# Closed form for $S(m) = \sum_{n=1}^\infty \frac{2^n \cdot n^m}{\binom{2n}n}$ for integer $m$?

What is the (simple) closed form for $\large \displaystyle S(m) = \sum_{n=1}^\infty \dfrac{2^n \cdot n^m}{\binom{2n}n}$ for integer $m$?

Notation: $\dbinom{2n}n$ denotes the central binomial coefficient, $\dfrac{(2n)!}{(n!)^2}$.

We have the following examples (all verified by WolframAlpha) for $m\geq 0$:

$$\begin{eqnarray} S(0) &=&\sum_{n=1}^\infty \dfrac{2^n }{\binom{2n}n} = 2+ \dfrac{\pi}2 \\ S(1) &=&\sum_{n=1}^\infty \dfrac{2^n \cdot n }{\binom{2n}n} = 3+ \pi \\ S(2) &=&\sum_{n=1}^\infty \dfrac{2^n \cdot n^2 }{\binom{2n}n} = 11+ \dfrac{7\pi}2 \\ S(3) &=&\sum_{n=1}^\infty \dfrac{2^n \cdot n^3 }{\binom{2n}n} = 55+ \dfrac{35\pi}2 \\ S(4) &=&\sum_{n=1}^\infty \dfrac{2^n \cdot n^4 }{\binom{2n}n} = 355 + 113\pi \approx \underline{709.9999}698 \ldots , \quad \text{So close to an integer? Coincidence?} \\ S(5) &=&\sum_{n=1}^\infty \dfrac{2^n \cdot n^5 }{\binom{2n}n} = 2807+ \dfrac{1787\pi}2 \\ S(6) &=&\sum_{n=1}^\infty \dfrac{2^n \cdot n^6 }{\binom{2n}n} = 26259+ \dfrac{16717\pi}2 \\ S(7) &=&\sum_{n=1}^\infty \dfrac{2^n \cdot n^7 }{\binom{2n}n} = 283623+ 90280\pi \\ \end{eqnarray}$$

And for $m<0$ as well (all verified by WolframAlpha as well):

$$\begin{eqnarray} S(-1) &=&\sum_{n=1}^\infty \dfrac{2^n \frac 1 n}{\binom{2n}n} = \dfrac{\pi}2 \\ S(-2) &=&\sum_{n=1}^\infty \dfrac{2^n \frac 1 {n^2} }{\binom{2n}n} = \dfrac{\pi^2 }8 \\ S(-3) &=&\sum_{n=1}^\infty \dfrac{2^n \frac 1 {n^3} }{\binom{2n}n} = \pi G - \dfrac{35\zeta(3)}{16} + \dfrac18 \pi^2 \ln2, \quad G \text{ denotes Catalan's constant} \\ \end{eqnarray}$$

So a natural question arise: Is there a closed form for $S(m)$ for all integers $m$?

For what it's worth, I couldn't get the (simple) closed form (using WolframAlpha alone) of $S(-4), S(-5), S(-6), \ldots$ and $S(8), S(9) , \ldots$ without expressing it in terms of hypergeometric functions.

My question is: Is there a closed form of $S(m)$ for all integers $m$ without using hypergeometric functions? And how do I compute all of these values?

Note that I don't consider hypergeometric functions to be a "legitimate" function because it defeats the purpose of this question.

My motivation: I was trying to solve this question and I decided to use the hint suggested by Mandrathrax, that is to use partial fractions to get

$$\dfrac{5n^5+5n^4+5n^3+5n^2-9n+9}{(2n+1)(2n+2)(2n+3)} = \dfrac{5n^2}8 - \dfrac{5n}8 - \dfrac9{n+1} + \dfrac{457}{64(2n+1)} + \dfrac{135}{64(2n+3)} + \dfrac{85}{32}$$

So if I can prove the values of $S(0), S(1), S(2)$ (which I failed to do so), then I'm pretty sure I'm halfway done with my solution.

Why do I still want to solve that question when it already has 21 upvotes? Because I think there's a simpler solution and I personally don't like to use polylogarithms.

My feeble attempt (with help from my friends Aareyan and Julian) to solve my own question: The Taylor series of $(\arcsin x)^2$ is $\displaystyle \dfrac12 \sum_{n=1}^\infty \dfrac1{n^2 \binom{2n}n} (2x)^n$. Differentiating with respect to $x$ then multiply by $x$ (repeatedly) gives some resemblance of $S(m)$, but these series only holds true for $|x| < 1$ and not $x=1$ itself. Now I'm stucked.

EDIT1 (14 May 2016, 1101 GMT): Twice the coefficients of $\pi$ for $S(m)$ with $m\geq0$ appears to follow this OEIS sequence, A014307.

EDIT2 (14 May 2016, 1109 GMT): The constant for $S(m)$ with $m\geq1$ appears to follow this OEIS sequence, A180875.

