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).

 A: 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 $a_{n+1}-a_n=\frac{1}{n+1}$.
A: Hint: Showing it doesn't converge (specifically that it goes to infinity) would help.
A: Starting from $e^x\ge 1+x$ and thus also $e^{-x}\ge 1-x$ one gets
$$
-\ln(1-x)\ge x \ge \ln(1+x)
$$
Now set $x=\frac1n$ to get
$$
\ln(n)-\ln(n-1)\ge\frac1n\ge\ln(n+1)-\ln(n)
$$
The left and right terms are telescoping in sums, so
$$
\ln(n)-\ln(m)\ge\sum_{k=m+1}^n\frac1k\ge\ln(n+1)-\ln(m+1)
$$
Since $\ln(n)\to\infty$, this shows that the sequence of partial sums can not be Cauchy.
