<s>[INST] Show that $a_n = 1 + \frac{1}{2} + \frac{1}{3} +\dotsb+ \frac{1}{n}$ is not a Cauchy sequence 
Let $$
a_n = 1 + \frac{1}{2} + \frac{1}{3} + \dotsb + \frac{1}{n} 
\quad (n \in \mathbb{N}).
$$
  Show that $a_n$ is not a Cauchy sequence even though 
  $$
\lim_{n \to \infty} a_{n+1}  - a_n = 0
$$
  (Therefore $a_n$ does not have a limit).

 [/INST] We use the definition of Cauchy sequence to show the sequence $(a_n)$ is not Cauchy.
Let $\epsilon=1/2$. We will show that there does not exist an $m$ such that for any $n\gt m$, we have  $|a_n-a_m|\lt \epsilon$. 
For let $m$ be given, and let $2^k$ be the smallest power of $2$ that is $\gt m$. Let $n=2^{k+1}-1$. Then
$$a_n-a_m\ge \frac{1}{2^k}+\frac{1}{2^{k}+1}+\cdots +\frac{1}{2^{k+1}-1}\gt \frac{1}{2}.$$
The fact that the above sum is $\gt \frac{1}{2}$ follows from the fact that the sum has at least $2^k$ terms, each $\gt \frac{1}{2^{k+1}}$
The fact that $\lim_{n\to \infty}(a_{n+1}-a_n)=0$ is clear, since $a_{n+1}-a_n=\frac{1}{n+1}$.
</s>