0
$\begingroup$

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 $\lim (a_0b_n+a_1b_{n-1}+...+a_nb_0)=0$.

$\endgroup$
  • 4
    $\begingroup$ 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
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.