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.

  • $\begingroup$ 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 $m=2,3$ $\endgroup$ May 14, 2016 at 10:58
  • $\begingroup$ Why is it that $S(1) = S(-2)$ and $S(2) = S(-3)$ in sum notation? $\endgroup$
    – giobrach
    May 14, 2016 at 11:04
  • $\begingroup$ 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 $\endgroup$ May 14, 2016 at 11:09
  • $\begingroup$ 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. $\endgroup$ May 14, 2016 at 11:17
  • 1
    $\begingroup$ @GohP.iHan: I updated my initial answer with rather simple generating functions that could interest you. Cheers, $\endgroup$ Oct 22, 2017 at 17:05

2 Answers 2


(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}

  • 2
    $\begingroup$ THANKYOUUUUU!! It will take me decades to understand everything here. $\endgroup$
    – GohP.iHan
    May 30, 2016 at 14:38
  • 1
    $\begingroup$ The $z=-1$ case: $$\small \, _5F_4\left(1,1,1,1,1;\frac{3}{2},2,2,2;-\frac{1}{4}\right)=-\text{Li}_4\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right)+16 \text{Li}_4\left(\frac{1}{2} \left(\sqrt{5}-1\right)\right)-8 \text{Li}_3\left(\frac{1}{2}-\frac{\sqrt{5}}{2}\right) \text{csch}^{-1}(2)+8 \text{Li}_3\left(\frac{1}{2} \left(\sqrt{5}-1\right)\right) \text{csch}^{-1}(2)-\frac{7 \pi ^4}{45}+\frac{2}{3} \pi ^2 \text{csch}^{-1}(2)^2+\frac{1}{3} \text{csch}^{-1}(2)^3 \left(4 \cosh ^{-1}\left(\frac{3}{2}\right)-17 \text{csch}^{-1}(2)\right)$$ $\endgroup$ Oct 1, 2020 at 5:44
  • 1
    $\begingroup$ 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. $\endgroup$ Oct 1, 2020 at 6:01
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    $\begingroup$ 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! $\endgroup$ Oct 4, 2020 at 14:51
  • 1
    $\begingroup$ Other arguments: $$\small \, _8F_7\left(1,1,1,1,1,1,1,1;\frac{3}{2},2,2,2,2,2,2;\frac{1}{4}\right)=\frac{17 \pi ^4 \zeta (3)}{810}+\frac{2 \pi ^2 \zeta (5)}{3}-\frac{493 \zeta (7)}{12}+\frac{\pi \left(\psi ^{(5)}\left(\frac{1}{3}\right)-\psi ^{(5)}\left(\frac{2}{3}\right)+\psi ^{(5)}\left(\frac{1}{6}\right)-\psi ^{(5)}\left(\frac{5}{6}\right)\right)}{311040 \sqrt{3}}$$ $\endgroup$ Oct 7, 2020 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".

  • 1
    $\begingroup$ From the last edit of the OP it seems that he already had the link to this paper making all this much less interesting... $\endgroup$ May 14, 2016 at 13:03
  • 1
    $\begingroup$ Way to make you sound less smart $\endgroup$ May 14, 2016 at 13:04
  • 1
    $\begingroup$ @SimpleArt: well the answer was already provided (by Dyson, Frankel and Glasser ) making this de facto less challenging... $\endgroup$ May 14, 2016 at 13:08
  • $\begingroup$ Extremely detailed. Now what about negative integers $m$? $\endgroup$
    – GohP.iHan
    May 18, 2016 at 0:27
  • $\begingroup$ @GohP.iHan: Well the pattern is rather different and increasingly difficult. Another answer should deserve these cases (I'll try...) $\endgroup$ May 19, 2016 at 22:18

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