Let $a_n$ and $b_n$ be sequences of real numbers. If $b_n$ is bounded and $\lim_{n \to \infty} a_n = 0$, then $\lim_{n\to\infty} a_{n}b_{n} = 0$

Question

Prove the below statement:

Let $a_n$ and $b_n$ be sequences of real numbers. If $b_n$ is bounded and $\lim_{n \to \infty} a_n = 0$, then $\lim_{n \to \infty} a_n b_n=0$

When I read this question, I read that $b_n$ may or may not converge, so taking the example $\dfrac{1}{n}$ and $3n$, $\lim_{n\to\infty} a_nb_n=3\neq0$. What am I getting wrong? It's a theorem from my book and the only hint is that $b_n$ being bounded is crucial. So I know it must be true but I can't understand why.

Any suggestions or further hints in the right direction would be greatly appreciated. This is the last problem I have and have been stumped by it all day.

$$\lim_{n\to\infty} \frac{1}{n}\frac{1}{3n} = 0$$ For the proof, try using the Squeezing theorem — Prahlad Vaidyanathan, Oct 22 '13 at 03:18
@david: For some basic information about writing math at this site see e.g. here, here, here and here. — The Chaz 2.0, Oct 22 '13 at 03:18
You misunderstand the question prahlad i meant the example to be $$\lim_{n\to \infty} \frac 1n 3n$$ — david stocker, Oct 22 '13 at 03:29
i'm showing my n00b skills now, is unbounded a finite sequence then? — david stocker, Oct 22 '13 at 03:30
Bounded means it never passes a certain point. For example, $\sin x$ is bounded by 1. Note that it doesn't converge — ndh, Oct 22 '13 at 03:35

score 5 · Accepted Answer · answered Oct 22 '13 at 03:40

5

Since $b_n$ is bounded, there is a number $M$ such that for all $n$, $|b_n|\leq M$. Now $$0 \leq |a_n b_n|\leq |a_n| M.$$ Take the limit as $n\to \infty$ and you are done.

answered Oct 22 '13 at 03:40

abnry

14,664

i think when defining these sequences you wouldn't use absolute values until after you take the limits right? since $a_n$ and $b_n$ are in the reals. thank you for the help, this helped clear it up for me tremendously. – david stocker Oct 22 '13 at 03:47
You must take the absolute value first unless you use $-a_n M \leq a_n b_n \leq a_n M$. Remember, $|x|\leq M$ means the same as $-M\leq x \leq M$. – abnry Oct 22 '13 at 03:48
roger. that makes sense – david stocker Oct 22 '13 at 03:48

Mael · Answer 2 · 2013-10-22T03:34:43.530

1

$|a_{n}b_{n}-ab|=|a_{n}b_{n}-ab_{n}+ab_{n}-ab|\leq |a_{n}b_{n}-ab_{n}| + |ab_{n}-ab|$

$=|b_{n}||a_{n}-a|+|a||b_{n}-b|=|b_{n}||a_{n}-a| + 0|b_{n}-b|=|b_{n}||a_{n}-a|$

The sequence $b_{n}$ is bounded i.e. $|b_{n}|<B<\infty$

then $|a_{n}b_{n}-ab|<B|a_{n}-a|$

$a_{n}$ is a convergent sequence so $\forall \varepsilon'>0$ $\exists N \in \mathbb{N}$ such that $\forall n \geq N$ $|a_{n}-a|<\varepsilon'$. Choose $\varepsilon'=\frac{\varepsilon}{B}$, then $|a_{n}b_{n}-ab|<\varepsilon$ for all $n\geq N$

$a$ is zero and so $|a_{n}b_{n}-0|<\varepsilon$ for all $n\geq N$ and so $a_{n}b_{n}\rightarrow 0$ as $n\rightarrow \infty$

edited Oct 22 '13 at 03:34

answered Oct 22 '13 at 03:28

Mael

741

Please explain what are $a$ and $b$. From context it seems that $\lim_{n \to \infty} a_{n} = a$, but what is $b$ is not clear. Its definitely not the limit of $ b_{n}$ as we don't know if $b_{n}$ converges. – Paramanand Singh Oct 22 '13 at 03:41
you are correct that a is the limit of $a_{n}$. $b$ can be anything you like, 42 say, it doesn't matter :) – Mael Oct 22 '13 at 13:05
For a condensed version: $|a_{n}b_{n}-0|=|a_{n}b_{n}|\leq |B||a_{n}|$ because $a_{n}$ converges there exists an $N$ such that for all $n\geq N$ $|a_{n}|<\frac{\varepsilon}{|B|}$ and so it follows that for all $n\geq N$ $|a_{n}b_{n}-0|<\varepsilon$ from which one can conclude that $\lim a_{n}b_{n}=0$ – Mael Oct 29 '13 at 16:17

score 1 · Answer 3 · answered Oct 22 '13 at 03:31

What does it means that $$\lim_{n\to\infty}a_n=0\,\text{?}$$

It means that for any arbitrarily small real $\varepsilon>0$, after some $n_\varepsilon\in\mathbb N$, all $|a_n|$ are less than $\varepsilon$, for $n>n_\varepsilon$.

What does it means the $\{b_n\}$ is bounded?

It means that there is a boundary $B\in\mathbb R$ so that $|b_n|<B$.

Hint: For a given $\varepsilon>0$, take $n_{\varepsilon/B}$.

score 0 · Answer 4 · answered Oct 22 '13 at 03:32

0

Suppose the upper bound of $b_n$ is the constant $c$. Now suppose that $b_n = c$.

$$\lim_{n\to\infty} ca_n=c\lim_{n\to\infty} a_n=0$$

answered Oct 22 '13 at 03:32

ndh

430

But we do not want to assume that $b_n=c$, do we? – Did Oct 26 '13 at 09:20
@Did: It would be correct because c is the upper bound of b. I could have worded my answer better – ndh Oct 26 '13 at 09:31
No it would not. Yes you could have. – Did Oct 26 '13 at 10:11

Let $a_n$ and $b_n$ be sequences of real numbers. If $b_n$ is bounded and $\lim_{n \to \infty} a_n = 0$, then $\lim_{n\to\infty} a_{n}b_{n} = 0$

4 Answers4