Let $n,m \in \mathbb{N}$, $n>m,$ $\varepsilon>0$ then prove that exist some $N\in \mathbb{N}$ such that if $n>m>N ,\dfrac{n-m}{(m+1)^2}< \varepsilon$

Question

I'm trying to prove that a sequence is a Cauchy sequence, and I got stuck, in this part:

$d(x_n,x_m)\leq\cdots\leq \dfrac{1}{(m+1)^2}+\cdots+\dfrac{1}{n^2}\leq\dfrac{n-m}{(m+1)^2}$.

In this part i did the following, $\dfrac{n-m}{(m+1)^2}<\dfrac{n}{m}$, then by the Archimedean property exist $N \in \mathbb{N}$ such that $N>\dfrac{n}{\varepsilon}$, and for $n>m>N$ it follows.

Is this a good way to go?

Answer 1

hint

Your argument is not valid since the $ N $ you find depends on $ n $.

You could observe that $$\forall k\in\{m+1,m+2,...,n\}$$ $$\frac{1}{k^2}\le \frac{1}{k-1}-\frac 1k$$ and telescoping.