Prove that for every positive real number $\epsilon$, there exists a natural number $N$ such that if $n > N$, then $\frac{1}{n^2+1} < \epsilon$. Initially, I had:

Let $N$ be an arbitrary natural number and $n > N$. Assume $N < \sqrt{\frac{1}{\epsilon}-1}$. Since $n > N$, $n > \sqrt{\frac{1}{\epsilon}-1}$. Therefore, $ n > \sqrt{\frac{1}{\epsilon}-1}$, then $ n^2 > \frac{1}{\epsilon}-1$ , then $n^2 + 1 > \frac{1}{\epsilon}$, then $\epsilon > \frac{1}{n^2}+1$.

Obviously, this is wrong. My professor wrote comments that N should not be arbitrary, and an X next to my assumption of $N < \sqrt{\frac{1}{\epsilon}-1}$. However, I can't think of many other ways to write this?
Currenty, I have:

Let $\epsilon$ be an arbitrary positive real number. Choose $N > \sqrt{\frac{1}{\epsilon}-1}$. Assume $n > N$. Since $n > N$, $\sqrt{\frac{1}{\epsilon}-1} < N < n$. Therefore, $ n > \sqrt{\frac{1}{\epsilon}-1}$, then $ n^2 > \frac{1}{\epsilon}-1$ , then $n^2 + 1 > \frac{1}{\epsilon}$, then $\epsilon > \frac{1}{n^2+1}$.

Does this new version look right?

Thing is, you are asked to choose a specific $N$. So instead of taking any $N>\sqrt{\frac{1}{\epsilon}-1}$, choose a specific one, for example $N=\lfloor \sqrt{\frac{1}{\epsilon}-1} \rfloor+1$, where $\lfloor \ \ \rfloor$ is the floor function.
Also, there is one annoying, but existing problem. Note that if $\epsilon$ is really large then $\frac{1}{\epsilon}-1$ is negative, and so its square root is not even defined. Not sure I would take off points for this (probably I would, but not much), but you should be aware of this problem. So to make it a perfect answer you should add that if $\epsilon>1$ then you can choose $N=1$ for it. If $\epsilon<1$ then just as you did.
