# Difficult series

by m00se
Tags: difficult, series
 P: 1 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.
 P: 1,060 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