• It is odd, but I just noticed that the coefficient of $\pi$ for $m=3,5$, when you divide them, it is almost the constant added to the $\pi$ terms divided. Same for $m=5,6$ and $m=2,3$ May 14 '16 at 10:58
• Why is it that $S(1) = S(-2)$ and $S(2) = S(-3)$ in sum notation? May 14 '16 at 11:04
• Suppose we solved the formula for the partial sums of $$\sum_{n=1}^p\frac{2^ne^{nx}}{\binom{2n}n}$$If we differentiate this and the partial sum formula, we may be able to produce your problem. Integrate for negative values of course, and take the limit to infinity to solve May 14 '16 at 11:09
• I do not think this question has anything to do with the riemann-zeta function. And I don't think pattern-recognition is the right tag either. Perhaps the sequences and series tag may be more useful. May 14 '16 at 11:17
• @GohP.iHan: I updated my initial answer with rather simple generating functions that could interest you. Cheers, Oct 22 '17 at 17:05

(this answer is about negative values of $$m$$: the power of $$n$$)

We want for $$m$$ any positive integer : $$\tag{1}S_{-m}(x) \equiv \sum_{n=1}^\infty \frac{(2x)^{2n}}{n^m\binom{2n}{n}}$$

Let's start again with : $$\tag{2}S_{-2}(x)=2\,\arcsin(x)^2=\sum_{n=1}^\infty \frac{(2x)^{2n}}{n^2\binom{2n}{n}}$$ Differentiation and multiplication by $$\dfrac x2$$ returns : $$\tag{3}S_{-1}(x)=\frac x2\,S_2'(x)=2x\,\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=1}^\infty \frac{(2x)^{2n}}{n\binom{2n}{n}}$$ While multiplication of $$(1)$$ by $$\dfrac 2x$$ and integration gives us : $$\tag{4}S_{-3}(x)=\int_0^x \frac 4t\,\arcsin(t)^2\,dt=\sum_{n=1}^\infty \frac{(2x)^{2n}}{n^3\binom{2n}{n}}$$ This integral may be expressed using polylogarithms but it will be more convenient to write it using (real) Clausen functions : $$\tag{5}\operatorname{Cl}_{2m-1}(x):=\sum_{n=1}^{\infty }\frac{\cos(n\,x)}{n^{2m-1}},\;\operatorname{Cl}_{2m}(x):=\sum_{n=1}^{\infty }\frac{\sin(n\,x)}{n^{2m}}$$ These functions appear in the Fourier series from jump discontinuities (their variants are Bernoulli polynomials) and are obtained from successive integrations of $$\;\displaystyle\operatorname{Cl}_1(x)=-\log\left(2\sin\frac x2\right)\;$$ using $$\tag{6}\displaystyle\operatorname{Cl}_{2m}(x)=\int_0^x \operatorname{Cl}_{2m-1}(t)\,dt,\ \;\operatorname{Cl}_{2m+1}(x)=\zeta(2m+1)-\int_0^x \operatorname{Cl}_{2m}(t)\,dt$$

Let's set $$\;t=\sin(u/2)\,$$ and integrate by parts $$\,\log(2\sin(u/2))\,$$ to get : \begin{align} S_{-3}(x)&=\int_0^x \frac 4t\,\arcsin(t)^2\,dt\\ &=4\int_0^{2\arcsin(x)} (u/2)^2\,\frac{\cos(u/2)}{2\,\sin(u/2)}\,du\\ &=\left.u^2\log(2\sin(u/2))\right|_0^{2\arcsin(x)}-\int_0^{2\arcsin(x)} 2\,u\log(2\,\sin(u/2))\,du\\ \end{align}

This becomes $$\quad\displaystyle S_{-3}(x)=\,-2\log(2x)\,\operatorname{Ls}_2^{(1)}(2\arcsin(x))+2\,\operatorname{Ls}_3^{(1)}(2\arcsin(x))\;$$

using Leonard Lewin's notation $$(7.14)$$ for the generalized log-sine integral $$\;\operatorname{Ls}$$ :
(Lewin $$1981$$ "Polylogaritms and associated functions" and Kalmykov and Sheplyakov's $$2004$$)

$$\tag{7}\operatorname{Ls}_j^{(k)}(x)=-\int_0^x t^k\,\left(\log\left(2\sin\frac t2\right)\right)^{j-k-1}dt,\quad k\ge 0,\ j\ge k+1\\ \text{(or simply }\;\operatorname{Ls}_j(x)\;\text{for k=0)}$$

What makes the rewriting of $$S_{-3}$$ interesting is that it may be generalized to any $$\,S_{-m}\,$$ as shown by Borwein, Broadhurst and Kamnitzer $$2001$$ "Central binomial sums, multiple Clausen values, and zeta values". The derivation is rather clever (little typo : $$\Gamma(n)$$ should be $$\Gamma(k)$$) and will be reproduced in details and slightly generalized as in Kalmykov and Veretin $$2000$$ :

The gamma function is defined by $$\;\displaystyle\Gamma(m):=\int_0^\infty t^{m-1}e^{-t}\,dt=\int_0^\infty (nt)^{m-1}e^{-nt}\,d(nt),\;(n>0)$$
The substitution $$\;t=-\log u\;$$ gives $$\;\displaystyle\Gamma(m)=n^m\int_0^1 (-\log u)^{m-1}\,u^{n-1}\,du\;$$ so that for $$\;u:=y^2$$ :

\begin{align} S_{-m}(x)&=\sum_{n=1}^\infty \frac{(2x)^{2n}}{\binom{2n}{n}\,n^m}\\ &=\sum_{n=1}^\infty \frac{(2x)^{2n}}{\binom{2n}{n}\,\Gamma(m)}\int_0^1 (-2\log y)^{m-1}y^{2n-2}\,d(y^2)\\ &= \frac{(-2)^{m-1}}{(m-1)!}\int_0^1 (\log y)^{m-1}\,\sum_{n=1}^\infty (2x)^{2n}\frac{2\,y^{2n-1}}{\binom{2n}{n}}\,dy\\ &= -\frac{(-2)^{m-1}}{(m-2)!}\int_0^1 \frac{(\log y)^{m-2}}y\,\sum_{n=1}^\infty \frac{(2xy)^{2n}}{n\,\binom{2n}{n}}\,dy,\quad\text{(by parts)}\\ &=-\frac{(-2)^{m-1}}{(m-2)!}\int_0^1 (\log y)^{m-2}\frac{(2x)\,\arcsin(xy)}{\sqrt{1-(xy)^2}}\,dy,\quad\text{(using}\;S_{-1}(xy)\;\\ &=-\frac{(-2)^{m-1}}{(m-2)!}\int_0^{2\arcsin x} \left(\log\left(\frac 1x\sin\frac t2\right)\right)^{m-2}\frac t2\,dt,\quad\text{(setting}\;xy=\sin\frac t2\;\text{)}\\ \tag{8}S_{-m}(x)&= \frac 1{(m-2)!}\int_0^{2\arcsin x}\left[2\log(2x)-2\log\left(2\sin\frac t2\right)\right]^{m-2}\;t\;dt\\ \tag{9} S_{-m}(x)&= - \sum_{j=0}^{m-2} \frac{(-2)^j}{(m-2-j)!j!} \,\left(2 \log(2x)\right)^{m-2-j}\, \operatorname{Ls}_{j+2}^{(1)}\left(2\arcsin x\right)\\ \end{align}

