Sum of Square using Cauchy Product

Question

Some background on the question:
I am trying to compute the expectation of $\mathbb E[(Y|X)^2]$ of Y conditional on X which follows an exponential distribution ~(λ/k)

So the question boils down to solving the following summation

$$\sum_{k=1}^\infty k^2\frac{μ^k}{k!} $$

Which can be split into two sums using the Cauchy Product transforming it into

$$\sum_{k=1}^\infty k^2\sum_{k=1}^\infty\frac{μ^k}{k!} $$

The second sum is easily identified as the exponent of μ. Now the second sum should be equal to $μ^2+μ$ but I cant get my head around why this is.

Any help is greatly appreciated.

Answer 1

Starting with

$$e^x=\sum_{n\ge 0}\frac{x^n}{n!}\;,$$

differentiate with respect to $x$ and multiply by $x$ to get

$$xe^x=\sum_{n\ge 0}\frac{nx^n}{n!}\;.$$

Repeat to get

$$xe^x+e^x=\sum_{n\ge 0}\frac{n^2x^n}{n!}\;.$$

Now let $x=\mu$.