0
$\begingroup$

Suppose I have two sequences, $a_n$ and $b_n$. I know that:

$\lim_{n\to\infty} a_n=1$ and that $\lim_{n\to\infty} a_nb_n=c$.

Does this mean that $\lim_{n\to\infty} b_n$ converges?

If so, by algebra of limits does it mean that $\lim_{n\to\infty} b_n=c$?

$\endgroup$
2
  • $\begingroup$ Yes, you are right. $\endgroup$ Commented Nov 6, 2013 at 0:43
  • $\begingroup$ Any ideas on a proof? $\endgroup$
    – adrug
    Commented Nov 6, 2013 at 0:44

1 Answer 1

0
$\begingroup$

Yes it does. We have that $$\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{a_nb_n}{a_n}=\frac{\lim_{n\rightarrow \infty}a_nb_n}{\lim_{n\rightarrow \infty}a_n}=\frac{c}{1}=c,$$ since the limit of a quotient is the quotient of the limits if they exist and the last is not $0$.

$\endgroup$
2
  • $\begingroup$ Thanks, this works. Last question here - what is the Tex syntax you used for the n->inf UNDER the limit, not next to it? It looks so much clearer! $\endgroup$
    – adrug
    Commented Nov 6, 2013 at 0:49
  • $\begingroup$ The same you used, but when you have a separated equation, the subscript for the $\lim$ is under it. $\endgroup$ Commented Nov 6, 2013 at 0:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .