# Difficult series

1. Dec 10, 2009

### m00se

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.

Last edited: Dec 10, 2009
2. Dec 10, 2009

### Gerenuk

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/ModifiedBesselFunctionoftheFirstKind.html