Nan-Yue and Williams gave $$(8)$$ for $$x=\dfrac 12$$ in $$1995$$ "Values of the Riemann zeta function and integrals involving $$\log\left(2\sinh\frac{\theta}2\right)$$ and $$\log\left(2\sin\frac{\theta}2\right)$$".
The general identities $$(8)$$ and $$(9)$$ (with a link to BBK's paper) were given in :

The KV $$2000$$ paper contains too an additional intriguing formula using Nielsen's generalized polylogarithm $$\;\displaystyle \operatorname{S}_{n,p}(z):=\frac {(-1)^{n+p-1}}{(n-1)!p!}\int_0^1 \log^{n-1}(t)\,\log^p(1-zt)\,\frac{dt}t\;$$ that I'll rewrite using $$\,\operatorname{S}_{m-2,1}(z)=\operatorname{Li}_{m-1}(z)\,$$ as : $$\tag{10}S_{-m}(x)=\int_0^1\operatorname{Li}_{m-1}\left((2x)^2\,s(1-s)\right)\,\frac {ds}s$$ This formula is interesting too to evaluate $$\;S_{+m}(x)\;$$ (rational functions are integrated).
$$-$$ Now that $$(9)$$ allows us to express $$\,S_{-m}(x)\,$$ as "generalized log-sine integrals" $$\operatorname{Ls}_m^{(1)}\,$$ what can we do with that?

$$\;\operatorname{Ls}_2^{(0)}(x)\;$$ is simply the "Clausen integral" $$\,\operatorname{Cl}_{2}(x)\,$$ while (from Lewin's book $$1981$$ "Polylogaritms and associated functions" p. $$200$$) $$\,\operatorname{Ls}_{n+2}^{(n)}(x)\,$$ may be written as a sum of Clausen functions :

\begin{align} \frac{(-1)^m}{(2m-2)!}\int_0^x t^{2m-2}\log\left(2\sin\frac t2\right)dt&=\operatorname{Cl}_{2m}(x)+\sum_{k=1}^{m-1}(-1)^{k}\left(\frac{x^{2k-1}}{(2k-1)!}\operatorname{Cl}_{2m-2k+1}(x)+\frac{x^{2k}}{(2k)!}\operatorname{Cl}_{2m-2k}(x)\right)\\ \frac{(-1)^{m-1}}{(2m-1)!}\int_0^x t^{2m-1}\log\left(2\sin\frac t2\right)dt&=\zeta(2m+1)+\sum_{k=0}^{m-1}(-1)^{k}\left(\frac{x^{2k}}{(2k)!}\operatorname{Cl}_{2m-2k+1}(x)+\frac{x^{2k+1}}{(2k+1)!}\operatorname{Cl}_{2m-2k}(x)\right)\\ \end{align}

Concerning your specific choice of $$\;x=\dfrac 1{\sqrt{2}}\;$$ (so that $$\,\displaystyle 2\arcsin(x)=\frac{\pi}2\,$$ and $$\,2\log(2x)=\log(2)$$) some explicit results were given in DK $$2001$$ (appendix A) that I complete here :

