I just learned about the Cauchy Criterion for convergence of a sequence:
Theorem: The sequence $\{ s_n \}^\infty_{n=1}$ converges if the difference $s_n-s_m$ approaches zero as both $m$ and $n$ approach infinity.
I tried to use this to test for the convergence of $$\{ s_n \}^\infty_{n=1}=\{\frac{\ln n}{n^p} \}^\infty_{n=1}$$ where $p$ is a positive number.
Following the theorem, I wrote $$\lim_{n\rightarrow\infty}\frac{\ln n}{n^p}-\lim_{m\rightarrow\infty} \frac{\ln m}{m^p}=0,$$ hence the sequence is convergent.
I realised that I am doing something wrong, since I'm subtracting the same things here. What is the correct way to do this?
One way to prove that the sequence converges is $$\lim_{n\to \infty} \frac{\ln n}{n^p} = \lim_{n\to\infty} \frac{1/n}{pn^{p-1}} =\lim_{n\to\infty} \frac{1}{pn^p} =0.$$
To use the Cauchy criterion, first suppose $p > 1$ and we fix $\epsilon > 0$ and choose $N$ large such that $N^{p-1} > 2/\epsilon$ and have $n\geq m > N$ so that $$\left\vert \frac{\ln n }{n^p} - \frac{\ln m}{m^p} \right \vert \leq \frac{\ln n }{n^p} + \frac{\ln m}{m^p} \leq \frac{n}{n^p} + \frac{m}{m^p} = \frac{1}{n^{p-1}} + \frac{1}{m^{p-1}} \leq \frac{2}{N^{p-1}} < \epsilon.$$
For $p \leq 1$, choose a $q$ such that $p \geq 1/q$. Now fix $\epsilon > 0$ and $N$ large such that $N^{p-1/q} > 2q/\epsilon$ and have $n\geq m > N$ so that $$\left\vert \frac{\ln n }{n^p} - \frac{\ln m}{m^p} \right \vert \leq \frac{\ln n }{n^p} + \frac{\ln m}{m^p} = \frac{q \ln n^{1/q}}{n^p} + \frac{q\ln m^{1/q}}{m^p} \leq \frac{qn^{1/q}}{n^p} + \frac{qm^{1/q}}{m^p} \leq \frac{2q}{N^{p-1/q}} < \epsilon.$$
Hence $\{\frac{\ln(n)}{n^p}\}$ is a Cauchy sequence.
Note that the second case doesn't really require the $p\leq 1$. We could remove this and realize the first case is a special case of the second argument where $q=1$.