Show that $\sum\limits_{i=0}^\infty iP[X>i]=\frac12( E[X^2]-E[X]) $ for every positive integer valued random variable $X$

Question

Let $X$ be a non-zero random variable with values in $\{1,2,3,...\}$. Proof the next equality: $$\displaystyle\sum_{i=0}^{\infty}iP\left[X>i\right]=\displaystyle\frac{1}{2}\left( E\left[X^{2}\right]-E\left[X\right]\right) $$I don't know how can I do it this proof.

Answer 1

Hint: write $$ \sum_{i=0}^{\infty}i\mathbb{P}(X>i)=\sum_{i=0}^{\infty}i\sum_{n=i+1}^{\infty}\mathbb{P}(X=n) $$ and interchange the sums.