\begin{align} \operatorname{Ls}_{2}\left(\frac{\pi}{2}\right) =&\;\beta(2)\\ \operatorname{Ls}_{3}\left(\frac{\pi}{2}\right) =&\;2\,\Im\,\operatorname{Li}_3\left(\tfrac {1+i}2\right)+\beta(2)\log(2)-\tfrac{23}{192}\pi^3-\tfrac 1{16}\pi\log^2(2)\\ \operatorname{Ls}_{4}\left(\frac{\pi}{2}\right) =&\;6\,\Im \,\operatorname{Li}_4\left(\tfrac {1+i}2\right) + 3\,\Im \,\operatorname{Li}_3\left(\tfrac {1+i}2\right)\log(2) - \tfrac 32\beta(4) + \tfrac 34\beta(2)\log^2(2) + \tfrac 34\pi\zeta(3) - \tfrac 1{16}\pi\log^3(2)\\ \hline \operatorname{Ls}_{2}^{(1)}\left(\frac{\pi}{2}\right) =&\;-\tfrac{1}8\pi^2\\ \operatorname{Ls}_{3}^{(1)}\left(\frac{\pi}{2}\right) =&\;\;\tfrac{1}2 \pi\,\beta(2)-\tfrac{35}{32}\zeta(3)\\ \operatorname{Ls}_{4}^{(1)}\left(\frac{\pi}{2}\right) =&\;-\tfrac{5}{96} \log^4(2)+\tfrac{5}{16} \zeta(2) \log^2(2) - \tfrac{35}{32} \zeta(3) \log(2) + \tfrac{125}{32} \zeta(4) + \tfrac{1}{2} \pi \operatorname{Ls}_{3}\left(\tfrac{\pi}{2}\right)- \tfrac{5}{4} \operatorname{Li}_{4}\left(\tfrac{1}{2}\right)\\ \operatorname{Ls}_{5}^{(1)}\left(\frac{\pi}{2}\right) =&\;-\tfrac{1}{16} \log^5(2) + \tfrac{5}{16} \zeta(2) \log^3(2)- \tfrac{105}{128} \zeta(3) \log^2(2) - \tfrac{15}{8} \operatorname{Li}_{4}\left(\tfrac{1}{2}\right) \log(2) - \tfrac{9}{8} \zeta(2) \zeta(3) \nonumber \\ &\quad + \tfrac{1}{2} \pi \operatorname{Ls}_{4}\left(\tfrac{\pi}{2}\right) - \tfrac{1209}{256} \zeta(5) - \tfrac{15}{8} \operatorname{Li}_{5}\left(\tfrac{1}{2}\right) \end{align} allowing us (thanks to Hypergeometric's powerful comments) to extend the OP's table using only the common special functions : \begin{align}S(-4)=&\;-2\pi\,\Im\,\operatorname{Li}_3\left(\tfrac {1+i}2\right) + \tfrac 52\operatorname{Li}_4\left(\tfrac 12\right) + \tfrac{19}{576}\pi^4 + \tfrac 1{48}\pi^2\log^2(2) + \tfrac 5{48}\log^4(2)\\ S(-5)=&\;4 \pi\,\Im\,\operatorname{Li}_4\left(\tfrac{1+i}{2}\right)-\tfrac{5}{2} \operatorname{Li}_5\left(\tfrac{1}{2}\right)+\tfrac 14{\pi ^2 \zeta (3)}-\tfrac{403}{64} \zeta (5)-\pi\,\beta\left(4\right)+\tfrac 1{48}{\log ^5(2)}+\tfrac{1}{144} \pi ^2 \log ^3(2)+\tfrac{19}{576} \pi ^4 \log (2)\\ \end{align} $$-$$ In Davydychev and Kalmykov $$2004$$ "Massive Feynman diagrams and inverse binomial sums" one finds some general results ($$\,z$$ is our $$(2x)^2\,$$) :

For $$\;\displaystyle\theta:=2\,\arcsin\left(\frac{\sqrt{z}}2\right),\ l_{\theta}:=\log\left(2\sin\frac{\theta}{2}\right)\;$$ we have : \begin{align} \sum_{n=1}^\infty \frac{1}{\left( 2n \atop n\right) } \frac{z^n}{n} &= \theta\,\tan\frac{\theta}{2}&\\ \sum_{n=1}^\infty \frac{1}{\left( 2n \atop n\right) } \frac{z^n}{n^2} &=\frac{1}{2}\,\theta^2 &\\ \sum_{n=1}^\infty \frac{1}{\left( 2n \atop n\right) } \frac{z^n}{n^4} &= - 2 \operatorname{Ls}_{4}^{(1)}(\theta)+ 4 l_{\theta} \left[\operatorname{Cl}_3(\theta) + \theta \operatorname{Cl}_2(\theta) - \zeta(3) \right] +\theta^2 l_{\theta}^2 &\\ \end{align}

For the conformal variable $$\;\displaystyle y:=\frac{\sqrt{z-4}-\sqrt{z}}{\sqrt{z-4}+\sqrt{z}}\;$$ we have : \begin{align} \sum_{n=1}^\infty \frac{1}{\left( 2n \atop n\right) } \frac{z^n}{n} & = \frac{1-y}{1+y}\;\log(y)\\ \sum_{n=1}^\infty \frac{1}{\left( 2n \atop n\right) } \frac{z^n}{n^2} & = -\frac{1}{2}\;\log^2(y)\\ \sum_{n=1}^\infty \frac{1}{\left( 2n \atop n\right) } \frac{z^n}{n^3} & = 2 \operatorname{Li}_3(y) - 2\;\log(y)\,\operatorname{Li}_{2}(y) - \log^2(y)\, \log(1-y) + \frac{1}{6}\,\log^3(y)- 2 \zeta(3) \\ \sum_{n=1}^\infty \frac{1}{\left( 2n \atop n\right) } \frac{z^n}{n^4} & = 4 \operatorname{S}_{2,2}(y) - 4 \operatorname{Li}_{4}(y) - 4 \operatorname{S}_{1,2}(y) \log(y) + 4 \operatorname{Li}_{3}(y) \log(1-y) \\ &+ 2 \operatorname{Li}_3(y) \log(y) - 4 \operatorname{Li}_{2}(y) \log(y) \log(1-y) - \log^2(y) \log^2(1-y) \\ &+ \frac{1}{3} \log^3(y)\log(1-y) - \frac{1}{24} \log^4(y) - 4 \log(1-y) \zeta(3) + 2 \log(y) \zeta(3)+ 3 \zeta(4) \end{align}

