Two series

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

Two series

Postby marco » Fri Sep 21, 2012 1:56 pm

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?
Thanks in advance.

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

Re: Two series

Postby blitzer6266 » Fri Sep 21, 2012 2:40 pm

For the first

\sum_{k=0}^\infty \frac{k^2}{k!} =  \sum_{k=1}^\infty \frac{k}{(k-1)!} = <br /> \sum_{k=0}^\infty \frac{k+1}{(k)!} =<br />e +  \sum_{k=1}^\infty \frac{1}{(k-1)!}

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

Re: Two series

Postby blitzer6266 » Fri Sep 21, 2012 2:43 pm

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




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