If $a_n \uparrow a$, then $\lim_{N\to \infty} \sum_{n=0}^{N-1} (a_N - a_n) = \sum_{n=0}^{\infty}(a -a_n)$ Suppose that $a_n$ is a sequence increasing to $a > 0$. Apparently, the following is true:
$$\lim_{N\to \infty} \sum_{n=0}^{N-1} (a_N - a_n) = \sum_{n=0}^{\infty}(a -a_n)$$
I am solving a problem and everything boiled down to proving the above equality. Can someone help me see this? (by hints or otherwise)
Thanks in advance
 A: If you can readily use Lebesgue integration: Let $\zeta$ be the counting measure on $\mathbb{N}$, and for $N \in \mathbb{N}$ define $f_N(n) = (a_N - a_n)^+ = \max \{ a_N - a_n, 0\}$, and $f(n) = a-a_n$. Then $0 \leqslant f_N \uparrow f$, so by the monotone convergence theorem
$$\sum_{n = 0}^{N-1} (a_N - a_n) = \int_{\mathbb{N}} f_N\,d\zeta \uparrow \int_{\mathbb{N}} f\,d\zeta = \sum_{n = 0}^\infty (a - a_n).$$
Without Lebesgue integration: It's clear that
$$\sum_{n = 0}^{N-1} (a_N - a_n)\leqslant \sum_{n = 0}^\infty (a - a_n),$$
and that the sequence of finite sums is monotonically increasing. So the limit of the finite sums exists, and is not larger than the infinite sum (which may be $+\infty$ or $< +\infty$, that doesn't matter). For every $K < \sum_{n = 0}^\infty (a - a_n)$, choose an $M$ such that
$$\sum_{n = 0}^M (a - a_n) > K.$$
Since $a_N \uparrow a$, we have
$$K < \sum_{n = 0}^M (a_N - a_n) \leqslant \sum_{n = 0}^{N-1} (a_N - a_n)$$
for all large enough $N$, so
$$\lim_{N\to\infty} \sum_{n = 0}^{N-1} (a_N - a_n) > K.$$
