Convergence of product of convergent series

Solution 1:

$\sum_m^na_kb_k = b_n\sum_m^na_k + \sum_{k=m}^{n-1} \big(\sum_{j=m}^{k}a_j \big)(b_k - b_{k+1})$.

This is the abels formula. You can now use the bound on b and convergence of sequence $a_n$ to get the convergence.

Since $b_n$ sequence is convergent you can easily show a bound M on $b_n$ and then use Cauchy criterion to get an N for any given $\epsilon$ such that $\sum_m^na_k < \epsilon / 4M \hspace{2mm} \forall m,n > N$ which will give you

$|\sum_{k=m}^{n}a_kb_k| \leq M \frac{\epsilon}{4M} + 2M \frac{\epsilon}{4M} < \epsilon$

Solution 2:

Take $N'\geq N$ such that $n>N'\implies a_n\leq 1.$ This is possible because the convergence of $\sum_n a_n$ implies that $a_n\to 0$ as $n\to \infty.$ For all $n>N'$ we have $0\leq |a_nb_n|=a_nb_n\leq b_n=|b_n|.$

The absolute values of the terms of the series $\sum_{n>N'}a_nb_n$ do not exceed the absolute values of the terms of the absolutely convergent series $\sum_{n>N'}b_n,$ so $\sum_{n>N'}a_nb_n $ converges.