# 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 $$N >\sqrt{\frac{\epsilon}{2}+\frac{1}{2}}$$

Easier:

$$x_n = \frac{1}{1-\frac{1}{2n^2}} \stackrel{n \to \infty}\longrightarrow 1$$ and convergent sequences are Cauchy sequences.

Your penultimate line is unnecessary. You actually want $$N>\sqrt{\frac{1}{2\epsilon}+\frac12}$$ to ensure $$|x_n-x_m|<2\epsilon$$. Usually, we'd instead prove $$|x_n-x_m|<\epsilon$$ is achievable, with $$\min\{m,\,n\}>\sqrt{\frac{1}{\epsilon}+\frac12}$$.

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 $$N, $$\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?

• so we need $N> \sqrt{1/ \epsilon}$ right?
– Bob
Jul 1, 2020 at 20:39
• @Bob correct :) Jul 2, 2020 at 14:09

In this case, you can easily conclude that it is Cauchy because it converges for $$n \rightarrow \infty$$