The Fermat prime 257 and binomial sum $\sum_{n=0}^\infty \frac{(-1)^n}{\binom {8n}{4n}}$?

We have,

$\begin{aligned} \sum_{n=0}^\infty \frac{(-1)^n}{\binom n{n/2}} &= \frac{4}{27}(9-\pi\sqrt{3}\,)\\[2.5mm] \sum_{n=0}^\infty \frac{(-1)^n}{\binom {2n}n} &= \frac{4}{5} - \frac{4\sqrt{5}}{25}\ln\left(\frac{1+\sqrt{5}}{2}\right)\\[2.5mm] \sum_{n=0}^\infty\frac{(-1)^n}{\binom{4n}{2n}}&=\frac{16}{17}+\frac{4\sqrt{34}\,(-2+\sqrt{17}\,)}{289\sqrt{-1+\sqrt{17}}}\arctan\left(\frac{\sqrt{2}}{\sqrt{-1+\sqrt{17}}}\right)\\&-\frac{2\sqrt{34}\,(2+\sqrt{17}\,)}{289\sqrt{1+\sqrt{17}}} \ln\left(\frac{\sqrt{1+\sqrt{17}}+\sqrt{2}}{\sqrt{1+\sqrt{17}}-\sqrt{2}}\right)\\[2.5mm] \sum_{n=0}^\infty \frac{(-1)^n}{\binom {8n}{4n}} &=\, ???\end{aligned}$

The third one was found by Renzo Sprugnoli. Question: Anybody knows a closed-form expression for the fourth one? And will the Fermat prime p = 257 appear? (Or is this the law of small numbers again?)

P.S. Strangely, the arguments of the logarithm can be expressed by the Dedekind eta function. Details in my blog.

POSTSCRIPT:

Thanks to Robert Israel's answer, I figured out how to extend it further. His answer (minus the imaginary part) can be expressed as, let,

$\begin{aligned} x_1&=\tfrac{1}{2}(-1)^{1/8}\\ x_2&=\tfrac{1}{2}(-1)^{7/8}\end{aligned}$

then,

$\begin{aligned}\sum_{n=0}^\infty \frac{(-1)^n}{\binom{8n}{4n}} &= \frac{256}{257}+\frac{1}{4}\left(\frac{x_1\arcsin(x_1)}{(1-x_1^2)^{3/2}}+\frac{x_2\arcsin(x_2)}{(1-x_2^2)^{3/2}}-\frac{x_1\rm arcsinh(x_1)}{(1+x_1^2)^{3/2}}-\frac{x_2\rm arcsinh(x_2)}{(1+x_2^2)^{3/2}} \right)\\ &= 0.985791\dots\end{aligned}$

(So the prime 257 does appear!) The form looked susceptible to a generalization so for the next level I tried,

$\begin{aligned} u_1&=\tfrac{1}{2}(-1)^{1/16}\\ u_2&=\tfrac{1}{2}(-1)^{3/16}\\ u_3&=\tfrac{1}{2}(-1)^{13/16}\\ u_4&=\tfrac{1}{2}(-1)^{15/16}\end{aligned}$

$\begin{aligned}\sum_{n=0}^\infty \frac{(-1)^n}{\binom{16n}{8n}} &= \frac{65536}{65537}+\frac{1}{8}\left( \sum_{k=1}^4 \frac{u_k \arcsin(u_k)}{(1-u_k^2)^{3/2}}-\sum_{k=1}^4 \frac{u_k \rm arcsinh(u_k)}{(1+u_k^2)^{3/2}} \right)\\ &=0.999223\dots\end{aligned}$

which worked, and so on. Since Sprugnoli’s version has all coefficients and arguments as real numbers, there might be a way to simplify these sums even further.

There is a connection between all of these expressions. Let $$ F_k(z) = \sum_{n=0}^\infty \frac{(2z)^{kn}}{{kn} \choose {kn/2}}$$ In particular $$F_1(z) = \frac{1}{1-z^2} + \frac{z \arcsin(z)}{(1-z^2)^{3/2}} + \frac{\pi z }{2(1-z^2)^{3/2}}$$ Now for $k > 1$, if $\omega_k$ is a primitive $k$'th root of unity we have $\sum_{j=0}^{k-1} \omega_k^n = k$ if $n$ is divisible by $k$, $0$ otherwise. So $$F_k(z) = \frac{1}{k} \sum_{j=0}^{k-1} \sum_{n=0}^\infty \omega_k^{jn} \frac{(2z)^{n}}{n \choose {n/2}}= \frac{1}{k} \sum_{j=0}^{k-1} F_1(\omega_k^j z)$$ Note that $1/(1-z^2) = \sum_{n=0}^\infty z^{2n}$ while $\pi z/(2 (1-z^2)^{3/2}$ is an odd function, so if $k$ is even we get $$F_k(z) = \frac{1}{1-z^{k}} + \frac{1}{k} \sum_{j=0}^{k-1} \frac{\omega_k^j z \arcsin(\omega_k^j z)}{(1 - \omega_k^{2j} z^2)^{3/2}}$$

