Suppose $|x_n-x_k|\le n/k^2$ for all $n$ and $k$. Show that $\{x_n\}$ is Cauchy.

Question

Suppose $|x_n-x_k|\le n/k^2$ for all $n$ and $k$. Show that $\{x_n\}$ is Cauchy.

Attempt: Find $M\in N$ s.t. for $n,k\ge M, n/k^2<\varepsilon \hspace{0.5cm}\forall\varepsilon>0.$

For given $\varepsilon$, let $M>1/\sqrt \varepsilon.$ Then, for $n,k\ge M$ $$|n/k^2|\le n/M^2<n\varepsilon.$$

I don't know how to deal with $n$ here. Could you give some hint?

Answer 1

Hint: Replace $k$ with $n+k$ to get the proper left-hand side of a Cauchy-like inequality.

Answer 2

If i understood correctly (what i doubt). Let's make $n$ constant. $$|x_n-x_k|\leq\frac{n}{k^2}$$ Holds for all $k$, so we can take a limit with both sides as $k\to \infty$. We get by the squezee theorem that $$\lim\limits_{k\to \infty} x_k = x_n$$ So the sequenence is convergent, so it's Cauchy

Answer 3

$\sum_k|x_n-x_k| < +\infty$

Thus $$\lim_{k \to + \infty}|x_n-x_k| \to 0 \Rightarrow \lim_{n,k \to + \infty}|x_n-x_k|=0$$

So $x_n$ is a Cauchy sequence.

Answer 4

Since $n \ge M$, you can say $\frac{n}{M^2} \le \frac{M}{M^2} = \frac1M$. So define $M > \frac1{\varepsilon}$ instead of $M > \frac1{\sqrt{\varepsilon}}$.