What is the sum of the series from k=1 to k=Infinity for k^2/k!?

[moo5003]

Homework Statement

Series k=1 to k=Infinity of:

k^2/k!

a) e
b) 2e
c) (1+e)(e-1)
d) e^2
e) Infinity

Homework Equations

e^x = Series from n=1 to n=Infinity of:
x^n / n!

The Attempt at a Solution

I was guessing this would end up being C but the answer is infact B. As far as I can tell 2e is equal to Series 2^k / k!, I'm unsure how this is equal.

 

[TD]

You don't need to get to x^n/n!, but to 1/n!, since this series gives e. A start:

<br /> \frac{{k^2 }}{{k!}} = \frac{{kk}}{{k\left( {k - 1} \right)!}} = \frac{k}{{\left( {k - 1} \right)!}}<br />