## Two series

Forum for the GRE subject test in mathematics.
marco
Posts: 263
Joined: Mon Jun 11, 2012 2:27 am

### Two series

Hello:

I have these two series which I can't handle:

\sum_{k=0}^\inf \frac{k^2}{k!} which according to the GRE exam is 2e. I can't see why.

The other one is:
\sum_{k=0}^\inf \frac{k}{4^(k+1)}. what is its value?

Could someone help me out?

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

### Re: Two series

For the first

$$\sum_{k=0}^\infty \frac{k^2}{k!} = \sum_{k=1}^\infty \frac{k}{(k-1)!} = \sum_{k=0}^\infty \frac{k+1}{(k)!} = e + \sum_{k=1}^\infty \frac{1}{(k-1)!}$$

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

### Re: Two series

for the second use

$$\sum_{k=0}^\infty r^k = \frac{1}{1-r} \Longrightarrow \sum_{k=0}^\infty k r^{k-1} = \frac{1}{(1-r)^2}$$
by taking a derivative. This works for any r in the open unit interval