Is this sequence Cauchy?
$$\left\{\frac{1}{n^{2}}\right\}$$
My attempt: Suppose that converges as it goes to $0$ and is therefore Cauchy, but I lack formality in my reply.
Is this sequence Cauchy?
$$\left\{\frac{1}{n^{2}}\right\}$$
My attempt: Suppose that converges as it goes to $0$ and is therefore Cauchy, but I lack formality in my reply.
In order to be Cauchy, it must be the case that for all $\epsilon\gt 0$ there exists $N\gt 0$ such that, for all $n,m\geq N$, we have $$\left|\frac{1}{n^2}-\frac{1}{m^2}\right|\lt \epsilon.$$ Let us assume without loss of generality that $n\geq m$. Then $$\left|\frac{1}{n^2}-\frac{1}{m^2}\right| = \frac{1}{n^2}-\frac{1}{m^2} \lt \frac{1}{n^2}.$$ If we can ensure that $\frac{1}{n^2}$ is small enough, provided $n$ is large enough, then that would suffice. Can we?
Added. Well, if we want $\frac{1}{n^2}\lt \epsilon$, then, since both $\epsilon$ and $n$ are positive, we need $$\frac{1}{\epsilon}\lt n^2,$$ which means we need $$\frac{1}{\sqrt{\epsilon}}\lt n.$$ So... what is a good $N$ to pick so that, if $n\geq N$, then $\frac{1}{n^2}\lt\epsilon$?
(What is behind this particular estimate is that: (i) every convergent sequence is Cauchy; and (ii) this sequence converges to $0$)