2
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ 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
  • 3
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
4
$\begingroup$

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 $$0\leq\sum\limits_{n=1} a_nb_n\leq M \left(\sum\limits_{n=1} a_n\right)<\infty$$

$\endgroup$
  • 1
    $\begingroup$ This is a perfectly reasonable answer... $\endgroup$ – goblin Jan 20 '18 at 11:04
1
$\begingroup$

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,$$ 0 < \sum_{k = 1}^\infty a_k b_k = \sum_{k = 1}^N a_k b_k + \sum_{k = N + 1}^\infty a_k b_k \leqslant \sum_{k = 1}^N a_k b_k + \sum_{k = N + 1}^\infty a_k < +\infty. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.