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?
Best Answer
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.