I've been stuck in this one for a while. Could anyone help by giving some hints on how to approach this problem?

Prove that if $\sum_{n=0}^\infty a_n$ is absolutely convergent and $\lim b_n=0$, then $\lim (a_0b_n+a_1b_{n-1}+...+a_nb_0)=0$.

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    $\begingroup$ use the fact that the tail of the series, i.e. $\sum_{k=n}^\infty |a_k| $ converges to $0$ as $n\to \infty$, and that the sequence $b_n$ is bounded. This allows you to split the desired sequence into two regimes, where in one you rely on convergence of $b_n$, and in the other one on convergence of the tail. $\endgroup$ – Hayk May 12 '18 at 13:58
  • $\begingroup$ @Hayk Please consider making it an answer. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 12 '18 at 14:18
  • $\begingroup$ @GNUSupporter, thanks; I've expanded on the comment to make it an answer. $\endgroup$ – Hayk May 12 '18 at 14:35
  • $\begingroup$ Slightly more general: $A = \sum_{n=0}^\infty a_n$ and $b_n \to B$ implies $\sum_{k=0}^n a_k b_{n-k} \to AB$. $\endgroup$ – Martin R May 12 '18 at 18:44

Fix any $\varepsilon >0$ small, and let $k\in \mathbb{N}$ be so that $$ \sum_{i=k}^\infty |a_i| < \frac{\varepsilon}{2M} , $$ where $M>0$ is fixed as an upper bound for $|b_n|$, i.e. we have $|b_n| \leq M$ for all $n \in \mathbb{N}$ ($b_n$ is bounded due to the fact that it converges). Let also $n \in \mathbb{N}$ be so large that $|b_{i-k}| < \frac{\varepsilon}{2 \sum_{i=0}^k |a_i|}$, for all $i=n, n+1,...$ .

We then get $$ |a_0b_n +...+a_n b_0| \leq |a_0 b_n +...+a_k b_{n-k}| + |a_{k+1} b_{n-k-1}+...+a_n b_0| \leq \\ \sum_{i=0}^k |a_i| \frac{\varepsilon}{2 \sum_{i=0}^k |a_i|} + M \frac{\varepsilon}{2M} \leq \\ \frac{\varepsilon}{2} + \frac{\varepsilon}{2}=\varepsilon. $$ Since $\varepsilon>0$ is arbitrary, we are done.

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