The following is a homework example:
For a nonnegative integer-valued random variable $X$, show that
$\sum^{\infty}_{i=0}{iP(X > i)} = 1/2(E(X²)-E(X)).$
The only way I could think of doing this proof is to rewrite the left side starting with:
$(1).$ $\sum^{\infty}_{i=0}{i} \sum^{\infty}_{j=i+1}{P(X=j}) = 1/2(E(X²)-E(X)).$
and continue from there to get to an expression similar to:
$(2).$ $1/2 * [\sum^{\infty}_{i=0}{i²*P(X=i)} - \sum^{\infty}_{i=0}{i*P(X=i))}] = 1/2(E(X²)-E(X)).$
However I don't know how to continue to transform $(1)$ into $(2)$. Could you give me a hint there? Is the approach even correct?
It seems that you are on the right track.
When looking at the left side of equation $(1)$, note that for every value of $j$, you are multiplying $P(X=j)$ by every integer $i$ from $0$ through $j-1$. So for every $j$, you will have $(0+1+2+...+j-1)*P(X=j)$.
Since $(1+2+...+n)={n(n+1)\over 2}$, this becomes${j(j-1)\over 2}*P(X=j)=1/2*(j^2*P(X=j)-j*P(X=j)).$
Summing this up for every possible value for $j$ gives you the left side of equation $(2)$.
As far as the last step, there is a general formula for discrete random variables that $E[f(x)] =\sum^{n}_{i=1}{f(x_i)*P(X=x_i)}$ where $x_1,...,x_n$ are all of the possible values for $x$.
Since in your example $x$ can be any non-negative integer, you would get $E[X] = \sum^{\infty}_{i=0}{i*P(X=i)}$
$E[X^2]=\sum^{\infty}_{i=0}{i²*P(X=i)}$
and the rest of the equation follows easily.