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.

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.
