I'm looking if there's a general closed form for the following function:$$f(k) = \sum_{n=1}^{\infty}\frac{n^k}{2n!},$$for $k\in \Bbb{N}$.For specific $k$ values I can utilize the fact that$$\cosh(x) = \sum_{n=0}^{\infty}\frac{x^{2n}}{2n!}.$$by repeated differentiation and solving for the desired sum.For example, for $f(3)$, we differentiate $\cosh(x)$ three times (and evaluate it at $x=1$) and obtain:$$\cosh(1) = \sum_{n=2}^{\infty}\frac{8n^3-12n^2+4n}{2n!} = 8\sum_{n=2}^{\infty}\frac{n^3}{2n!}-12\sum_{n=2}^{\infty}\frac{n^2}{2n!}+4\sum_{n=2}^{\infty}\frac{n}{2n!}.$$Rearranging and utilizing the following equalities:$$-12\sum_{n=2}^{\infty}\frac{n^2}{2n!} = -3(\cosh(1)+\sinh(1)-2) $$and$$4\sum_{n=2}^{\infty}\frac{n}{2n!} = 2\sinh(1)-2,$$we obtain that$$\sum_{n=2}^{\infty}\frac{n^3}{2n!} = \frac{3\cosh(1)+2\sinh(1)-4}{8}.$$Lastly, we add $\frac{1}{2}$ to both sides to finish:$$\sum_{n=1}^{\infty}\frac{n^3}{2n!} = \frac{3\cosh(1)+2\sinh(1)}{8}.$$
However, this technique really begins to get tedious and I am not seeing an obvious pattern to find a closed form. Any direction or advice would be appreciated. Alternatively, I really want to show that there exists some function $g(n)$ such that$$\sum_{i,j\in \Bbb{N}}\frac{(8\pi^2)^{i+j}\sum_{n=1}^{\infty}\frac{n^{i+j}}{g(n)}}{i!j!} < \infty.$$I've tried $g(n)=n!$ and that failed, therefore, the next natural guess would be $2n!$. However, that seems much more difficult at first glance.
