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.
Two series

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: Two series
For the first
$$\sum_{k=0}^\infty \frac{k^2}{k!} = \sum_{k=1}^\infty \frac{k}{(k1)!} =
\sum_{k=0}^\infty \frac{k+1}{(k)!} =
e + \sum_{k=1}^\infty \frac{1}{(k1)!}$$
$$\sum_{k=0}^\infty \frac{k^2}{k!} = \sum_{k=1}^\infty \frac{k}{(k1)!} =
\sum_{k=0}^\infty \frac{k+1}{(k)!} =
e + \sum_{k=1}^\infty \frac{1}{(k1)!}$$

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: Two series
for the second use
$$\sum_{k=0}^\infty r^k = \frac{1}{1r} \Longrightarrow \sum_{k=0}^\infty k r^{k1} = \frac{1}{(1r)^2}$$
by taking a derivative. This works for any r in the open unit interval
$$\sum_{k=0}^\infty r^k = \frac{1}{1r} \Longrightarrow \sum_{k=0}^\infty k r^{k1} = \frac{1}{(1r)^2}$$
by taking a derivative. This works for any r in the open unit interval