# Question about convergence of $\sum a_n \cdot b_n$ when $\sum a_n$ converges.

So i have some series, it is $\sum a_n$ and i know it converges absolutely. Is it true that for given: $$b_n = \frac{n^2+1}{n^2}$$ $\sum a_n \cdot b_n$ converges absolutely too?

What about $\sum a_n$ semi-converges, $b_n$ is the same. Does $\sum a_n \cdot b_n$ semi-converges too?

In general, i have given some, any sequence $a_n$ and $\sum a_n$ (semi)converges. For given $b_n$, what are basic assumptions? What do i have to know to solve this kind of tasks?

For my logic, if $\sum a_n$ converges absolutely, then $\sum a_n \cdot b_n$ we can rewrite as $$\sum \left( 1 + \frac{1}{n^2} \right) \cdot a_n = \sum \left( a_n + \frac{a_n}{n^2} \right) = \sum a_n + \sum \frac{a_n}{n^2}$$ And we know that $\sum a_n$ converges absolutely, and since $n$ are natural numbers then $\sum \frac{a_n}{n^2}$ have to converge absolutely too, is it good approach? I'd be very greatful if you could share some experience with such "theoretical" convergence. Thanks in advance

Notice that $$b_n\sim_\infty1$$ then $$|a_nb_n|\sim_\infty |a_n|$$ hence the series $$\sum_n a_nb_n$$ is absolutely convergent by asymptotic comparison. We have the same result for the semi-convergent series.

• Hey, thanks for your answer, it's nice. What if $b_n$ is asymptotic to something else than constant, like $\frac{1}{2^n}$ or $\frac{1}{n^2}$ ? That bugs me the most now. Commented Feb 25, 2014 at 21:12
• With your two examples there's no problem since in the first case for example we have $$|a_nb_n|\le \frac1{2^n} \quad \forall n\ge n_0$$ since the sequence $(a_n)$ is convergent to $0$.
– user63181
Commented Feb 25, 2014 at 21:16
• I don't fully understand, could you elaborate a bit more? Commented Feb 25, 2014 at 21:25
• If the series $\sum_n a_n$ is convergent then the sequence $(a_n)$ is convergent to $0$ hence if you give $b_n$ is asymptotic to $\frac1{2^n}$ then we have $$|a_nb_n|\le |b_n|\sim_\infty\frac1{2^n}$$
– user63181
Commented Feb 25, 2014 at 21:29
• Then when the problems starts or for what $b_n$ $\sum a_n*b_n$ diverges? It's not really clear for me cause for $b_n=\dfrac1n$ $\sum b_n$ diverges, but $\sum a_n*b_n$ doesn't. But for $b_n = n!$ $\sum a_n*b_n$ diverges too... Is there some rule or something? Thanks for your time though, i really appreciate it and i don't know if i'm not overthinking this stuff... Commented Feb 25, 2014 at 21:45

$\sum a_n*b_n$ converges absolutely if $\sum |a_n(1+\frac{1}{n^2})|$ converges.

We know, $\sum\limits_{n=1}^N |a_n(1+\frac{1}{n^2})|\le \sum\limits_{n=1}^N |a_n|+|\frac{a_n}{n^2}|\le \sum\limits_{n=1}^N |a_n|+ \sum\limits_{n=1}^N \frac{|a_n|}{n^2}$.

Since, $\sum\limits_{n=1}^N |a_n|$ converges, $|a_n|\rightarrow 0$, that is $\exists N\in \mathbb{N}, s.t. \forall n\ge N, |a_n|\le \epsilon$. That is $\sum\limits_{n=1}^{\infty} \frac{|a_n|}{n^2}\le \sum\limits_{n=1}^N \frac{|a_n|}{n^2} + \epsilon\sum\limits_{n=N+1}^{\infty} \frac{1}{n^2}$ , which converges.