Proving convergence of $a_{n} =1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$ by cauchy criterion

Question

Here is what I've tried can someone tell me if i did it right,or help me fix it if wrong. by cauchy criterion : $$ \begin{align}|a_{m+n} -a_n|&=\frac{1}{n+1} + \frac{1}{n+n} +\cdots + \frac{1}{n+m}\\ &\leq \frac{1}{n} +\frac{1}{n+1} +\cdots + \frac{1}{n+m-1}\\ &= \frac{1}{n} + (\frac{1}{n} - \frac{1}{n+1})+ (\frac{1}{n+1} - \frac{1}{n+2})+\cdots + (\frac{1}{n+m-2} - \frac{1}{n+m-1})\\ &= \frac{2}{n} - \frac{1}{n+m-1}\\ &< \frac{2}{n}. \end{align}$$

Let $\epsilon>0$ and let $N=\frac{2}\epsilon$ ,$\forall n\geq N$ we have $$|a_{n+m} -a_n|<\frac{2}{n}\leq \frac{2}{N} < \epsilon.$$ Did i succeed ?

Answer 1

You cannot succeed since the sequence is not convergent (Google harmonic series.) The third inequality is incorrect. You are using $$ \frac{1}{n+k}=\frac{1}{n+k-1}-\frac{1}{n+k}, $$ which is clearly false.

Answer 2

There're a couple of errors in your solution.

1. $|a_{m+1}-a_n| = \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{m+1} $
2. The 2nd line doesn't equal the 3rd line.
3. The series doesn't converge anyway. In fact, the series $\sum_{n=1}^{\infty}\frac{1}{n^p} $ converges if and only if $p>1.$