Concluding the convergence of a product of series

Question

Let $a(n)$ be a bounded sequence (not necessarily convergent) and assume $\lim b(n) = 0$. Prove that $\lim a(n)b(n) = 0$. Can we conclude anything about the convergence of $a(n)b(n)$ if $\lim b(n)=b$, where $b$ is not equal to $0$?

I know how to prove that first part, but in the convergence of $a(n)b(n)$, is there really anything you can definitively conclude since we don't know anything about $a(n)$ other than that it is bounded?

For the second part, think about the counterexample you used in your previous question. This should allow you to find an example where the product does not converge. Also, try to think of an example where the product does converge. — David Mitra, Feb 13 '13 at 15:15
Please write your mathematics with LaTeX, to make them more readable. In the FAQ section you can find directions. — DonAntonio, Feb 13 '13 at 15:18
OK thanks I've been trying to find out how to do that since I've started asking. — Student, Feb 13 '13 at 15:19
@Student: Welcome to MSE! It really helps readability if you format your questions using MathJax. Regards — Amzoti, Feb 13 '13 at 15:20
@Amzoti I'm working on it! Lol computers aren't my "strong-suit" ..thanks for that link — Student, Feb 13 '13 at 15:37

score 1 · Accepted Answer · answered Feb 13 '13 at 15:23

Take

$$a_n:=(-1)^n\;\;,\;\;b_n=\frac{n}{n+1}$$

Then

$$|a_n|\le 1\,\,\,\forall\,n\in\Bbb N\,\,,\,\,b_n\xrightarrow[n\to\infty]{}1$$

yet

$$\lim_{n\to\infty}a_nb_n=\lim_{n\to\infty}\frac{(-1)^nn}{n+1}$$

doesn't exist.

If instead the above bounded sequence you choose $\,a_n=k=\text{ constant}\,$ , then $\,a_nb_n\,$ does converge, so in general nothing can be said about the sequence $\,\{a_nb_n\}\,$

Concluding the convergence of a product of series

1 Answers1