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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.

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Hint: $|b_n - L| \leqslant |b_n - a_n| + |a_n - L|$

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  • $\begingroup$ Ahhhh Triangle Inequality. I can see what to do now! Thanks. $\endgroup$ – Eric Hansen Sep 24 '15 at 16:20
  • $\begingroup$ @Eric Hansen: You're welcome. $\endgroup$ – RRL Sep 24 '15 at 16:21
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You can rewrite your hypothesis as $$-\frac{1}{n} < a_{n} - b_{n} < \frac{1}{n}$$ and deduce $$a_{n} -\frac{1}{n} < b_{n} < a_{n} + \frac{1}{n}$$

Then you can apply squeeze theorem.

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  • $\begingroup$ I don't see why this approach doesn't work too, actually. $\endgroup$ – Eric Hansen Sep 24 '15 at 16:22

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