# Prove $\sum_{n=1}^\infty a_n b_n$ is convergent if $\sum_{n=1}^\infty (b_n -b_{n+1})$ is absolutely convergent , $\sum a_n$ convergent

Prove that $$\sum_{n=1}^\infty a_n b_n$$ is convergent if $$\sum_{n=1}^\infty a_n$$ is convergent and $$\sum_{n=1}^\infty (b_n -b_{n+1})$$ is absolutely convergent series. Since, $$\sum_{n=1}^\infty (b_n -b_{n+1})$$ converges absolutely i.e. $$\sum_{n=1}^\infty \vert (b_n -b_{n+1}) \vert$$ converges implies $$\sum_{n=1}^\infty (b_n -b_{n+1})$$ converges also.

Also $$\sum_{n=1}^\infty a_n$$ is also a convergent sequence.

Let, $$A_n= \sum_{k=0}^n a_k$$. Then for $$0 \leq p \leq q$$, we have $$\sum_{n=p}^q a_n b_n = \sum_{n=p}^{q-1} A_n (b_n-b_{n+1})+ A_q b_q - A_{p-1} b_p$$

I think I need to use the comparison test or I need to show $$\sum b_n$$ is bounded.

The fact that $$\sum_n(b_n-b_{n+1})$$ converges means $$b_1-b_{n+1}=\sum_{k=1}^n(b_k-b_{k+1})$$ converges, hence $$b_n$$ converges. Similarly, $$A_n:=\sum_{k=1}^na_k$$ converges. Hence both sequences $$A_n$$ and $$b_n$$ are bounded by, say, $$A$$ and $$B$$ respectively, and both are Cauchy sequences.

$$\sum_{n=p}^qa_nb_n=\sum_{n=p}^qA_n(b_n-b_{n+1})+A_qb_{q+1}+b_pA_{p-1}$$ Hence \begin{align}|\sum_{n=p}^qa_nb_n|&\le |\sum_{n=p}^qA_n(b_n-b_{n+1})|+ |A_q||b_{q+1}-b_p|+|b_p||A_q-A_{p-1}|\\ &\le A\sum_{n=p}^q|b_n-b_{n+1}|+A|b_{q+1}-b_p|+B|A_q-A_{p-1}|\to0\end{align} as $$p,q\to\infty$$ since all terms are Cauchy. Hence the series is Cauchy and converges.

Step 1: $$\displaystyle \exists \lim_{n \to \infty} b_n = b$$.

Let $$B = \sum_{n = 0}^\infty |b_n - b_{n+1}|$$. Then we have $$|b_n - b_0| \leq \sum_{k = 0}^{n-1} |b_{k+1} - b_k| \leq B,$$ i.e. $$b_n \in |b_0-B, b_0+B|$$ for every $$n$$. By compactness, we extract a subsequence $$b_{n_k}$$ such that $$b_{n_k} \to b$$. By the Cauchy property of $$|b_{n+1} - b_n|$$ we deduce that in fact the whole sequence $$b_n$$ converges to $$b$$, as if $$n \geq n_k$$ we have $$|b_n-b_{n_k}| \leq \sum_{j = n_k}^{n-1} |b_{j+1} - b_{j}|.$$

Step 2: Using your notation, $$\sum_{n=0}^q a_nb_n = \sum_{n=0}^{q-1} A_n(b_{n+1} - b_n) + A_qb_q.$$ In the limit $$q \to \infty$$, the first sum defines a convergent series and the second term converges to $$B\sum_{n = 0}^\infty a_n$$, so you get a finite limit of the partial sums.

• Are you sure that $\sum_\limits{n=0}^{q-1} A_n(b_{n+1} + b_n)$ is convergent? Commented Jul 26, 2020 at 8:24
• It will be $\sum_{n=0}^{q-1} A_n(b_{n+1}-b_n)$. Commented Jul 26, 2020 at 16:51
• Thank you, edited
– Hugo
Commented Jul 26, 2020 at 17:57
• how to prove the converse of this? i.e. $\sum a_nb_n$ converges for all convergent series $\sum b_n$ then $\sum|a_n-a_{n+1}|$converges. Commented May 4, 2022 at 21:16