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.
Best Answer
(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}