Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(a_n)^\infty_{n=1}$ and $(b_n)^\infty_{n=1}$ be two sequence of real numbers such that $|a_n - b_n|<{1\over{n}}$.

Suppose that $L=\lim_{n\to\infty}a_n$ exists. Show that $(b_n)^\infty_{n=1}$ converges to L also.

My thought:

Let the limit of $(b_n)^\infty_{n=1}=M$ and then show $L=M$ or $L-M = 0$at last.

$L=\lim_{n\to\infty}a_n$ and $M=\lim_{n\to\infty}b_n$

By limit arithmetic,



In order to make use of the inequality give, I squared both sides.


Again by limit arithemetic,



then... Im stuck... I probably did a wrong approach from the very first step...

share|cite|improve this question
You can't start by assuming $b_n$ converges. – 1015 Feb 4 '13 at 19:33
You begin assumingwhat you need to prove: that $\,\lim b_n\,$ exists...! This is wrong, of course. – DonAntonio Feb 4 '13 at 19:36
up vote 2 down vote accepted

Hint: $$ 0\leq |b_n-L|=|b_n-a_n+a_n-L|\leq |a_n-b_n|+|a_n-L|. $$ Now use the squeeze theorem, for instance.

share|cite|improve this answer
Thanks!! let me try. how do you come up with this? – Paul Feb 4 '13 at 19:35

By the squeeze theorem, $\left|a_n-b_n\right|\to 0\iff b_n-a_n\to 0$. But $$b_n=(b_n-a_n)+a_n\to 0+L=L$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.