I am trying to prove the variance of the Poisson distribution. To do that, I first work out $E(X^2)$. I know there are easier ways to do this.
Let $$X \texttt{~} Poisson(\lambda)$$
Then,
$$ E(X^2) = \sum_{x=0}^{\infty} x^2\dfrac{e^{-\lambda}\lambda^x}{x!} $$
and since the first term is 0, $$= \sum_{x=1}^{\infty} x^2\dfrac{e^{-\lambda}\lambda^x}{x!} $$
$$= \dfrac{e^{-\lambda}\lambda^1}{1!} + \sum_{x=2}^{\infty} x^2\dfrac{e^{-\lambda}\lambda^x}{x!} $$
So all I have done is taken the $x=1$ term out of the summation. But I end up with the $e^{-\lambda}\lambda$, and after more manipulation end up with $e^{-\lambda}\lambda + \lambda^2$ instead of the correct $\lambda + \lambda^2$.
Any ideas on where I have gone wrong would be greatly appreciated.
Separating the first term from the sum only makes it more troublesome, but going with what you have, you should do this $$\displaystyle\sum_{x=2}^{\infty} x^2 \dfrac{e ^{-\lambda} \lambda^{x}}{x!}=\displaystyle\sum_{x=2}^{\infty} x\dfrac{e ^{-\lambda} \lambda^{x}}{(x-1)!}$$ $$\qquad \qquad \qquad \qquad =\displaystyle\sum_{y=1}^{\infty} (y+1)\dfrac{e ^{-\lambda} \lambda^{y+1}}{y!}$$ $$\qquad \qquad \qquad \qquad \qquad \qquad =\lambda\displaystyle\sum_{y=1}^{\infty} y\dfrac{e ^{-\lambda} \lambda^{y}}{y!}+\lambda\displaystyle\sum_{y=1}^{\infty} \dfrac{e ^{-\lambda} \lambda^{y}}{y!}$$ $$\qquad \qquad \qquad \ =\lambda^2+\lambda(1-e^{-\lambda})$$ to get the desired manipulation.