• THANKYOUUUUU!! It will take me decades to understand everything here. May 30 '16 at 14:38
• The $z=-1$ case: $$\small \, _5F_4\left(1,1,1,1,1;\frac{3}{2},2,2,2;-\frac{1}{4}\right)=-\text{Li}_4\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right)+16 \text{Li}_4\left(\frac{1}{2} \left(\sqrt{5}-1\right)\right)-8 \text{Li}_3\left(\frac{1}{2}-\frac{\sqrt{5}}{2}\right) \text{csch}^{-1}(2)+8 \text{Li}_3\left(\frac{1}{2} \left(\sqrt{5}-1\right)\right) \text{csch}^{-1}(2)-\frac{7 \pi ^4}{45}+\frac{2}{3} \pi ^2 \text{csch}^{-1}(2)^2+\frac{1}{3} \text{csch}^{-1}(2)^3 \left(4 \cosh ^{-1}\left(\frac{3}{2}\right)-17 \text{csch}^{-1}(2)\right)$$ Oct 1 '20 at 5:44
• A Logsine integral example: $$\small \, _6F_5\left(1,1,1,1,1,1;\frac{3}{2},2,2,2,2;\frac{1}{2}\right)=4 \pi \Im\left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{5 \text{Li}_5\left(\frac{1}{2}\right)}{2}+\frac{\pi ^2 \zeta (3)}{4}-\frac{403 \zeta (5)}{64}-\frac{1}{256} \pi \zeta \left(4,\frac{1}{4}\right)+\frac{1}{256} \pi \zeta \left(4,\frac{3}{4}\right)+\frac{\log ^5(2)}{48}+\frac{1}{144} \pi ^2 \log ^3(2)+\frac{19}{576} \pi ^4 \log (2)$$ See more in the link. Oct 1 '20 at 6:01
• Two neat results indeed! The first one may be nicely rewritten using the golden ratio $o$ and its powers (and $\operatorname{csch}^{-1}(2)=\log(o)$) with the last term simply $-3\;\operatorname{csch}^{-1}(2)^4=-3\log(o)^4$. In your second result you may replace the two Hurwitz zeta terms simply by $-\pi\,\beta(4)$ (the Dirichlet beta function which could be useful for higher order generalizations!). Your summary of MZV identities is very interesting too! Oct 4 '20 at 14:51
• Other arguments: $$\small \, _8F_7\left(1,1,1,1,1,1,1,1;\frac{3}{2},2,2,2,2,2,2;\frac{1}{4}\right)=\frac{17 \pi ^4 \zeta (3)}{810}+\frac{2 \pi ^2 \zeta (5)}{3}-\frac{493 \zeta (7)}{12}+\frac{\pi \left(\psi ^{(5)}\left(\frac{1}{3}\right)-\psi ^{(5)}\left(\frac{2}{3}\right)+\psi ^{(5)}\left(\frac{1}{6}\right)-\psi ^{(5)}\left(\frac{5}{6}\right)\right)}{311040 \sqrt{3}}$$ Oct 7 '20 at 2:26

UPDATE 2017/10

We want the closed form for $\,\displaystyle S_m(x):=\sum_{n=1}^\infty \frac{n^m (2x)^{2n}}{\binom{2n}{n}}\,$ for nonnegative $m$ and will start with the well known expression for $m=-2$ :

$$\tag{1}F(x):=2\,\arcsin(x)^2=\sum_{n=1}^\infty \frac{(2x)^{2n}}{n^2\binom{2n}{n}}=S_{-2}(x)$$ From this we deduce : \begin{align} F'(x)&=4\,\frac{\arcsin(x)}{\sqrt{1-x^2}}&=4\sum_{n=1}^\infty \frac{(2x)^{2n-1}}{n\binom{2n}{n}}\\ (x\,F'(x))'&=4\,\frac{\frac{\arcsin(x)}{\sqrt{1-x^2}}(1)+x}{1-x^2}&=8\sum_{n=1}^\infty \frac{(2x)^{2n-1}}{\binom{2n}{n}}\\ (x\,(x\,F'(x))')'&=4\,\frac{\frac{\arcsin(x)}{\sqrt{1-x^2}}(1+2x^2)+3x}{(1-x^2)^2}&=16\sum_{n=1}^\infty \frac{n\;(2x)^{2n-1}}{\binom{2n}{n}}\\ &4\,\frac{\frac{\arcsin(x)}{\sqrt{1-x^2}}(1+10x^2+4x^4)+7x+8x^3}{(1-x^2)^3}&=32\sum_{n=1}^\infty \frac{n^2\;(2x)^{2n-1}}{\binom{2n}{n}}\\ &4\,\frac{\frac{\arcsin(x)}{\sqrt{1-x^2}}(1+36x^2+60x^4+8x^6)+15x+70x^3+20x^5}{(1-x^2)^4}&=64\sum_{n=1}^\infty \frac{n^3\;(2x)^{2n-1}}{\binom{2n}{n}}\\ &\cdots \end{align} Of course the idea is to take $\;x=\dfrac 1{\sqrt{2}}\,$ and obtain your results after multiplication by $\dfrac{\sqrt{2}}{2^{m+3}}\,$ if $m$ is the power of $n$ at the numerator.

A pattern seems to emerge from these laborious computations. Let's begin with : $$\tag{2}F_{m-1}(x):=\frac{a(x)P(x)+Q(x)}{(1-x^2)^m}$$ where $\,P(x),\;Q(x)\,$ are two polynomials while $\ a(x):=\dfrac {\arcsin(x)}{\sqrt{1-x^2}}\;$ so that $\ a(x)'=\dfrac{x\,a(x)+1}{1-x^2}$.

