# Convergence of $\sum_{n=0}^{\infty}a_nb_n$ if $a_n>0 \ \forall n\in \mathbf{N}$, $\sum_{n=0}^{\infty}a_n$ converges and $b_n$ bounded

Let $$(a_n)$$ a sequence such that $$a_n>0 \ \forall n\in \mathbf{N}$$. Suppose that $$\sum_{n=0}^{\infty}a_n$$ converges and the sequence $$(b_n)$$ is bounded. Show that $$\sum_{n=0}^{\infty}a_nb_n$$ converges.

I would like to have a feedback on my proof and to know if everything holds. Thanks in advance.

Proof.

First of all, as $$\sum_{n=0}^{\infty}a_n$$ converges, we can note that $$\lim_{n\to \infty}a_n=0$$. So, $$\exists N \ \forall n\ge N$$: $$0 .

Then, as $$(b_n)$$ is a bounded sequence, we have by definitiot that:

$$\exists M>0: \ |b_n|\le M \ \forall n\in \mathbf{N}$$.

With all these preliminary results we can write that:

$$|b_n|\le M \iff |b_n|\cdot a_n\le M\cdot a_n \underbrace{\iff}_{a_n>0 \ \forall n} |b_na_n|\le M|a_n|$$.

As $$\sum_{n=0}^{\infty}a_n$$ converges and $$a_n>0 \ \forall n \in \mathbf{N}$$ (so there is no difference between $$\sum_{n=0}^{\infty}a_n$$ and $$\sum_{n=0}^{\infty}|a_n|$$), $$\sum_{n=0}^{\infty}|a_n|$$ converges and so $$\sum_{n=0}^{\infty}M|a_n|$$ too.

By comparaison we conclude that $$\sum_{n=0}^{\infty}a_nb_n$$ converges absolutely so it converges.

• You don't need the two lines just after Proof. Commented Feb 16, 2021 at 19:19
• @Ryan Shesler but as $a_n>0$ it holds no? If i didn't have that hypothesis i should have written $0<|a_n|<1$ Commented Feb 16, 2021 at 19:21
• @archuser Sorry I missed where it said $a_n > 0$, my bad. Your argument looks good to me. Commented Feb 16, 2021 at 19:21
• @hamam_Abdallah Oh yes of course... First i was trying to resolve using the inequality with $a_n$ so forgot to remove that from my paper x) Thank you! Commented Feb 16, 2021 at 19:22
• @Ryan Shesler Thank you for your feedback! Commented Feb 16, 2021 at 19:23

Assume $$M>0$$. Given $$\epsilon>0$$,
For $$m$$ and $$n$$ large enough $$|\sum_{k=n}^ma_k|=\sum_{k=n}^ma_k< \frac{\epsilon}{M}$$ and $$|\sum_{k=n}^ma_kb_k|\le M\sum_{k=n}^ma_k<\epsilon$$
This proves the convergence of the series $$\sum a_nb_n$$.