We have,

For the next step, Renzo Sprugnoli gave the Ramanujan-like identity,

(The sign of the third term has been changed by this author.) However, to make it more symmetrical, we can express the *arctan* in terms of the *log function*. Since,

then,

In this manner, it reduces to the concise,

where, and are the appropriate roots of,

I found that, curiously, *the argument of the log can be expressed in terms of the Dedekind eta function*, . Let,

then,

*Is this coincidence?* Furthermore, using these as the argument of the *polylogarithm*,

one can find a *polylogarithm ladder* to express *Apery’s constant.* For example, getting the square root and reciprocal of so that *z* < 1,

then,

A simpler one exists for the other argument. The next step, of course, is,

Since the first three *Fermat primes* 3, 5, 17 have already appeared, it should be interesting to conjecture if 257 will be next.

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Posted by Nikos Bagis on May 4, 2013 at 3:44 am

Set $p_{s,k}(z)$ to be the $k$-root of the equation $-1+zX^{s-1}+zX^s=0$, then

in general

\begin{equation}

A_s(z)=\sum^{\infty}_{m=1}\frac{z^m}{\left(^{sm}_m\right)}=

\frac{1}{z}\sum^{s}_{k=1}\frac{\log\left(\frac{1+p_{s,k}(z)}{p_{s,k}(z)}\right)}{p_{s,k}(z)^{s-2}\left(s-1+sp_{s,k}(z)\right)}

\end{equation}

Posted by Nikos Bagis on May 4, 2013 at 3:46 am

I have a proof of the above theorem. It is quite elementary.What do you think about it?

Posted by tpiezas on May 5, 2013 at 4:37 am

Elegant, though I had to paste your LaTeX onto Wikipedia to see it clearly. (It seems WordPress does not support LaTeX in the responses.)

You may want to see this post by Robert Israel who answered my question at http://math.stackexchange.com/questions/153504.