Power Series involving Double Factorials

Question

This comes from my partial solution to another question. I need to find a closed form for the following summation

$$\sum_{n=1}^{\infty}\frac{n!}{(2n+1)!!}n^x$$ Where $x$ is a fixed integer

This appears similar to many arcsine series and formulas

---

Particular values: $$\begin{array}{c|c|c|} & 0 & 1 & 2 & 3 & 4 \\ \hline x & \frac{\pi}{2}-1 & \frac{0\pi}{2}+1 & \frac{\pi}{2}+1 & \frac{3\pi}{2}+5 & \frac{16\pi}{2}+25\\ \hline \end{array}$$

$$\begin{array}{c|c|c|} & 5 & 6 & 7 & 8 \\ \hline x & \frac{105\pi}{2}+165 & \frac{841\pi}{2}+1321 & \frac{7938\pi}{2}+12469 & \frac{86311 π}{2}+135577\\ \hline \end{array}$$

---

Update 1: @RobertIsrael found a closed form for this sum's Exponential Generating Function, but I have been unable to find any expression for the Maclaurin series.

Update 2: Grant B. has found a simple recurrence, but a complete closed form has not been found

Answer 1

Why just even powers of $n$? Let $$ F(j) = \sum_{n=1}^\infty \frac{n! n^j}{(2n+1)!!} = \sum_{n=1}^\infty \frac{2^{-1-n} \sqrt{\pi} n! n^j}{\Gamma(n+3/2)} $$ The exponential generating function is

$$\eqalign{E(z) &= \sum_{j=0}^\infty \frac{F(j)}{j!} z^j\cr &= \sum_{n=1}^\infty \frac{2^{-1-n} \sqrt{\pi} n! e^{zn}}{\Gamma(n+3/2)}\cr &= -1 + \frac{2\; e^{-z/2}}{\sqrt{2-e^z}} \arcsin\left(e^{z/2}/\sqrt{2}\right)}$$

EDIT: The last line could be obtained (in hindsight) as follows. As you know, (for $|x| < 1$)

$$\arcsin(x) = \sum _{k=0}^{\infty }{\frac { \left( 2\,k \right) !\,{4}^{-k}{x}^{2\,k +1}}{ \left( k! \right) ^{2} \left( 2\,k+1 \right) }} $$ Taking $x = e^{z/2}/\sqrt{2}$, we want to multiply this by $$ \frac{2 e^{-z/2}}{\sqrt{2-e^z}} = \frac{1}{x \sqrt{1-x^2}} = \sum_{k=0}^\infty \frac{(2k)!\, 4^{-k} x^{2k-1}}{(k!)^2}$$

The Cauchy product is

$$ \sum_{n=0}^\infty \sum_{k=0}^n \frac{(2k)! 4^{-k}}{(k!)^2} \frac{(2(n-k))! 4^{-n+k}}{((n-k))!^2(2n-2k+1)} x^{2n}$$

And it turns out that

$$ \sum _{k=0}^{n}{\frac { \left( 2\,k \right) !\, \left( 2\,n-2\,k \right) !}{ \left( k! \right) ^{2} \left( \left( n-k \right) ! \right) ^{2}(2n-2k+1)}}={\frac {\sqrt {\pi}{4}^{n}n!}{2\,\Gamma \left( n+3/2 \right) }} $$

Answer 2

Let $$F(n) = \sum_{j=1}^\infty \frac{j! j^n}{(2j+1)!!}$$ Then, following from Robert Israel's answer, the exponential generating function is $$E(z) = \sum_{n=0}^\infty \frac{F(n)}{n!}z^n = -1 + \frac{2 e^{-z/2}}{\sqrt{2-e^z}}\arcsin(e^{z/2}/\sqrt{2})$$ Using this, we see that $$E'(z) = \frac{1}{2-e^z} + \frac{1}{2}\left(\frac{1}{2e^{-z}-1} -1\right) \frac{2 e^{-z/2}}{\sqrt{2-e^z}}\arcsin(e^{z/2}/\sqrt{2})$$ $$=\frac{1}{2-e^z} + \frac{1}{2}\left(\frac{1}{2e^{-z}-1} -1\right) (E(z)+1)$$ $$=\frac{1}{2-e^z} + \left(\frac{1}{2-e^z} -1\right) (E(z)+1)$$ Now we crack out our generating function tools. $1/(2-e^z)$ is the exponential generating function for the Fubini numbers $a_n$, given by $$a_n = \sum_{k=0}^n k! {n\brace k}$$ where ${n\brace k}$ is a Stirling number of the second kind, itself given by $${n\brace k}=\frac{1}{k!}\sum_{j=0}^k(-1)^{k-j}{k\choose j}j^n$$ We can also see that $1/(2-e^z)-1$ is the e.g.f. of the sequence $b_n$, where $b_0=0$ and $b_n=a_n$ for $n\geq 1$.

Thus from our differential equation (and the product rule for exponential generating functions), $$F(n+1) = a_n + b_n + \sum_{k=0}^n{n\choose k} b_k F(n-k)$$ $$=2a_n + \sum_{k=1}^n{n\choose k}a_k F(n-k)$$ which is a recursive formula valid for $n\geq 1$ (if $n=0$, $b_n\neq a_n$). From the generating function we see that $F(0)=\pi/2-1$ and $F(1)=1$, and we can use this relation to generate further $F(n)$.

For example, $F(3) = 5+3\pi/2$, $F(4)=25+8\pi$, $F(8) = 135577+86311\pi/2$, $F(12)=5791940089+3687263581\pi/2$, etc.

Edit: Scratch that. A much much better recurrence is given by $$F(n+1)=1-F(n)+\sum_{k=0}^{n}{n+1\choose k}F(k)$$ which can be found from the same differential equation, multiplied through by $2-e^z$.

Edit 2: If the coefficient of the $\pi$ term is given by $G(n)$, then it satisfies the recurrence $$G(n+1)=-G(n)+\sum_{k=0}^{n}{n+1\choose k}G(k)$$ since the $\pi$ term is unaffected by the $1$ in the original recurrence. This leads to the e.g.f. $\frac{e^{-z/2}}{2\sqrt{2-e^z}}$, which gives $G(n)$ as $$\frac{1}{2}\sum_{k=0}^n (2k-1)!!^2 {n \brace 2k}$$

We can also find the non-$\pi$ term in a closed form, though it is a bit more complicated. Ignoring the $\pi/4$ constant term (since we just took care of it), we have the $n+1$-th order term of $\arcsin(e^{z/2}/\sqrt{2})$ is $$s_{n+1}=\sum_{j=0}^{n-1}\sum_{k=j+1}^n {n\brace k}k!{k-1\choose j}\frac{(2j-1)!!}{(j+1)!}(-1)^{j}$$ with $s_1=1$. Now we multiply the series for our two functions together, giving $$F(n)=\sum_{i=1}^n 2G(n-i) s_i$$ Note the recurrence is probably more useful for calculations, while the direct form may be helpful for theoretical work.