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.

Thanks in advance!

  • $\begingroup$ It is not necessarily true that $a_nb_n=c$ for $n\ge N.$ But you are in the right way. $\endgroup$ – mfl Jun 15 '14 at 17:41
  • $\begingroup$ @mfl, that is a given, an assumption, not a deduction... $\endgroup$ – DonAntonio Jun 15 '14 at 17:42
  • $\begingroup$ We know that $\lim_{n\to \infty} a_nb_n=c,$ not that $a_nb_n=c$ for any $n.$ $\endgroup$ – mfl Jun 15 '14 at 17:45
  • $\begingroup$ Oh, I see @mfl: you meant in the OP's work, not in the question's info. You're right, of course. $\endgroup$ – DonAntonio Jun 15 '14 at 17:54

Without much words:

$$b_n=\frac1{a_n}a_nb_n\stackrel{\text{arithm. of limits}}{\xrightarrow[n\to\infty]{}}\frac1{c_1}\cdot c=\frac c{c_1}$$

  • 1
    $\begingroup$ ah ok, that solves all doubts :D Thanks $\endgroup$ – Thorben Jun 15 '14 at 17:47

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