In [this question](https://stats.stackexchange.com/questions/305031/expectation-when-cumulative-distribution-function-is-given/305035#305035 ) 

whuber comments that the technique used in this answer is summation by parts:

> The discrete case, assume that $X \ge 0$ takes non-negative integer
> values. Then we can write the expectation as  $$\DeclareMathOperator{\E}{\mathbb{E}}
 \DeclareMathOperator{\P}{\mathbb{P}} \E X = \sum_{k=0}^\infty k \P(X=k) $$ Now, we will first write this as a double sum, and then
> change the order of summation. Observe that $k = \sum_{j=0}^{k-1} 1$
> (the case $k=0$ gives an upper limit that is smaller than the lower
> limit, we take that as the empty sum, which is zero). This gives $$   
\E X = \sum_{k=0}^\infty \sum_{j=0}^{k-1} 1 \cdot \P(X=k) $$ Now, in
this double sum we sum first on $j$, which clearly goes to $\infty$.
> Observe that in the inner summation the indices satisfy the inequality
> $$
 0 \le j \le k-1 $$ Solving that for $k$ gives $ k \ge j+1$, which then gives the limits of summation in the new inner sum: $$ \E X = \sum_{j=0}^\infty \sum_{k=j+1}^\infty \P(X=k) = \sum_{j=0}^\infty \P(X > j) $$ which is the result.  The continuous case is similar.



I went to the wikipedia to try to understand his comment but I do not get it.


> Suppose $\{f_k\}$ and $\{g_k\}$ are two sequences. Then,
> 
> $$
\sum_{k=m}^{n} f_k (g_{k+1} - g_k) = (f_{n+1} g_{n+1} - f_m g_m) - \sum_{k=m}^{n} g_{k+1} (f_{k+1} - f_k). 
$$

In what way is rearranging the sums related to summation by parts?