Set $s_n$ to be a sequence so that $|s_{n+1}-s_n|\le\frac{1}{n^p}$ for all $n \in \mathbb{N}$

Question

If $p>1$ prove $s_n$ is convergent

Following the formula, set $\epsilon>0$.

$$|s_m-s_n|=|s_m-s_{m-1}+s_{m-1}-s_{m-2}+...+s_{n+1}+s_n|\le|=|s_m-s_{m-1}|+|s_{m-1}-s_{m-2}|+\cdots+|s_{n+1}+s_n|\le\frac{1}{(m-1)^p}+\frac{1}{(m-2)^p}+\cdots+\frac{1}{n^p}$$

From this point, I could be taking this in the wrong direction, but I get:

above $\lt\dfrac{(m-1-n+1)}{n^p}\lt\dfrac{m}{n^p}$. This needs to be $\lt\epsilon$, but I'm not sure how...

I can pick $\dfrac{1}{N^p}=\dfrac{\epsilon}{m}$. Then $n>N$ implies above$\lt\dfrac{m}{N^p}=\epsilon$, which doesn't work, because $N$ can't depend on $m$. Help me get out of this circular reasoning!

Answer 1

Hint: Let $S_{N}=\sum_{q=1}^{N-1}\frac{1}{q^{p}}$. You proved

$$\left|s_{m}-s_{n}\right|\leq \left|S_{m}-S_{n}\right|$$

What can you say about the sequence $S_{N}$ for $p>1$?