Find the expectation of this mass function

Question

$p_x$ is defined as follows:\begin{cases} \dfrac{2^x}{(e^2-1)x!} \quad &\text{if} \, x =1,2,3,\dots \\ 0 \quad &\text{if} \, o/w \\ \end{cases}

Find $E(X)$.

$$E(X)=\sum_{x=1}^{\infty}xp_x==\frac{1}{e^2-1}\sum_{x=1}^{\infty}\frac{x*2^x}{x!}=\frac{2}{e^2-1}\sum_{x=1}^{\infty}\frac{2^{x-1}}{(x-1)!}$$

I don't know how to evaluate the series.

Wolfram tells me that the series alone (not the constants on the outside) evaluates to $e^2$

Is there some sort of trick to see this?

Answer 1

You can re-index the final sum to get: $$ E(X) = \frac{2}{e^2 - 1} \sum_{k=0}^\infty \frac{2^k}{k!} $$ As mentioned in @T. Bongers' comment, this summation is just the evaluation of the Taylor series $e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$ at $x = 2$. So we get: $$ E(X) = \frac{2e^2}{e^2 - 1} $$