1\!\left(\tfrac23\right)=\frac{4\pi^2}3-\psi_1\!\left(\tfrac13\right),$$
we can put the result in this form:
$$S=\frac{2\pi}{9\sqrt3}(2-\log3)-\frac{8\pi^2}{81}+\frac4{27}\psi_1\!\left(\tfrac13\right).$$

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