Nice method, G.

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

Board index **‹** Contests **‹** Series Contest (SC2)
## Series Contest (SC2)

I have some question.

why the integeal tend to 0 As \(\displaystyle N\to \infty\), and what is N? and Integral Path?

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:

$$S=\frac{2\pi}{9\sqrt3}(2-\log3)-\frac{8\pi^2}{81}+\frac4{27}\psi_1\!\left(\tfrac13\right).$$

**Moderators:** Random Variable, sos440

Nice method, G.

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

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

baran

I have a question.

why the integral tend to zero when N tend to infinite?

and what is the pass of the integral?

why the integral tend to zero when N tend to infinite?

and what is the pass of the integral?

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 \(\displaystyle \frac{\pi}{8}\)

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?

sos440 wrote:

Putting together, we obtain

\[S = \frac{2}{27}\left\{ \sqrt{3}\pi (2 - \log 3) + \psi_{1}\left( \frac{1}{3} \right) - \psi_{1}\left( \frac{2}{3} \right) \right\}.\]

\[S = \frac{2}{27}\left\{ \sqrt{3}\pi (2 - \log 3) + \psi_{1}\left( \frac{1}{3} \right) - \psi_{1}\left( \frac{2}{3} \right) \right\}.\]

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:

$$S=\frac{2\pi}{9\sqrt3}(2-\log3)-\frac{8\pi^2}{81}+\frac4{27}\psi_1\!\left(\tfrac13\right).$$