The derivative of $\,(x\;F_{m-1}(x))\,$ will then be given by : \begin{align} &=\frac{a(x)P(x)+Q(x)}{(1-x^2)^m}+x\left(\frac{a(x)P(x)+Q(x)}{(1-x^2)^m}\right)'\\ &=\frac{(a(x)P(x)+Q(x))\left(1+\dfrac{2mx^2}{1-x^2}\right)+x\left(a(x)'P(x)+a(x)P(x)'+Q(x)'\right)}{(1-x^2)^m}\\ &=\frac{(a(x)P(x)+Q(x))\left(1+\dfrac{2mx^2}{1-x^2}\right)+\dfrac{x^2\,a(x)+x}{1-x^2}P(x)+x(a(x)P(x)'+Q(x)')}{(1-x^2)^m}\\ &=\frac{(a(x)P(x)+Q(x))\left(1+(2m-1)x^2\right)+(x^2\,a(x)+x)P(x)+(x-x^3)(a(x)P(x)'+Q(x)')}{(1-x^2)^{m+1}}\\ &=\frac{a(x)\left[P(x)\left(1+2mx^2\right)+(x-x^3)P(x)'\right]+xP(x)+\left(1+(2m-1)x^2\right)Q(x)+(x-x^3)Q(x)'}{(1-x^2)^{m+1}}\\ \end{align}

which follows clearly our $(2)$ pattern with the recurrence for the polynomials (starting with $\;P_0(x)=1,\;Q_0(x)=x$) given by : \begin{align} \tag{3}P_m(x)&=P_{m-1}(x)\left(1+2mx^2\right)+(x-x^3)P_{m-1}(x)'\\ Q_m(x)&=x\,P_{m-1}(x)+\left(1+(2m-1)x^2\right)Q_{m-1}(x)+(x-x^3)Q_{m-1}(x)'\\ \end{align} From this recurrence we obtain : $$\tag{4}\boxed{\displaystyle\frac{a(x)P_m(x)+Q_m(x)}{(2\,(1-x^2))^{m+1}}=\sum_{n=1}^\infty \frac{n^m\;(2x)^{2n-1}}{\binom{2n}{n}}=\frac {S_m(x)}{2x}},\quad a(x)=\frac{\arcsin(x)}{\sqrt{1-x^2}}$$ that you may apply to your specific case $\;x=\dfrac 1{\sqrt{2}}\,$ for which $\;a(x)=\dfrac {\pi}{2\sqrt{2}}\,$ and $\,(2\,(1-x^2))=1\,$ (of course the $\sqrt{2}$ terms disappear after multiplication by $\,2x$ ). $$-$$ We didn't use the fact that $P(x),\;Q(x)$ were polynomials. Let's do that and suppose that $P_m(x)=\sum_{k=0}^m p_k\,x^{2k},\;Q_m(x)=\sum_{k=0}^m q_k\,x^{2k+1}$ then $(3)$ becomes :

\begin{align} P_m(x)&=\left(\sum_{k=0}^{m-1} p_k\, x^{2k}\right)\left(1+2mx^2\right)+(x-x^3)\left(\sum_{k=0}^{m-1} p_k\, x^{2k}\right)'\\ Q_m(x)&=x\,\left(\sum_{k=0}^{m-1} p_k\, x^{2k}\right)+\left(1+(2m-1)x^2\right)\left(\sum_{k=0}^{m-1} q_k\, x^{2k+1}\right)+(x-x^3)\left(\sum_{k=0}^{m-1} q_k\, x^{2k+1}\right)'\\ \end{align} Starting with $P_m(x)$ : \begin{align} P_m(x)&=\sum_{k=0}^{m-1} p_k\, x^{2k}+2m\sum_{k=1}^{m} p_{k-1}\, x^{2k} +\sum_{k=0}^{m-1} p_k\, 2k\,x^{2k}-\sum_{k=1}^{m} p_{k-1}\, 2(k-1)\,x^{2k}\\ &=p_0+(2m-2(m-1))\,p_{m-1}x^{2m}+\sum_{k=1}^{m-1} \left(p_k+2m\,p_{k-1}+p_k 2k-p_{k-1}\, 2(k-1)\right)\,x^{2k}\\ &=p_0+2\,p_{m-1}x^{2m}+\sum_{k=1}^{m-1} \left((2k+1)\,p_k+2(m-k+1)\,p_{k-1}\right)\,x^{2k}\\ \end{align}

The coefficients $p_k(m)$ of $P_m(x)$ may thus be obtained from the coefficients $p_k(m-1)$ of $P_{m-1}(x)$ (beginning with $\,p_0(m)=1\,$ since $p_0$ is the only $x^0$ term) : $$\tag{5}p_k(m)=\begin{cases} k=0 & 1 \\ 0<k<m& 2(m-k+1)\,p_{k-1}(m-1)+(2k+1)\,p_k(m-1)\\ k=m & 2\,p_k(m-1)\\ \text{else}& 0 \end{cases}$$ We may illustrate this by showing how the first coefficients $p_k(m)$ (in blue) were obtained
($k$ is indicated as $\,(k)\,$ in its appropriate diagonal) : $$\begin{array} {c|ccccccccccc} m&&&&&&&\color{blue}{p_k}\\ \hline &&&&&&&&(0)\\ 0&&&&&&&\color{blue}{1}\\ &&&&&&\rlap{\LARGE{\swarrow}}\scriptsize{\ \ \ \times 1}&&\rlap{\LARGE{\searrow}}\scriptsize {\times 2}&&(1)\\ 1&&&&&\color{blue}{1}&&&&\color{blue}{2}\\ &&&&\rlap{\LARGE{\swarrow}}\scriptsize{\ \ \ \times 1}&&\rlap{\LARGE{\searrow}}\scriptsize {\times 4}&&\rlap{\LARGE{\swarrow}}\scriptsize{\ \ \ \times 3}&& \rlap{\LARGE{\searrow}}\scriptsize {\times2}&&(2)\\ 2&&&\color{blue}{1}&&&&\color{blue}{10}&&&&\color{blue}{4}\\ &&\rlap{\LARGE{\swarrow}}\scriptsize{\ \ \ \times 1}&&\rlap{\LARGE{\searrow}}\scriptsize {\times6}&&\rlap{\LARGE{\swarrow}}\scriptsize{\ \ \ \times 3}&&\rlap{\LARGE{\searrow}}\scriptsize {\times 4}&&\rlap{\LARGE{\swarrow}}\scriptsize{\ \ \ \times 5}&&\rlap{\LARGE{\searrow}}\scriptsize {\times2}&&(3)\\ 3&\color{blue}{1}&&&&\color{blue}{36}&&&&\color{blue}{60}&&&&\color{blue}{8}\\ \end{array}$$ $\qquad\qquad\quad p_2(3)=\color{blue}{60}\,$ for example was obtained as $\;4\times \color{blue}{10}+5\times \color{blue}{4}$.

