In deriving some approximate solutions, I've come across the following sum I need to evaluate
$$S=\sum_{n=0}^\infty(-1)^nI_n(\alpha)e^{-n^2\beta},$$
where $I_n$ is the modified Bessel function of the first kind and $\alpha,\beta>0$ are arbitrary parameters. As far as I've found, there isn't a known closed form for this sum (please let me know if there is), so I've attempted to find a simplified form by expanding in power series: $$I_n(\alpha)=\sum_{m=0}^\infty\frac{\left(\frac{\alpha}{2}\right)^{2m+n}}{m!(n+m)!}$$ $$e^{-n^2\beta}=\sum_{k=0}^\infty(-1)^k\frac{n^{2k}\beta^k}{k!}$$
So the sum above can be written as a triple sum of the form $$S=\sum_{n=0}^\infty\sum_{m=0}^\infty\sum_{k=0}^\infty(-1)^n\frac{\left(\frac{\alpha}{2}\right)^{2m+n}}{m!(n+m)!}(-1)^k\frac{n^{2k}\beta^k}{k!}$$
In an effort to make some progress, I changed the order of summation to attempt to remove the summation over $n$: $$S=\sum_{m=0}^\infty\sum_{k=0}^\infty(-1)^k\frac{\beta^k\left(\frac{\alpha}{2}\right)^{2m}}{m!k!}\sum_{n=0}^\infty(-1)^n\frac{\left(\frac{\alpha}{2}\right)^nn^{2k}}{(n+m)!}$$
The innermost sum here seems to be a power series of some function evaluated at some multiple of $\alpha$, but I can't seem to find anything that matches. I would appreciate any help on simplifying this sum. I am also interested in the sum
$$S_2=\sum_{n=0}^\infty(-1)^nn^2I_n(\alpha)e^{-n^2\beta},$$
which should be similar to the sum above. Thanks!
You can use this integral representation to derive a formula in terms of the Jacobi theta function: \begin{align*} \sum\limits_{n = 0}^\infty {( - 1)^n \frac{1}{\pi }\int_0^\pi {e^{\alpha \cos t} \cos (nt)dt} e^{ - n^2 \beta } } & = \frac{1}{\pi }\int_0^\pi {e^{\alpha \cos t} \left( {\sum\limits_{n = 0}^\infty {( - 1)^n \cos (nt)e^{ - n^2 \beta } } } \right)dt} \\ & = \frac{1}{{2\pi }}\int_0^\pi {e^{\alpha \cos t} \left( {1 + \vartheta \left( {\frac{{t + \pi }}{{2\pi }};i\frac{\beta }{\pi }} \right)} \right)dt} . \end{align*} Does this help?