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.

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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...