an identity regarding the summation Assuming necessary convergence, why does the following identity hold?
$$\sum_{n=1}^\infty\sum_{k=0}^{n-1} a_n b_k = \sum_{k=0}^\infty\sum_{n=k+1}^{\infty} a_n b_k$$
Intuitively this is clear, but what's the formal justification?
Thanks!
 A: Let $S_n = \sum_{k=1}^n \sum_{i=0}^{k-1} a_kb_i$ and $T_n = \sum_{i=0}^{n-1} \sum_{k=i+1}^{n} a_kb_i$.
Now, it's true that
$$
S_1 = a_1b_0 = T_1
$$
Let's assume that $S_n=T_n$. Now,
$$
S_{n+1} = \sum_{k=1}^{n+1}\sum_{i=0}^{k-1} a_kb_i = \sum_{k=1}^n \sum_{i=0}^{k-1} a_k b_i + \sum_{i=0}^n a_{n+1}b_i = S_n + \sum_{i=0}^n a_{n+1}b_i
$$
And
$$
T_{n+1} = \sum_{i=0}^{n} \sum_{k=i+1}^{n+1} a_kb_i = \sum_{i=0}^{n-1}\sum_{k=i+1}^na_kb_i + \sum_{i=0}^na_{n+1}b_i = T_n + \sum_{i=0}^na_{n+1}b_i
$$
But we assumed $S_n=T_n$. Therefore, $S_{n+1}=T_{n+1}$, and thus $S_n=T_n$ for integer $n\geq1$.
And if it's true for all positive integers, then it's also true in the limit, $S_\infty=T_\infty$.
Note that the final statement, "if it's true for all positive integers, then it's also true in the limit", is only correct for situations where the summation is absolutely convergent. This arises due to the inner sums in $T_n$ depending on $n$. Having said this, I suspect that there aren't any cases where the summations are well-defined and fail to agree in the case of sums of the form provided in the question.