You're interested in the case $z = \omega_k^{1/2}/2$ so that $(2z)^{kn} = (-1)^n$, thus when $k$ is even $$F_k(\omega_k^{1/2}/2) = \frac{2^k}{2^k + 1} + \frac{1}{k} \sum_{j=0}^{k-1} \frac{\omega_k^{j+1/2} \arcsin(\omega_k^{j+1/2}/2)}{2 (1- \omega_k^{2j+1}/4)^{3/2}}$$

It seems to be the real part of $$ -{\frac {1}{ \left( -1+{\frac {1}{256}}\,({-1})^{1/4}{256}^{3/4} \right) \left( 1+{\frac {1}{256}}\,({-1})^{1/4}{256}^{3/4} \right) }}+{\frac {1}{512}}\,{\frac {({-1})^{1/8}{256}^{{ {7}/{8}}}\sqrt {1-{\frac {1}{256}}\,({-1})^{1/4} {256}^{3/4}}\arcsin \left( {\frac {1 }{256}}\,({-1})^{1/8}{256}^{{ {7}/{8}}} \right) }{ \left( -1+{ \frac {1}{256}}\,({-1})^{1/4}{256}^{3/4} \right) ^{2}}}-{\frac {1}{ 512}}\,{\frac {({-1})^{1/8}{256}^{{ {7}/{8}}}{\rm arcsinh} \left( {\frac {1}{256}}\,({-1})^{1/8}{256}^{{ {7}/ {8}}} \right) }{ \left( 1+{\frac {1}{256}}\,({-1})^{1/4} {256}^{3/4} \right) ^{3/2}} } $$

EDIT: Here's what I did (in Maple 16):

S0:= sum((-1)^n/binomial(8*n,4*n),n=0..infinity);

$$S0 := {\mbox{$_5$F$_4$}(1/4,1/2,3/4,1,1;\,1/8,3/8,5/8,{\frac {7}{8}};\,-{\frac {1}{256}})}$$

S1:= subs(-1/256=z,S0);

S2:= convert(S1,FPS,z);

$$S2 := \sum _{k=0}^{\infty }{\frac { \left( 4\,k \right) !\,{16}^{-k}{z}^{k}} {{\it pochhammer} \left( 1/4,2\,k \right) {\it pochhammer} \left( 3/4, 2\,k \right) }} $$

S3:= value(S2);

$$S3 := -{\frac {1}{ \left( -1+\sqrt [4]{z} \right) \left( 1+\sqrt [4]{z} \right) }}+1/2\,{\frac {\sqrt [8]{z}\sqrt {1-\sqrt [4]{z}}\arcsin \left( \sqrt [8]{z} \right) }{ \left( -1+\sqrt [4]{z} \right) ^{2}}}- 1/2\,{\frac {\sqrt [8]{z}{\it arcsinh} \left( \sqrt [8]{z} \right) }{ \left( 1+\sqrt [4]{z} \right) ^{3/2}}} $$

S4:= eval(S3, z=-1/256);

Somehow this acquired an imaginary part (I suspect because a choice of branch was made in going from S2 to S3).

Denote our sum by $S_4$. \begin{equation} S_4 := \sum\limits_{n=0}^\infty \frac{(-1)^n}{\binom{8 n}{4 n}} \end{equation} We have: \begin{eqnarray} S_4 &= & 1-4 \int\limits_0^1 \frac{t^3 (1-t)^4}{[1+t^4 (1-t)^4]^2} dt\\ &=& 1-8 \int\limits_0^{\frac{1}{4}}\left(\frac{1}{1-\sqrt{1-4 u}} + \frac{1}{1+\sqrt{1-4 u}}\right) \cdot \frac{u^4}{(1+u^4)^2} \frac{1}{\sqrt{1-4 u}} du \\ &=&1-4 \int\limits_0^{\frac{1}{4}} \frac{u^3}{(1+u^4)^2} \frac{1}{\sqrt{1-4 u}} du \\ &=& \frac{256}{257} + \frac{2}{257} Re[\left(16-i\right) (-1)^{3/4} \cdot \\&&\left(\frac{\left(1+4 \sqrt[4]{-1}\right) \tanh ^{-1}\left(\frac{1}{\sqrt{1-4 \sqrt[4]{-1}}}\right)}{\sqrt{1-4 \sqrt[4]{-1}}}+\frac{\left(4 \sqrt[4]{-1}-1\right) \tanh ^{-1}\left(\frac{1}{\sqrt{1+4 \sqrt[4]{-1}}}\right)}{\sqrt{1+4 \sqrt[4]{-1}}}\right) ] \end{eqnarray} In the first line we used the integral representation of the beta function and we resummed the series under the integral. In the second line we substituted for $u=t(1-t)$. In the third line we simplified the result and finally in the last line we substituted for $\sqrt{1-4 u}$.