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...

  • $\begingroup$ You can't start by assuming $b_n$ converges. $\endgroup$ – Julien Feb 4 '13 at 19:33
  • $\begingroup$ You begin assumingwhat you need to prove: that $\,\lim b_n\,$ exists...! This is wrong, of course. $\endgroup$ – DonAntonio Feb 4 '13 at 19:36

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.

  • $\begingroup$ Thanks!! let me try. how do you come up with this? $\endgroup$ – 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 $$b_n=(b_n-a_n)+a_n\to 0+L=L$$


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