Looking for a closed form for $a_m =\sum_{k=0}^{\infty}\binom{k+m}{m}\frac{1}{4^{k}(2(k+m))!}$

Question

I have this sequence $$ a_m =\sum_{k=0}^{\infty}\binom{k+m}{m}\frac{1}{4^{k}(2(k+m))!} $$ and there seems to exist a patern arising when it is evaluated by WA. It involves $\cosh(1/2)$ and $\sinh(1/2)$. Bellow are the first $$ \begin{align*} a_0 &= \cosh\left(\frac{1}{2}\right)\\ a_1 &= \sinh\left(\frac{1}{2}\right)\\ a_2 &= \frac{1}{2}\left(\cosh\left(\frac{1}{2}\right )-2\sinh\left(\frac{1}{2} \right ) \right )\\ &\vdots\\ \end{align*} $$ but $\cosh(1/2)$ and $\sinh(1/2)$ keeps showing up for $a_3$, $a_4$, $\cdots$

Could anyone find a general expression for $a_m$ involving $\cosh(1/2)$ and $\sinh(1/2)$.

I'd apprecaite any help, thanks.

Answer 1

As Guilherme Thompson commented, $$a_m =\sum_{k=0}^{\infty}\binom{k+m}{m}\frac{a^k}{(2(k+m))!}=\sqrt{\pi }\frac{ 2^{-m-\frac{1}{2}} a^{\frac{1}{4}-\frac{m}{2}} }{m!} I_{m-1/2}\left(\sqrt{a}\right)$$ where appears the modified Bessel function of the first kind.

So, for $a=\frac 14$, this reduces to $$a_m=\frac{\sqrt{\pi } }{2 }\frac 1{m!}\,I_{m-\frac{1}{2}}\left(\frac{1}{2}\right)$$ which results, as you observed, in linear combinations of $\cosh \left(\frac{1}{2}\right)$ and $\sinh \left(\frac{1}{2}\right)$.

Answer 2

$$a_m = \sum_{k\geq 0}\frac{(k+m)! 4^{-k}}{m! k! (2k+2m)!}=\frac{4^m}{m!}\cdot\left.\frac{d^m}{dx^m}\sum_{k\geq 0}\frac{(x/4)^{k+m}}{(2k+2m)!}\right|_{x=1} $$ and now you may recognize in the last series part of the Taylor series of $\cosh(\sqrt{x}/2)$.