## Difficult series

I know:

$$\sum_{n=0}^\infty \frac{x^n}{n!}=e^x$$

However, is there a similar solution for:

$$\sum_{n=0}^\infty \left(\frac{x^n}{n!}\right)^2$$

Thanks in advance; I'm not very good at this kind of maths (I teach statistics ), and I've been struggling with this one for a while.
 I'm afraid for that you need the Bessel function. I looked up the series and the answer is $$\sum_{k=0}^\infty \frac{x^{2k}}{k!(k+n)!}=x^{-n}I_n(2x)$$ I suppose the function I_n is http://mathworld.wolfram.com/Modifie...FirstKind.html