## Fermat primes and Binomial sums

We have,

\begin{aligned} \sum_{n=0}^\infty \frac{(-1)^n}{\binom n{n/2}} &= \frac{4}{3}-\frac{4\pi\sqrt{3}}{27}\\[2.5mm] \sum_{n=0}^\infty \frac{(-1)^n}{\binom {2n}n} &= \frac{4}{5} - \frac{4\sqrt{5}}{25}\ln\left(\frac{1+\sqrt{5}}{2}\right) \end{aligned}

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

\begin{aligned}\sum_{n=0}^\infty\frac{(-1)^n}{\binom{4n}{2n}}&=\frac{16}{17}+\frac{4\sqrt{34}\,(-2+\sqrt{17}\,)}{17^2\,\sqrt{-1+\sqrt{17}}}\arctan\left(\frac{\sqrt{2}}{\sqrt{-1+\sqrt{17}}}\right)\\&-\frac{2\sqrt{34}\,(2+\sqrt{17}\,)}{17^2\,\sqrt{1+\sqrt{17}}} \ln\left(\frac{\sqrt{1+\sqrt{17}}+\sqrt{2}}{\sqrt{1+\sqrt{17}}-\sqrt{2}}\right) = 0.846609\dots \end{aligned}

(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,

\begin{aligned}&\arctan(z) = \frac{i}{2}\ln\left(\frac{1-i z}{1+i z}\right)\end{aligned}

then,

\begin{aligned}\sum_{n=0}^\infty\frac{(-1)^n}{\binom{4n}{2n}}&=\frac{16}{17}-\frac{2\sqrt{-34}\,(-2+\sqrt{17}\,)}{17^2\,\sqrt{-1+\sqrt{17}}}\ln\left(\frac{\sqrt{-1+\sqrt{17}}+\sqrt{-2}}{\sqrt{-1+\sqrt{17}}-\sqrt{-2}}\right)\\&-\frac{2\sqrt{34}\,(2+\sqrt{17}\,)}{17^2\,\sqrt{1+\sqrt{17}}} \ln\left(\frac{\sqrt{1+\sqrt{17}}+\sqrt{2}}{\sqrt{1+\sqrt{17}}-\sqrt{2}}\right) =0.846609\dots \end{aligned}

In this manner, it reduces to the concise,

\begin{aligned}\sum_{n=0}^\infty\frac{(-1)^n}{\binom{4n}{2n}}&=\frac{16}{17}-\frac{x_1}{17}\ln(y_1)-\frac{x_2}{17}\ln(y_2)\end{aligned}

where, $x_1,x_2$ and $y_1,y_2$ are the appropriate roots of,

\begin{aligned} &289x^4-799x^2-676 = 0\\ &y^4-5y^3+4y^2-5y+1 = 0\end{aligned}

I found that, curiously, the argument of the log can be expressed in terms of the Dedekind eta function, $\eta(z)$.  Let,

\begin{aligned} t_1 &=\frac{1+\sqrt{-5}}{2}\\ t_2 &= \frac{1+\sqrt{-17}}{2}\\ \zeta_{48} &=\exp(\pi i/24)\end{aligned}

then,

\begin{aligned} &\frac{1}{2}\left(\frac{\zeta_{48}\, \eta(t_1)}{\eta(2t_1)}\right)^4 = \frac{1+\sqrt{5}}{2}\\ &\frac{1}{2}\left(\frac{\zeta_{48}\, \eta(t_2)}{\eta(2t_2)}\right)^4 = \frac{\sqrt{1+\sqrt{17}}+\sqrt{2}}{\sqrt{1+\sqrt{17}}-\sqrt{2}}\end{aligned}

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

\begin{aligned} &L_s(z) = \text{Li}_s (z) = \sum_{k=1}^\infty\frac{z^k}{k^s}\end{aligned}

one can find a polylogarithm ladder to express Apery’s constant.  For example, getting the square root and reciprocal of $y_2$ so that z < 1,

\begin{aligned} z &= \sqrt{\frac{\sqrt{1+\sqrt{17}}-\sqrt{2}}{\sqrt{1+\sqrt{17}}+\sqrt{2}}} = 0.480533\dots\end{aligned}

then,

$-12L_3(z)+75L_3(z^2)-68L_3(z^3)-33L_3(z^4)+43L_3(z^6)+12L_3(z^8)-7L_3(z^{12})+2\log^3(1/z) = 3\zeta(3)$

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

\begin{aligned}&\sum_{n=0}^\infty \frac{(-1)^n}{\binom{8n}{2n}}=\,?\end{aligned}

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

### 3 responses to this post.

1. 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

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)}