# Series Limit

1. Oct 13, 2008

### moo5003

1. The problem statement, all variables and given/known data

Series k=1 to k=Infinity of:

k^2/k!

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

2. Relevant equations

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

3. 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.

2. Oct 13, 2008

### TD

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

$$\frac{{k^2 }}{{k!}} = \frac{{kk}}{{k\left( {k - 1} \right)!}} = \frac{k}{{\left( {k - 1} \right)!}}$$