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...