Solution 1:

Let $p_i = P[X=i]$. You know that $$E[X]=\sum_{i=1}^\infty ip_i$$ and that $$P[X\ge i] = \sum_{j=i}^\infty p_j,$$ so you want to show that $$\sum_{i=1}^\infty ip_i = \sum_{i=1}^\infty \sum_{j=i}^\infty p_j\;.$$ Try reversing the order of summation in the double summation.