Abel criterion proof

For another approach, we can show that the sequence of partial sums converges without using the Cauchy criterion. We have existence of the limits $\lim_{n \to \infty}A_n = \lim_{n \to \infty}\sum_{k=1}^na_k =A$ and $\lim_{n\to \infty}b_n = b .$

Summing by parts, we get

$$S_n =\sum_{k=1}^n a_kb_k = a_1b_1+\sum_{k=2}^n (A_k - A_{k-1})b_k = a_1b_1+\sum_{k=2}^{n} A_k b_k- \sum_{k=2}^{n} A_{k-1} b_k \\ = \sum_{k=1}^{n} A_k b_k- \sum_{k=1}^{n-1} A_{k} b_{k+1} = A_nb_{n+1} + \sum_{k=1}^{n} A_k (b_k - b_{k+1})$$

The series $\sum(b_k - b_{k-1}) $ converges since $\sum_{k=1}^n (b_k - b_{k+1}) = b_1 - b_{n+1} \to b_1 - b$ as $n \to \infty$. Since $(A_k)$ is a bounded sequence and the terms $(b_k- b_{k+1})$ are all of the same sign, it follows that $\sum A_k(b_k - b_{k+1})$ is convergent.

Therefore, the series $\sum a_kb_k$ converges since

$$\sum_{k=1}^\infty a_kb_k = \lim_{n \to \infty}A_nb_{n+1} + \lim_{n \to \infty}\sum_{k=1}^{n} A_k (b_k - b_{k+1}) = Ab + \sum_{k=1}^\infty A_k(b_k - b_{k+1})$$