Convergence of product of convergent series

Question

Given $a_n, b_n$ such that $\exists N \forall n>N a_n, b_n \geq 0$ I want to show that if $\sum_{n=0}^{\infty}a_n$ and $\sum_{n=0}^{\infty}b_n$ converge then $\sum_{n=0}^{\infty}a_n b_n$ also converges.

I tried to show it in the following way, if $\sum_{n=0}^{\infty}b_n$ converges, then $\lim_{n\rightarrow\infty}b_n=0$, so there exists $n_0$ such that $b_{n_0}\geq b_n$ for all $n\gt n_0$, therefore $a_n b_{n_0}\geq a_n b_n$, so that $\sum_{n=0}^{\infty}a_n b_n\leq \sum_{n=0}^{n_0}a_n b_n+b_{n_0}\sum_{n=n_0+1}^\infty a_n$, and as my series is bounded by sum of finitely many terms and tail of convergent series multiplied by a constant it's also convergent. Is my proof correct, or are there any flaws in my reasoning?

Answer 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

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

Answer 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.

Answer 3

There's another proof, involving partial sums:

By hypothesis you know that, $\sum_{n=0}^{\infty}a_n$ and $\sum_{n=0}^{\infty}b_n$ converge, then by definition, if $(S_n)_{n\in\mathbb{N}}$ and $(T_n)_{n\in\mathbb{N}}$ are respectively the partial sums then they converge. So, by a previous result, it is known that $\lim_{n \to \infty}(S_n \cdot T_n)$ exists, ie, $(S_n \cdot T_n)_{n \in \mathbb{N}}$ converges and that implies that $\sum_{n=0}^{\infty}a_nb_n$ converges.