Show that $\{x_n\}_{n=1}^{\infty}$ is cauchy.

Question

Question: Suppose $|x_n - x_k| \le n/k^2$ for all $n$ and $k$.Show that $\{x_n\}_{n=1}^{\infty}$ is cauchy.

Attempt : To prove this, I have to find $M \in N$ that for $\varepsilon >0$, $n/k^2 < \varepsilon$ for $n,k \ge M$.

Let $\varepsilon > 1/M$.

Then, $n/k^2 \le M/M^2$ (#) $= 1/M < \varepsilon$ for $n,k \ge M$.

I feel (#) is not necessarily true. Is there any way to show (#) is correct? or could you give me some any hint regarding this question?

Answer 1

No, $(\#)$ is not necessarily true. $M=1$, $n=5$, $k=2$ provides a counterexample.

Note that since $|x_n-x_k|=|x_k-x_n|$, we have $|x_n-x_k|\leq\min\{n/k^2,k/n^2\}$. Given $\varepsilon>0$, choose $M\in\mathbb N$ such that $\varepsilon>1/M$. Suppose $n,k\geq M$ and $k\geq n$. Then we have $$|x_n-x_k|\leq\frac{n}{k^2}\leq\frac{1}{k}\leq\frac{1}{M}<\varepsilon.$$

Answer 2

Hint:

With $n$ fixed, the map $k\mapsto\dfrac n{k^2}$ is strictly decreasing with limit $0$ as $k\to\infty$. This means, given $\epsilon\gt 0$, for a fixed $n$, there exists $N\in\Bbb N$ such that $\dfrac n{k^2}\lt\epsilon~\forall~k\geq N$

Answer 3

To prove a sequence is Cauchy we have to study the behavior of terms of the form $|x_n-x_k|$, where $n, k$ are greater than some positive integer $N$.

Let's fix some $N$ to get a better feeling for what is happening. Say $N=10$, and we concerned ourselves with terms of the form $|x_n-x_k|$, where $n, k \ge 10$. The question is, what's the supremum of all possible values of these terms? Well, if we fix $n=10$, then $|x_{10}-x_k| \le \dfrac {10}{k^2}$, so the largest value in this case is at most $\dfrac {10}{11^2} = \dfrac {10}{121} \approx 0.0826$. What if we move $n$ up to $11$? Then $|x_{11}-x_k| \le \dfrac {11}{k^2}$, so the largest value of $|x_{11}-x_k|$ is $\le \dfrac {11}{12^2} = \dfrac {11}{144} \approx 0.0764$, which is samaller than $0.0826$. Now the question is, will the pattern hold? If we fix $n=12$, will our new value be less than $ 0.0764$? You can check that it will be. So when we fix $N=10$, everything seems to be smaller than the original $n=10, k=11$ case.

Can you try to generalize it from here?