# Question about limit of a product of two real sequences

Let $(a_n)_{n\in\mathbb N},(b_n)_{n\in\mathbb N}\subseteq\mathbb R$. Moreover assume that $\lim_{n\to \infty} a_n=c_1\in\mathbb R$ with $c_1\neq 0$ and that $\lim_{n\to \infty} a_nb_n=c$ for some $c\in\mathbb R$.

Then I was asking myself if under the upper assumptions one can conlude that there exists some $c_0\in \mathbb R$ such that $b_n$ converges to $c_0$.

I composed a proof to verify the statement.

The proof is,

Since $c_1\neq 0$ we can wlog assume that there exists some $N\in\mathbb N$ such that $\forall n\geq N:\; a_m>0$. Hence for $m\geq N$ we have $a_nb_n=c\quad\Leftrightarrow \quad b_n=\frac{c}{a_n}$ and consequently $\lim_{n\to \infty}b_n=\lim_{n\to \infty}\frac{c}{a_n}=\frac{c}{c_1}$ i.e $b_n$ is convergent.

Since I somehow can not remember that I saw this statement in my basic Analysis courses I somehow doubt that it is true... Is there a mistake in my proof? I Also doubt that it is true because if someone relaxes the assumption that $c_1\neq 0$ and allows $c_1$ to be zero then for every bounded sequence $b_n$ we have $a_nb_n=0$ and in particular this holds also for not converging bounded sequences.

• It is not necessarily true that $a_nb_n=c$ for $n\ge N.$ But you are in the right way. – mfl Jun 15 '14 at 17:41
• We know that $\lim_{n\to \infty} a_nb_n=c,$ not that $a_nb_n=c$ for any $n.$ – mfl Jun 15 '14 at 17:45
$$b_n=\frac1{a_n}a_nb_n\stackrel{\text{arithm. of limits}}{\xrightarrow[n\to\infty]{}}\frac1{c_1}\cdot c=\frac c{c_1}$$