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Series Contest (SC2)

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Moderators: Random Variable, sos440

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

Posts: 902
Nice method, G.

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

Post Wed Dec 10, 2014 4:45 am

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

Post Wed Dec 10, 2014 4:56 am

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 \(\displaystyle z=-1/2, \;\ z=n, \;\ z=\frac{(2n+1)i}{2}-\frac{1}{2}\)

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

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

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


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

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?

Post 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
we can put the result in this form:


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