I want to solve this:
$E[X^2]=\sum_{k=0}^{\infty} k^2 e^{-\lambda}\frac{\lambda^k}{k!}$
My try:
$E[X^2]=\sum_{k=0}^{\infty} k^2 e^{-\lambda}\frac{\lambda^k}{k!}=\sum_{k=1}^{\infty} k^2 e^{-\lambda}\frac{\lambda^k}{k!}$
Indexshift and reduction:
$\sum_{k=0}^{\infty} (k+1) e^{-\lambda}\frac{\lambda^{k+1}}{k!}$
Now I am not sure if this is legal:
$\sum_{k=0}^{\infty} \lambda ke^{-\lambda}\frac{\lambda^k}{k!}+\lambda e^{-\lambda}\frac{\lambda^k}{k!}$
$\lambda\lim_{n\to\infty}\sum_{k=0}^n ke^{-\lambda}\frac{\lambda^k}{k!}+\sum_{k=0}^n e^{-\lambda}\frac{\lambda^k}{k!}$
I know that $\sum_{k=0}^n ke^{-\lambda}\frac{\lambda^k}{k!}=\lambda$ and
$\sum_{k=0}^n e^{-\lambda}\frac{\lambda^k}{k!}=1$
Thus:
$\lambda \lim_{n\to\infty} \lambda+1=\lambda(\lambda+1)$
I know that the result should be right, but I am not sure if it is legal to split the sum, like I did after inculding the limit.
Thanks in advance.
Yes, as long as every $a_n$ and $b_n$ is nonnegative, one has $$\sum_na_n+b_n=\sum_na_n+\sum_nb_n.$$ Only, the identity holds in $[0,\infty]$, for example this could mean that $\infty=42+\infty$.