How to prove that a sequence is Cauchy How do I show if $x_n = \frac{n^2}{n^2 -1/2}$ is a Cauchy sequence? (using the definition of Cauchy sequence)
My attempt: A sequence is Cauchy if $ \forall \epsilon>0$ $ \exists N \in \mathbb N$ $\forall m,n \geq N$ :|$x_n -x_m$|$\leq \epsilon$
|$x_n -x_m|=|\frac{n^2}{n^2 -1/2} -\frac{m^2}{m^2 -1/2}$| $\leq|\frac{n^2}{n^2 -1/2}-1|+|1-\frac{m^2}{m^2 -1/2}|= 1/2|\frac{1}{n^2 -1/2}|+1/2|\frac{1}{m^2 -1/2}|$
$1/2|\frac{1}{n^2 -1/2}| \leq \epsilon$ and also $1/2|\frac{1}{m^2 -1/2}| \leq \epsilon$
So $1/2|\frac{1}{n^2 -1/2}|+1/2|\frac{1}{m^2 -1/2}| \leq 2 \epsilon$
if we choose $N \in \mathbb N$ such that $N >\sqrt{\frac{\epsilon}{2}+\frac{1}{2}}$

The definition you posted is correct. Note though that
$$
x_n
 = \frac{n^2}{n^2-1/2}
 = 1 + \frac{1/2}{n^2-1/2},
$$
and it's clear this is a sequence which decreases to $1$, so $x_{n+1} < x_n$ for all $n$. (You could at this point claim that since the sequence is convergent, it is also Cauchy, but if you need a proof from the fundamentals, read on).
Assuming $N<m<n$,
$$
\begin{split}
\left|x_m - x_n\right|
 &= x_m - x_n \\
 &= \left(1 + \frac{1/2}{m^2-1/2} - 1 - \frac{1/2}{n^2-1/2}\right) \\
 &= \frac12 \left(\frac{1}{m^2-1/2} - \frac{1}{n^2-1/2}\right) \\
 &= \frac12
    \left(\frac{n^2 - m^2}
               {\left(m^2-1/2\right)\left(n^2-1/2\right)}\right) \\
 &\le \frac12
    \left(\frac{n^2}
               {\left(m^2-1/2\right)\left(n^2-1/2\right)}\right) \\
 &\le \frac{1/2}{m^2-1/2} \left(\frac{n^2}{n^2-1/2}\right)\\
 &\le \frac{1/2}{m^2-1/2} \left(1 + \frac{1/2}{n^2-1/2}\right)\\
 &\le \frac{1}{m^2-1/2}\\
 &\le \frac{1}{N^2}.
\end{split}
$$
Can you find what $N$ you need to pick in terms of $\epsilon$ to have that expression $< \epsilon$ in the end?
