Proving Infinite Sum of n^2 a^n/n! = a(1+a)e^a

[Repetit]

Hey!

Can someone tell me or just give a hint on how to show that:

$$\sum_n \frac{n^2 a^n}{n!}=a(1+a)e^a$$

when n goes to infinity? I know how to show that:

$$\sum_n \frac{n a^n}{n!}=a e^a$$

by using the facts that n/n! = 1/(n-1)! and a^n = a a^(n-1). But how can I prove the other one?

Thanks!

 

[StatusX]

After doing what you did in the second example, you'll be left with something like n a^n/(n-1)!. Rewrite the n in the numerator as (n-1) +1, and you'll get two familiar sums. Another way to do both problems would be to differentiate both sides, the sum term by term and the exponential as usual (using the chain rule in this case).

 

[Repetit]

Ahh, of course... thanks a lot! :-)