Concerning $Q_m(x)$ : \begin{align} Q_m(x)&=\sum_{k=0}^{m-1} p_k\, x^{2k+1}+\sum_{k=0}^{m-1} q_k\, x^{2k+1}+(2m-1)\sum_{k=1}^{m} q_{k-1}\, x^{2k+1}\\&\quad+\sum_{k=0}^{m-1} q_k\, (2k+1)\,x^{2k+1}-\sum_{k=1}^{m} q_{k-1}\, (2k-1)\,x^{2k+1}\\ &=(p_0+2q_0)x+\sum_{k=1}^{m-1}\left(p_k+q_k+(2m-1)\,q_{k-1}+q_k\, (2k+1)-q_{k-1}\,(2k-1)\right)\,x^{2k+1}\\ &=(1+2q_0)x+\sum_{k=1}^{m-1}\left(p_k+2(m-k)\,q_{k-1}+2(k+1)\,q_k\right)\,x^{2k+1}\\ \end{align} $$\tag{6}q_k(m)=\begin{cases} k=0 & \begin{cases} m=0 & 1 \\m>0&1+2\,q_0(m-1) \\ \end{cases}\\ 0<k<m& p_k(m-1)+2(m-k)\,q_{k-1}(m-1)+2(k+1)\,q_k(m-1)\\ \text{else}&0 \end{cases}$$

These results were obtained earlier by D.H. Lehmer in "Interesting Series Involving the Central Binomial" (pdf here "télécharger" Lehmer_binom.pdf).

The $p_k(m)$ triangle appears too in OEIS A156919 and in Savage and Viswanathan's paper "The $1/k$-Eulerian Polynomials" (for $k:=2$) and following generating function is provided (Alpha) : \begin{align} \tag{7}\sqrt{\frac {1-y}{\exp(2z(y-1))-y}}&=\sum_{n\ge 0}A_n^{(2)}(y)\frac {z^n}{n!}\\ &=1+z+(1+2y)\frac{z^2}{2!}+(1+10y+4y^2)\frac{z^3}{3!}+\cdots \end{align}

In your specific case $\,y=x^2=\dfrac 12$ this becomes this exponential generating function (e.g.f.) : $$\tag{8}\frac 1{\sqrt{2\exp(-z)-1}}=1+1z+2\frac{z^2}{2!}+7\frac{z^3}{3!}+35\frac{z^4}{4!}+226\frac{z^5}{5!}+\cdots$$ (OEIS A014307: exactly your numerators $\;1,2,7,35,226,\cdots\;$ for the $\dfrac {\pi}2$ terms!) $$-$$

Your integer sequence $\;1,3,11,55,355,\cdots$ is known too (OEIS A180875) but with no indicated generating function. An idea to get this one is to obtain the exponential generating function for the complete $S_m(x)$ terms first (subtracting $\frac {\pi}2$ times $(8)$ should then return the wished e.g.f. and sequence).

Let's multiply $(4)$ by $\;\displaystyle (2x)(2\,(1-x^2))^{m+1}\frac{z^m}{m!}\;$ and sum over $m$ to get : \begin{align} G_x(z)&:=(2x)\sum_{m=0}^\infty \left(a(x)P_m(x)+Q_m(x)\right)\frac{z^m}{m!}\tag{9}\\ &=\sum_{m=0}^\infty S_m(x) \left(2\left(1-x^2\right)\right)^{m+1}\frac{z^m}{m!}\\ &=\sum_{m=0}^\infty\sum_{n=1}^\infty \frac{n^m\;(2x)^{2n}}{\binom{2n}{n}} \left(2\left(1-x^2\right)\right)^{m+1}\frac{z^m}{m!}\\ &=2\left(1-x^2\right)\sum_{n=1}^\infty \frac{(2x)^{2n}}{\binom{2n}{n}} \sum_{m=0}^\infty\frac{\left(2n\left(1-x^2\right)z\right)^m}{m!}\\ &=2\left(1-x^2\right)\sum_{n=1}^\infty \frac{(2x)^{2n}\,\exp\left(2n\left(1-x^2\right)z\right)}{\binom{2n}{n}}\\ &=2\left(1-x^2\right)\sum_{n=1}^\infty \frac{\left(2x\exp\left(\left(1-x^2\right)z\right)\right)^{2n}}{\binom{2n}{n}}\\ &=2\left(1-x^2\right)S_0(u),\quad\text{for}\ \,u:=x\exp\left(\left(1-x^2\right)z\right)\\ G_x(z)&=2\left(1-x^2\right)u\,\frac{\large{\frac{\arcsin(u)}{\sqrt{1-u^2}}}+u}{1-u^2}\tag{10}\\ \end{align} since in our third equation $\,\displaystyle\frac x4(x\,F'(x))'\,$ gives $\;\displaystyle S_0(u)=\sum_{n=1}^\infty \frac{(2u)^{2n}}{\binom{2n}{n}}=u\,\frac{\large{\frac{\arcsin(u)}{\sqrt{1-u^2}}}+u}{1-u^2}$.

