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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.


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,

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


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

(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.

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Could you figure out the reference for such formulas? – Frank Science Jun 4 '12 at 2:52
@frank: The first two are well known. The third can be found at – Tito Piezas III Jun 4 '12 at 5:53
up vote 4 down vote accepted

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 $$F_k(\omega_k^{1/2}/2) = \frac{2^k}{2^k + 1} + \frac{1}{k} \sum_{j=0}^{k-1} \frac{\omega_k^{j+1/2} \arcsin(\omega_k^{j+1/2}/2)}{2 (1- \omega_k^{2j+1}/4)^{3/2}}$$

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Thanks for completing the analysis. Now if only we can convert the complex terms into real ones like Sprugnoli's, as it is connected to an observation I made in my blog about a certain "coincidence" between a term and the Dedekind eta function. – Tito Piezas III Jun 5 '12 at 19:40
If $0 < t < \pi$, then according to Maple $$\eqalign{\arcsin \left({{\rm e}^{it}}/2 \right) &=-\arcsin \left( -1/4\, \sqrt {5+4\,\cos \left( t \right) }+1/4\,\sqrt {5-4\,\cos \left( t \right) } \right)\cr +i &\left( -2\,\ln \left( 2 \right) \right.\cr &\left.+\ln \left( \sqrt {5+4\,\cos \left( t \right) }+\sqrt {5-4\,\cos \left( t \right) }+\sqrt {-6+2\,\sqrt {5+4\,\cos \left( t \right) }\sqrt {5-4\,\cos \left( t \right) }} \right) \right) \cr} $$ – Robert Israel Jun 5 '12 at 21:32
Ah, the terms will now come as pairs of $(a+bi)(c+di),(a-bi)(c-di)$ and their sum, of course, is the real number $2(ac-bd)$. I know how to take it from here. – Tito Piezas III Jun 6 '12 at 9:14

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).

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So, Tito, is 256 close enough for you? – Gerry Myerson Jun 4 '12 at 0:46
@gerrymyerson: well, the minimal polynomial of the arcsinh's and arcsin's argument is $256x^4-1024x^3+1536x^2-1024x+257 = 0$, so $257$ does appear, hehe. – Tito Piezas III Jun 4 '12 at 5:34
@Robert Just curious. How did you come up with such an expression? – user17762 Jun 4 '12 at 5:40
@robert: thanks, dr israel! However, there is a typo in the third term. The numerator should start with $(-1)^{1/8}$, not $(-1)^{1/4}$. The expression evaluates to $a+bi$ where $a$ is the desired value. To seal the answer, what is the expression for its conjugate $a-bi$? – Tito Piezas III Jun 4 '12 at 5:42
Typo corrected, thanks. – Robert Israel Jun 4 '12 at 7:43

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