Let ${a_n}$ and ${b_n}$ sequences with positive terms, such that $\sum a_n$ and $\sum b_n$ converges. Prove that $\sum a_nb_n$ converges as well.

What I did: $\sum a_nb_n \le \left(\sum a_n\right) \left(\sum b_n\right)$ and by comparison test we are done.

However, when I asked my teacher regarding that solution, he told me that I can assume that inequality for finite numbers but for infinite numbers, it is not valid. It seems to me that $\left(\sum a_n\right) \left(\sum b_n\right)$ is absolute converges so that are no Riemann issues with this.

Can any one enlighten me and explain why is this invalid?

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    $\begingroup$ You know the inequality is true for finite sums. Here, you are dealing with limits of sums, so you have to make that explicit. $\endgroup$ – user370967 Jan 20 '18 at 10:08
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    $\begingroup$ Use $a_nb_n\leq \frac{a_n^2+b_n^2}{2}$, and that $a_n^2\leq a_n$ and $b_n^2\leq b_n$ given the convergence hypotheses, for $n$ large. $\endgroup$ – orole Jan 20 '18 at 10:08
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    $\begingroup$ As all is positive, boundedness would have sufficed for convergence. With $a=\sum a_n$, $b=\sum b_n$, we have $a_n\le a$, hence $\sum a_nb_n\le \sum ab_n\le a\sum b_n\le ab$. You do not even have to resort to absolute convergence for this $\endgroup$ – Hagen von Eitzen Jan 20 '18 at 10:13
  • $\begingroup$ Why is that $a_n^2\leq a_n$ true? @orole $\endgroup$ – Moshe King Jan 20 '18 at 10:48
  • $\begingroup$ @GADI Because when you square a number that is in $[0,1)$ it gets smaller, and the $a_n$ eventually get small. $\endgroup$ – orole Jan 20 '18 at 10:55

The way you started and since $\left(a_n\right)_{n\in\mathbb{N}}$ and $\left(b_n\right)_{n\in\mathbb{N}}$ are positive then for partial sums we have

$$0\leq\sum\limits_{n=1}^{k} a_nb_n\leq \left(\sum\limits_{n=1}^{k} a_n\right)\left(\sum\limits_{n=1}^{k} b_n\right)$$ The limit has the property to preserve the inequality (number 7 down the list), then $$0\leq\lim\limits_{k\rightarrow\infty}\sum\limits_{n=1}^{k} a_nb_n\leq \lim\limits_{k\rightarrow\infty}\left(\sum\limits_{n=1}^{k} a_n\right)\left(\sum\limits_{n=1}^{k} b_n\right) =\lim\limits_{k\rightarrow\infty} \left(\sum\limits_{n=1}^{k} a_n\right)\cdot \lim\limits_{k\rightarrow\infty} \left(\sum\limits_{n=1}^{k} b_n\right) $$ which is

$$0\leq \sum\limits_{n=1} a_nb_n\leq \left(\sum\limits_{n=1} a_n\right)\cdot \left(\sum\limits_{n=1}b_n\right) $$

Since we know that $0\leq\sum\limits_{n=1}a_n < \infty$ and $0\leq\sum\limits_{n=1}b_n < \infty$, thus we have $0\leq \sum\limits_{n=1} a_nb_n <\infty$

Another way, since $0\leq\sum\limits_{n=1}b_n < \infty$ then $\lim\limits_{n\rightarrow\infty}b_n=0$ which also means $\left(b_n\right)_{n\in\mathbb{N}}$ is bounded or $0\leq b_n < M, \forall n\in\mathbb{N}$. Then $$0\leq\sum\limits_{n=1}^{k} a_nb_n\leq M \left(\sum\limits_{n=1}^{k} a_n\right)$$ Taking the limit $$0\leq\sum\limits_{n=1} a_nb_n\leq M \left(\sum\limits_{n=1} a_n\right)<\infty$$

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    $\begingroup$ This is a perfectly reasonable answer... $\endgroup$ – goblin Jan 20 '18 at 11:04

Since $\sum\limits_{k = 1}^\infty b_k$ converges, then $b_n \to 0 \ (n \to \infty)$. Note that $\{b_n\}$ is a sequence of positive numbers, thus there exists $N \in \mathbb{N}_+$ such that$$ 0 < b_n < 1. \quad \forall n > N $$ Therefore,$$ 0 < \sum_{k = 1}^\infty a_k b_k = \sum_{k = 1}^N a_k b_k + \sum_{k = N + 1}^\infty a_k b_k \leqslant \sum_{k = 1}^N a_k b_k + \sum_{k = N + 1}^\infty a_k < +\infty. $$


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