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.


2 Answers 2


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.

  • $\begingroup$ Are you sure that $\sum_\limits{n=0}^{q-1} A_n(b_{n+1} + b_n)$ is convergent? $\endgroup$
    – Angelo
    Commented Jul 26, 2020 at 8:24
  • $\begingroup$ It will be $\sum_{n=0}^{q-1} A_n(b_{n+1}-b_n)$. $\endgroup$
    – Sharlin
    Commented Jul 26, 2020 at 16:51
  • $\begingroup$ Thank you, edited $\endgroup$
    – Hugo
    Commented Jul 26, 2020 at 17:57
  • $\begingroup$ 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. $\endgroup$
    – aritracb
    Commented May 4, 2022 at 21:16

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