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

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).

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

• If $\epsilon < b$ then $|b_n| = b_n$ and $b-\epsilon < b_n < b+\epsilon$. – fleablood Sep 22 at 15:38
• so $1/\varepsilon |b_n|>1/\varepsilon \cdot (b-\varepsilon)=b/\varepsilon -1$ why does that make it etter? – ParabolicAlcoholic Sep 22 at 15:46
• What if $\frac b\epsilon - 1 > \epsilon$? – fleablood Sep 22 at 16:00
• 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. – fleablood Sep 22 at 16:05
• actually, it says that it is unbounded... – 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.

• I think you missed 2 absolute value at the end. – Botond Sep 22 at 15:33
• @Botond I've edited my answer. Thank you. – 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}

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.