# Proving limit using Epsilon-N definition

So I'm trying to prove the limit of $a_{n} = 1/(n^{2}+1)$

Where: $$\lim_{n\rightarrow\infty}\left(\frac{1}{n^{2}+1}\right)=0$$

My solution so far:

$\forall\epsilon>0\ \ \exists N\in\mathbb{N} \ \ s.t. |a_{n}-0|<\epsilon \ \forall \ n \geq N$

Which implies that:

$|\left(\frac{1}{n^{2}+1}\right)|<\epsilon$

Taking the reciprocal:

$|n^{2}+1|> \epsilon^{-1}$

This is where I've gotten to, and I can't seem to evaluate the inequality for a suitable N.

Any help would be appreciated.

• Algebra: $$|n^2+1| > \epsilon^{-1} \implies n > \sqrt{1- \frac{1}{\epsilon}}.$$ So what should you choose for $N$...? Jun 14 '18 at 13:23
• You should have $|n^2+1| > \frac{1}{\epsilon}$? Jun 14 '18 at 13:24
• @Dzoooks where $\epsilon>1$. Jun 14 '18 at 13:29
• @mathphys Yeah I realised I made an error there now. Jun 14 '18 at 13:32
• @Dzoooks Thank you, that makes sense! Jun 14 '18 at 13:33

Observation: $\frac{1}{n^2+1} < \frac{1}{n^2}.$ So, if we set $$\frac{1}{n^2} < \epsilon$$ then it is clear that a suitable $N$ (in terms of $\epsilon$) would satisfy $$N > \frac{1}{\sqrt{\epsilon}}.$$

Indeed, let $\epsilon > 0$ and choose $N \in \mathbb{N}$ such that $$N > \frac{1}{\sqrt{\epsilon}}.$$ Then for any $n \geq N$, we have $$\frac{1}{n^2+1} < \frac{1}{n^2} \leq \frac{1}{N^2} < \epsilon$$ and the result follows.

Note 1: working with something similar but “easier” is a useful trick. $\frac{1}{n^2}$ is “easier” to deal with than $\frac{1}{n^2+1}.$

Note 2: As pointed out in the comments, choosing $N > \frac{1}{\epsilon}$ is also an option since $$\frac{1}{n^2+1} < \frac{1}{n^2} \leq \frac{1}{n} \leq \frac{1}{N} < \epsilon.$$ There are many ways of picking a suitable $N$.

• Or even $\frac{1}{n^2+1} < \frac{1}{n^2} \le \frac{1}{n}$
– lhf
Jun 14 '18 at 13:47
• Existing of a fixed $N$ between two variable makes a problem here! Jun 14 '18 at 13:53
• @user108128 I don’t follow. The goal is to pick a suitable $N$ as a response to a given $\epsilon.$ Where do I fall short on that? Jun 14 '18 at 14:21

You're going backwards. Fix $\varepsilon>0$; then you must find $N$ such that, for $n>N$, $$\left|\frac{1}{n^2+1}\right|<\varepsilon$$ This is equivalent to $$n^2>\frac{1}{\varepsilon}-1$$ The $N$ you look for is $0$ if $\varepsilon^{-1}-1<0$, it is any integer such that $N\ge\sqrt{\varepsilon^{-1}-1}$ otherwise.