# limit of product exists and one limit exists

Question is to check :

If $\lim_{n\rightarrow \infty}a_nb_n$ exists and $\lim_{n\rightarrow \infty}a_n$ exists implies $\lim_{n\rightarrow \infty}b_n$ exists.

Considering $a_n=\frac{1}{n}$ and $b_n=n$ then we see that $\lim_{n\rightarrow \infty}a_nb_n$ exists, equals to $1$ and $\lim_{n\rightarrow \infty}a_n$ exists and equals to $0$. In this case $\lim_{n\rightarrow \infty}b_n$ does not exists..

So, the answer to the question is Not always..

Now, what if $\lim_{n\rightarrow \infty}a_n$ exists and is non zero and $(b_n)$ is bounded?

Suppose that $\lim_{n\rightarrow \infty}a_nb_n=M$ with $\lim_{n\rightarrow \infty}a_n=P\neq 0$ and $|b_n|\leq A$ for all $n\in \mathbb{N}$.

I claim that $\lim_{n\rightarrow \infty}b_n=\frac{M}{P}$

Consider $|b_n-\frac{M}{P}|$.. We estimate this. Given $\epsilon>0$ there exists $N\in \mathbb{N}$ such that $|a_nb_n-M|<\epsilon$ and $|a_n-P|<\epsilon$ for all $n\geq N$.

$$|b_n-\frac{M}{P}|=\frac{1}{P}|Pb_n-M|=\frac{1}{P}|Pb_n-a_nb_n+a_nb_n-M|\leq \frac{1}{P}|b_n||a_n-P|+\frac{1}{P} \epsilon$$ As $(b_n)$ is bounded, we have for all $n\geq N$ $$|b_n-\frac{M}{P}|\leq \frac{1}{P}A\epsilon+\frac{1}{P} \epsilon=\epsilon\left(\frac{1}{P}(A+1)\right)$$

Thus, we are done.

I am just wondering if i can relax any of the conditions that i have assumed. Help me to know more about this.

• You have a typo where you said $g_n$ instead of $b_n$. Otherwise I don't think you can relax any assumption. Apr 10 '16 at 8:46
• Can you give a counter example if you relax the bounded condition? Apr 10 '16 at 9:11
• In the example you gave after "Considering...", we have that $\;\lim b_n\;$ does exist. Not finitely but it surely exists. If you want a simple example of a non-existing limit in this case, take $\;a_n=\frac1n\;,\;\;b_n=(-1)^n\;$ Apr 10 '16 at 9:25
• @PatrickStevens : Edited. Thanks
– user312648
Apr 10 '16 at 10:01
• @Kyson : As the answers below say that boundedness comes immediately, there is no question of non bounded sequences...
– user312648
Apr 10 '16 at 10:02

If

$$\;\lim_{n\to\infty}a_n=L\neq0\;,\;\;\lim_{n\to\infty}a_nb_n= K\;,\;\;\text{then since for almost all indexes}\;\;a_n\neq0\,,$$

we get that for all indexes except a finite number of them, from arithmetic of limits:

$$b_n=\frac{a_nb_n}{a_n}\xrightarrow[n\to\infty]{}\frac KL$$

and all this is well-defined and always finite since $\;L\neq0\;$ . No need to require a priori boundedness for $\;\{b_n\}\;$ .

• I do not have to assume it is bounded as it is already bounded... THanks..
– user312648
Apr 10 '16 at 10:11
• @cello Exactly, you get boundedness for free. It was a pleasure. Apr 10 '16 at 10:14

You get the condition that $(b_n)$ is bounded for free. Let us first show that by contradiction. Assume that $(b_n)$ is unbounded. Then there exists subsequence $(b_{p(n)})$ of $(b_n)$ such that $|b_{p(n)}|>n$, for all $n$. But, then we have $$|na_{p(n)} |\leq |a_{p(n)}b_{p(n)}| \leq M$$ where $M$ is such that $|a_nb_n|\leq M$, which exists by convergence of $(a_nb_n)$. It follows that $$0\leq |a_{p(n)}| \leq \frac M n \implies \lim_na_n = \lim_n a_{p(n)} = 0$$ Contradiction.

On the other hand, you could easily prove that $\lim_na_n\neq 0$ implies convergence of $(b_n)$ just by noting the general rule: $$\lim_nb_n\neq 0\implies\lim_n\left(\frac{a_n}{b_n}\right) = \frac{\lim_na_n}{\lim_nb_n}$$ for convergent sequences $(a_n)$ and $(b_n)$.

• This was useful.. Thaks
– user312648
Apr 10 '16 at 10:03
1. If $\lim_{n}a_{n}=0$, the conclusion is not correct. Take as a counter example: $$a_{n}=1/n$$ $$b_{n}=\sin(n\pi).$$ By squeeze theorem $a_{n}b_{n}\rightarrow0$, but $b_{n}$ has no limit.

2. If $\lim_{n}a_{n}\not=0$ Then the result is true. Because, for $n>N$ (ultimately), we have $a_{n}\not=0$ and so we can write $$b_{n}=\frac{1}{a_{n}}.a_{n}b_{n}$$ and since the limit of the right side exists, the left side must have also a limit.