In an exercise I have to calculate the following expected value: $$ \sum_{i=4}^\infty500\frac{e^{-4}4^i}{i!} $$ Leaving $500e^{-4}$ outside and using D'Alambert's criterion (with $a_n = \frac{4^n}{n!}$) we have:
$\lim\limits_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|=\lim\limits_{n\rightarrow\infty}|\frac{4^{n+1}}{(n+1)n!}\frac{n!}{4^n}|=\lim\limits_{n\rightarrow\infty}|\frac{4}{n+1}|$
Thus we can say that the series converges, but now the problem is that I need to calculate the expected value, but I don't know how, because they have just taught me how to determine whether a series converges, but not how to get the value at which converges.
Hint
$$\sum_{i=4}^{\infty}500 e^{-4}\frac{4^i}{i!}=500 e^{-4}\sum_{i=4}^{\infty}\frac{4^i}{i!}=500 e^{-4}\Big(\sum_{i=0}^{\infty}\frac{4^i}{i!}-\sum_{i=0}^{3}\frac{4^i}{i!}\Big)$$