# Using Squeeze Theorem to prove two sequences converge to same limit

In a Real Analysis book without solutions, I came across the question

Let $(a_{n})_{n=1}^{\infty}$ and $(b_{n})_{n=1}^{\infty}$ be two sequences of real numbers such that $|a_{n} - b_{n}| < \frac{1}{n}$. Suppose that $L = \lim\limits_{n \rightarrow \infty} a_{n}$ exists. Show that $(b_{n})_{n=1}^{\infty}$ converges to $L$ also.

My attempt:

We claim that for the sequence $\{c_{n}\} = \frac{1}{n}$, the limits as $n \rightarrow \infty$ is $0$. To show this, it is simple to choose $N > \frac{1}{\epsilon}$ and use epsilon-delta definition of a limit.

Now this is where I'm stuck. It's super intuitively easy to see that this statement is true and $b_{n}$ also converges to $L$ because if not then $\{c_{n}\}$ cannot converge to $0$, but I don't know how to describe/show this formally.

Hint: $|b_n - L| \leqslant |b_n - a_n| + |a_n - L|$
You can rewrite your hypothesis as $$-\frac{1}{n} < a_{n} - b_{n} < \frac{1}{n}$$ and deduce $$a_{n} -\frac{1}{n} < b_{n} < a_{n} + \frac{1}{n}$$