Let $(a_n), (b_n)\subset \mathbb{R}$. Show:

$1/a_n \to 0$ and $b_n \to b>0$ implies that $(a_nb_n)$ diverges.

Here's my progess (ignore the first two lines, they repeat the task).

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How do I proceed? How can I get rid of $|b_n|$?

  • $\begingroup$ If $\epsilon < b$ then $|b_n| = b_n$ and $b-\epsilon < b_n < b+\epsilon$. $\endgroup$ – fleablood Sep 22 at 15:38
  • $\begingroup$ so $1/\varepsilon |b_n|>1/\varepsilon \cdot (b-\varepsilon)=b/\varepsilon -1$ why does that make it etter? $\endgroup$ – ParabolicAlcoholic Sep 22 at 15:46
  • $\begingroup$ What if $\frac b\epsilon - 1 > \epsilon$? $\endgroup$ – fleablood Sep 22 at 16:00
  • $\begingroup$ What's your definiton of "diverge"?. $|a_nb_n| \to \infty$ there is no way of determining which $a_nb_n$ are positive and which are negative. $\endgroup$ – fleablood Sep 22 at 16:05
  • $\begingroup$ actually, it says that it is unbounded... $\endgroup$ – ParabolicAlcoholic Sep 22 at 16:06

For any $\epsilon: 0 < \epsilon < b$ you have have an $N_1$ where $n > N_1$ implies $|\frac 1{a_n}| < \epsilon$ and $|a_n| > \frac 1{\epsilon}$ and an $N_2$ where $n > N_1$ means $|b-b_n| < \epsilon$ so $0 < b-\epsilon < b_n < b+\epsilon$.

So if $n > \max (N_1, N_2)$ then $|a_n*b_n| =|a_n|*|b_n| > \frac 1{\epsilon}(b-\epsilon)= \frac b{\epsilon} - 1$.

So it's a matter of making sure that $\epsilon < b$ and $\epsilon <\frac b{\epsilon} -1$. (i.e. $ \epsilon^2 +\epsilon < b$) which we can assure if $\epsilon < \min(\frac b2, 1)$.

Redo properly:

For any $\epsilon' > 0$ let $\epsilon: 0 < \epsilon < \min(\epsilon', \frac b2, 1)$.

The $\epsilon^2 + \epsilon < 2\epsilon < b$ and $0< \epsilon < \frac b\epsilon -1$

Let $N =\max(N_1,N_2)$ so that $n > N_1 \implies |a_n| > \frac 1\epsilon$ and $n > N_2 \implies b_n > b-\epsilon > 0$ and so $n>N\implies |a_n*b_n| >\frac 1{\epsilon}(b-\epsilon)=\frac b{\epsilon} -1 > \epsilon >\epsilon'$ and therefore $a_n*b_n$ does not converge.


Since $\lim_{n\to\infty}\frac1{a_n}=0$, you have $\lim_{n\to\infty}\frac1{\lvert a_n\rvert}=0$ and therefore $\lim_{n\to\infty}\lvert a_n\rvert=\infty$. And, since $\lim_{n\to\infty}b_n=b>0$, there is a $N\in\mathbb N$ such that $n\geqslant N\implies b_n>\frac b2$. But then$$\lim_{n\to\infty}\lvert a_nb_n\rvert\geqslant\lim_{n\to\infty}\frac{\lvert a_n\rvert b}2=\infty,$$and so $(a_nb_n)_{n\in\mathbb N}$ diverges.

  • $\begingroup$ I think you missed 2 absolute value at the end. $\endgroup$ – Botond Sep 22 at 15:33
  • 1
    $\begingroup$ @Botond I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Sep 22 at 16:29
  • Since $\frac{1}{a_n}\to 0$, for all $n$, there is $N_n\geq n$ s.t. $|a_{N_n}|\geq n.$

  • Since $b_n\to b>0$, there is $M\in\mathbb N$ s.t. $b_n>\frac{b}{2}>0$ for all $n\geq M$.

Therefore, if $n\geq M$, $$|a_{N_n}b_{N_n}|\geq \frac{b}{2}n\to \infty .$$


The condition that $b>0$ is very much used in the following proof.

As $\frac{1}{a_n}\rightarrow 0$, given $\epsilon>0$ there exists $N_1\in \mathbb{N}$ such that $\frac{1}{|a_n|}<\epsilon$, in other words, $|a_n|>\frac{1}{\epsilon}$ for all $n\geq N_1$.

Let $M>0$. Set $\epsilon=\frac{1}{M}$. Then, there exists $N_1\in \mathbb{N}$ such that $|a_n|>M$ for all $n\geq N_1$.

As $(b_n)\rightarrow b$, given $\epsilon'>0$ there exists $N'\in \mathbb{N}$ such that $|b_n-b|<\epsilon'$. This $\epsilon'$ can be any positive real number. As $b$ is positive, we can take $\epsilon'=\frac{b}{2}$ and see that $$|b_n-b|<\frac{b}{2}\Rightarrow -\frac{b}{2}<b_n-b\Rightarrow \frac{b}{2}<b_n$$

Choose $N=\max{N_1,N_2}$. For $n\geq N$, we have $|a_n|>M$ and $b_n>\frac{b}{2}$. As $b_n>0$, multiplying both sides of the equality $|a_n|>M$ with $b_n$ keeps the inequality unchanged; that is $|a_n|b_n>Mb_n$ for each $n\geq N$. As $b_n>\frac{b}{2}$, for each $n\geq N$, we see that $|a_nb_n| >Mb_n>\frac{Mb}{2}$ for each $n\geq N$. See that, as $b_n>0$ I can take it inside the modulus.

So, given $M>0$, we have found $N\in \mathbb{N}$ such that $|a_nb_n|>\frac{Mb}{2}$ for each $n\geq N$. Thus, $(a_nb_n)$ diverges.


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