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.
On
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.
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 $N$.