From the definition $(9)$ we may then use $(10)$ to evaluate for any $m\in \mathbb{N}$ the terms : $$\tag{11}(2x)\left(a(x)P_m(x)+Q_m(x)\right)=S_m(x)\left(2\left(1-x^2\right)\right)^{m+1}=\left.\left(\frac d{dz}\right)^m\right|_{z=0} G_x(z)$$ (the e.g.f. of the sole $\,S_m(x)$ terms is obtained by preferring $\;u:=x\,\exp(z/2)\;$)

In our specific case $\,x=\dfrac 1{\sqrt{2}}\,$ we have $\,u=\dfrac {\exp(z/2)}{\sqrt{2}}$ and $(10)$ and $(11)$ become : \begin{align} \tag{12}G_{\large{\frac 1{\sqrt{2}}}}(z)&=\frac{\large{u\frac{\arcsin(u)}{\sqrt{1-u^2}}}+u^2}{1-u^2}=\frac{{2\exp(z/2)}\frac{\arcsin(\exp(z/2)/\sqrt{2})}{\sqrt{2-\exp(z)}}+\exp(z)}{2-\exp(z)}\\ S_m\left(\frac{\pi}2\right)&=\frac{\pi}2 P_m\left(\dfrac 1{\sqrt{2}}\right)+\sqrt{2}\,Q_m\left(\dfrac 1{\sqrt{2}}\right)\\ \tag{13}&=\left.\left(\frac d{dz}\right)^m\right|_{z=0} \frac{{2\exp(z/2)}\frac{\arcsin(\exp(z/2)/\sqrt{2})}{\sqrt{2-\exp(z)}}+\exp(z)}{2-\exp(z)}\\ \tag{14}&=\left.\left(\frac d{dz}\right)^{m+1}\right|_{z=0} \frac{{2\exp(z/2)}\arcsin(\exp(z/2)/\sqrt{2})}{\sqrt{2-\exp(z)}}\\ \tag{15}&=\left.\left(\frac d{dz}\right)^{m+2}\right|_{z=0} 2\arcsin^2\left(\frac{\exp(z/2)}{\sqrt{2}}\right)\\ &\text{(integrating twice at the end)}\\ \end{align}

The last equation provides a very nice method to obtain all the $S(m)$ terms in the question
(btw $\,S(0)=1+\frac {\pi}2\,$ rather than $\,2+\frac {\pi}2\,$) :

• expand $\;\displaystyle 2\arcsin^2\left(\frac{\exp(z/2)}{\sqrt{2}}\right)$ in series : $$2\arcsin^2\left(\frac{\exp(z/2)}{\sqrt{2}}\right)=\frac{\pi^2}8+\frac{\pi}2z+\left(\frac {\pi}2+1\right)\frac{z^2}{2!}+\left(\frac{2\pi}2+3\right)\frac{z^3}{3!}+\left(\frac{7\pi}2+11\right)\frac{z^4}{4!}+\left(\frac{35\pi}2+55\right)\frac{z^5}{5!}+\cdots$$
• and compute the second derivative or simply ignore the two first terms and shift by $2$ the remaining ones!

Concerning an e.g.f. for $\;1,3,11,55,\cdots$ we will rewrite $(14)$ and combine it with $(8)$ to get : $$\frac{{2}\arcsin(\exp(z/2)/\sqrt{2})-\large{\frac {\pi}2}}{\sqrt{2\exp(-z)-1}}=1z+3\frac{z^2}{2!}+11\frac{z^3}{3!}+55\frac{z^4}{4!}+355\frac{z^5}{5!}+2807\frac{z^6}{6!}+\cdots$$

For an explicit general solution (instead of the recursive method provided) see this neat paper by the masters :

• Dyson, Frankel and Glasser "Lehmer's Interesting Series"
where they obtain following expression $(20)$ for $\;\displaystyle S_k(z) = \sum_{n=1}^{\infty} \frac{n^kz^n}{{2n \choose n}}\;$ :

$$\tag{16}S_k(z)=\sum_{n=1}^{k+1} n! \left({\frac{z}{4-z}}\right)^n\ E(k,n)\\ \left[\frac{1}{n} + \sum_{p=0}^{n-1} (-1)^p \frac{({\frac{1}{2}})_p}{(p+1)!}\ C^{n-1}_p \left({\frac 4z}\right)^{p+1} \left\{ \sqrt{\frac{z}{4-z}}\arcsin\left({\frac{\sqrt{z}}{2}}\right)- \frac{1}{2} \sum_{l=1}^p \frac{\Gamma(l)}{\left({\frac{1}{2}}\right)_l} \left(\frac{z}{4}\right)^l\right\}\right]$$

with $\;\displaystyle E(k,n) := \frac{(-1)^n}{n!} \sum_{m=1}^n (-1)^m\ C^{n}_m\ m^{k+1}$ the Stirling numbers of the second kind
and $\left({\dfrac{1}{2}}\right)_l$ the Pochhammer symbol.
Comparing this to $(1)$ we see that $z=(2x)^2$ while $k=m$ (of course you want $z=2$ here).

This paper includes too your interesting observation about the ratios approaching $\pi$ (page $12$).

A subsequent paper by Glasser ($2012$) is "A Generalized Apery Series".

• From the last edit of the OP it seems that he already had the link to this paper making all this much less interesting... May 14 '16 at 13:03
• Way to make you sound less smart May 14 '16 at 13:04
• @SimpleArt: well the answer was already provided (by Dyson, Frankel and Glasser ) making this de facto less challenging... May 14 '16 at 13:08
• Extremely detailed. Now what about negative integers $m$? May 18 '16 at 0:27
• @GohP.iHan: Well the pattern is rather different and increasingly difficult. Another answer should deserve these cases (I'll try...) May 19 '16 at 22:18