Prove, using the definition, that $(2 + \frac{1}{n^2})$ is a Cauchy sequence.

Question

"Prove, using the definition, that $(a_n)_{n \in \mathbb{N}} = (2 + \frac{1}{n^2})$ is a Cauchy sequence."

My answer:

Let $\epsilon > 0$ and choose $N = \frac{1}{\epsilon}$. Then for all $n, m \in \mathbb{N}$ such that $n, m > N$, one has that

|$a_n - a_m$| = |$\frac{1}{n^2} - \frac{1}{m^2}$| < |$\frac{1}{n^2}$| < $\frac{1}{N^2} < \frac{1}{N}$ = $\epsilon$

However, the actual solution chooses $N = \sqrt{\frac{2}{e}}$

I can see how $N = \sqrt{\frac{2}{e}}$ would work. However my question is - does my answer work too, or have I done something wrong in my working?

Answer 1

So technically it doesn't have to be less than $\frac{1}{n^2}$, since if $n$ is really big, $\frac{1}{n^2}$ will be basically 0, and $$\left|\frac{1}{n^2}-\frac{1}{m^2}\right|\approx \left|-\frac{1}{m^2}\right|\not < \left| \frac{1}{n^2}\right| \approx 0$$

But you can assume WLOG that $n\le m$, and then your reasoning goes through.