Convergence of product of convergent series 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?
 A: 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.
A: $\sum_m^na_kb_k = b_n\sum_m^na_k + \sum_{k=m}^{n-1} \big(\sum_{j=m}^{k}a_j \big)(b_k - b_{k+1})$.
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
$|\sum_{k=m}^{n}a_kb_k| \leq M \frac{\epsilon}{4M} + 2M \frac{\epsilon}{4M} < \epsilon$
A: Since $\sum_{n=1}^\infty a_n$ converges, $\{a_n\}$ converges to $0$, hence it's bounded. Let $|a_n|<M$ for all $n\in\mathbb{N}$. Then $|a_nb_n|\leq Mb_n$ for all $n$ (because $b_n\geq0$) and since $\sum_{n=0}^\infty b_n$ converges, $\sum_{n=1}^\infty a_nb_n$ also converges.
(Alternative proof idea : Show that $a_n^2
\leq a_n$ and $b_n^2\leq b_n$ when $n$ is large enough and use AM-GM inequality $|a_nb_n|\leq\frac12(a_n^2+b_n^2)$ )
A: The easiest way to see the answer to this is that since each series is positive valued and converges, eventually each series will be strictly less than 1. The sum of the terms before that point is finite. And the product of the terms in each series after this point will obviously be less than each of the corresponding terms of the two individual series. And because the sums of those series are finite, and the new series terms will be, after some finite point, less than all the terms of the both of those series, it’s sum is also finite.
