Find $\sum_{m=1}^{\infty}\frac{(2m-1)!!}{2^m \, m! \, m^{n+1}} $

Question

I need to find the value of $$ S(n)=\sum_{m=1}^{\infty}\frac{(2m-1)!!}{2^m \, m! \, m^{n+1}} $$ where $n$ is an integer greater than or equal to $0$. Mathematica can do individual cases, $S(0) = \ln(4)$ for example, but it can't do the general case.

Answer 1

Now I have an answer that supports my initial feelings: $S(n)$ can be handled through polygamma identities (see $(4)$ and $(5)$). We have:

$$ S(n) = \sum_{m=1}^{+\infty}\binom{2m}{m}\frac{1}{4^m m^{n+1}}, $$ and, for any $x\in(0,1)$, $$ f(x) = \sum_{m=1}^{+\infty}\binom{2m}{m}\frac{x^{m-1}}{4^m} = \frac{1}{x\sqrt{1-x}}-\frac{1}{x}=\frac{1}{(1+\sqrt{1-x})\cdot\sqrt{1-x}}\tag{1}.$$ Using now the identity: $$\int_{0}^{1}x^{m-1}\log^n(x)\,dx=\frac{n!}{m^{n+1}}\tag{2}$$ we can state: $$S(n)=\frac{1}{n!}\int_{0}^{1}f(x)\log^n(x)\,dx =\frac{2}{n!}\int_{0}^{1}\frac{\log^n(1-u^2)}{1+u}\,du\tag{3}$$ but we can see the LHS also as: $$S(n)=\frac{1}{n!}\frac{\partial}{\partial a_1}\ldots\frac{\partial}{\partial a_n}\left.\left(\int_{0}^{1}f(x)\cdot x^{\sum a_i}dx\right)\right|_{(a_1,\ldots,a_n)=(0^+,\ldots,0^+)},\tag{4}$$ but since: $$\int_{0}^{1}\frac{x^{k-1}}{\sqrt{1-x}}\,dx = B(1/2,k) = \frac{\Gamma(1/2)\cdot\Gamma(k)}{\Gamma(k+1/2)}\tag{5}$$ we have: $$\int_{0}^{1}f(x)\,x^{\sum a_i}dx = \frac{\Gamma(1/2)\cdot\Gamma(\sum a_i)}{\Gamma(1/2+\sum a_i)}-\frac{1}{\sum a_i}=g\left(\sum a_i\right).\tag{6}$$ By $(4)$ and $(6)$, now we can read any $S(n)$ in the Taylor series coefficients of $$ g(x) = \frac{2^{2x}\Gamma(x)^2}{2\cdot\Gamma(2x)}-\frac{1}{x}$$ or $$ x\cdot g(x) = \frac{2^{2x}\Gamma(x+1)^2}{\Gamma(2x+1)}-1 \tag{7}$$ around $x=0$. By the Weierstrass product for the Gamma function we know that: $$ \log\Gamma(z+1)=-\gamma z+\sum_{k=2}^{+\infty}\frac{\zeta(k)}{k}(-z)^k, \tag{8}$$ and by $(7)$: $$ x\cdot g(x) = -1+\exp\left(2x\log 2+2\log\Gamma(x+1)-\log\Gamma(2x+1)\right),$$ $$ x\cdot g(x) = -1+\exp\left(2x\log 2-2\sum_{k=2}^{+\infty}\frac{\zeta(k)(2^k-1)}{k}(-z)^k\right).\tag{9}$$ This gives that we can compute $S(n)$ by exponentiating a relatively simple Taylor series whose coefficients are $\log(4)$ and rational multiples of the values of the Riemann zeta function in the integers greater than one.