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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.

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

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    $\begingroup$ Or even $\frac{1}{n^2+1} < \frac{1}{n^2} \le \frac{1}{n}$ $\endgroup$
    – lhf
    Jun 14 '18 at 13:47
  • $\begingroup$ Existing of a fixed $N$ between two variable makes a problem here! $\endgroup$
    – Nosrati
    Jun 14 '18 at 13:53
  • $\begingroup$ @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? $\endgroup$
    – user328442
    Jun 14 '18 at 14:21
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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.

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