0
$\begingroup$

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,

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

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

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

$(\lim_{n\to\infty}a_n-b_n)^2=(L-M)^2$

Again by limit arithemetic,

$\lim_{n\to\infty}(a_n-b_n)^2=(L-M)^2$

$\lim_{n\to\infty}(|a_n-b_n|)^2=(L-M)^2$

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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks!! let me try. how do you come up with this? $\endgroup$ – Paul Feb 4 '13 at 19:35
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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