How do I find an $N$ such that for $n > N$, the following term is less than $\epsilon$?

Question

How do I find an $N$ such that for $n > N$, the following term is less than $\epsilon$?

\begin{align*} \sqrt{\left(\frac{1}{n + 1}\right)^2 + \left(\frac{1}{n + 2}\right)^2} \end{align*}

The model answer lets $N = \left\lceil\frac{\epsilon}{\sqrt{2}} \right\rceil$, but I am not sure how this will help in proving that for any $n > N$ the inequality will hold. Any help will be appreciated.

Answer 1

Since $f(n) =\sqrt{\left(\frac{1}{n + 1}\right)^2 + \left(\frac{1}{n + 2}\right)^2} \lt \sqrt{\left(\frac{1}{n + 1}\right)^2 + \left(\frac{1}{n + 1}\right)^2} =\sqrt{\frac{2}{(n + 1)^2}} =\frac{\sqrt{2}}{n+1}$, to make $f(n) < \epsilon$ it is sufficient to make $\frac{\sqrt{2}}{n+1} \lt \epsilon $ or $n > \frac{\sqrt{2}}{\epsilon}-1 $.

This is not the best possible bound, but it is good enough.

Answer 2

You could be as accurate as you wish since using long division $$\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}=\frac 14 \sum_{k=2}^\infty(-1)^n \frac{\left(2^k+4\right) (k-1)}{n^k}$$ Using power series reversion $$n=\frac{\sqrt{2}}{\epsilon }-\frac{3}{2}+\frac{3 \epsilon }{8 \sqrt{2}}-\frac{25 \epsilon ^3}{256 \sqrt{2}}+O\left(\epsilon ^5\right)$$ gives tight bounds for $n$