How do I prove that $a_n=\sum_{k=1}^{n} \frac{1}{k}$ is not a Cauchy sequence? First, I have to say that it does not make sense to me, because I know for sure that $\sum_{k=1}^{n}\frac{1}{k^2}$ is a Cauchy sequence so naturally in my mind the discussed one "should" converge too. But this is a homework so I guess my intuition is wrong this time.
This is what I have done so far:
There exists $N\in \mathbb{N}$. So,
$$
\begin{aligned}
|a_{n+N}-a_n| =|\sum_{k=1}^{n+N} \frac{1}{k}-\sum_{k=1}^{n} \frac{1}{k}|=
|\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+N}|=\frac{1}{n+1}+\\
\frac{1}{n+2}+...+\frac{1}{n+N} \geq \frac{N}{n+N} \geq \epsilon
\end{aligned}
$$
I probably not skilled enough in algebra to continue from here. 
Edit: I have to find an $\epsilon$ that satisfies this expression.
 A: Good start.  Now observe that $$\frac{1}{n+1}+\cdots+\frac{1}{n+N}\ge \frac{N}{n+N}$$
With appropriate choice of $N$, this will be larger than $0.5$ (say).
A: These are great answers. I will try to write a detailed direct proof.
Pre-proof:
The Cauchy sequence definition: a sequence $a_n$ is a Cauchy sequence if for any $\epsilon>0$ there exists $N\in \mathbb{N}$ such that for any pair $n,m>N$ we have $|a_m-a_n|<\epsilon$.

So to prove that a sequence is not a Cauchy sequence we have to show that there exists an $\epsilon>0$ such that for any $N\in \mathbb{N}$ there exist $n,m>N$ such that $|a_m-a_n|\ge \epsilon$.
The given sequence: $a_n = \sum_{k=1}^n \frac{1}{k}$.
Lets try to find this $\epsilon$:
Without loss of generality, we can assume $m>n$. So, 
$
|a_m-a_n| = |\sum_{k=1}^m \frac{1}{k}-\sum_{k=1}^n \frac{1}{k}|=|\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{m}|
$
$=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{m} \ge (m-n)\frac{1}{m}=\frac{m-n}{m}$
We want a constant $\epsilon$, we can choose $m=2n$ to get $\epsilon=\frac{2n-n}{2n}=\frac{n}{2n}=\frac{1}{2}$(for example).  
Formally:
$|a_m-a_n| = |\sum_{k=1}^m \frac{1}{k}-\sum_{k=1}^n \frac{1}{k}|=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{m}\ge (m-n)\frac{1}{m}=\frac{m-n}{m} $
$\Leftrightarrow |a_{2(N+1)}-a_{N+1}|\ge \frac{2N+2-(N+1)}{2N+2}=\frac{N+1}{2(N+1)}=\frac{1}{2}.$
We define $\epsilon=\frac{1}{2}$, such that for a given $N$ we choose $n=N+1$ and $m=2(N+1)$
$\Rightarrow (a_n)_{n=1}^\infty$ is not a Cauchy sequence. $\Box$ 
A: For every $n$ we have $$a_n:=\sum_{k=1}^n\frac{1}{k}\in\mathbb{R}$$
Suppose the sequence $\{a_n\}_n$ is Cauchy sequence. Since $\mathbb{R}$ is complete (i.e. every Cauchy sequence in $\mathbb{R}$ converges to some element in $\mathbb{R}$) then there exists some $a\in\mathbb{R}$ such that $\lim_na_n=a$. But $$\lim_na_n=\sum^{\infty}_{k=1}\frac{1}{k}=+\infty$$
And $\infty\notin\mathbb{R}$. 
A: Assume that $a_n$ is a Cauchy sequence.
Then for $\epsilon=\frac{1}{2}$ we have that $\exists n_0 \in \Bbb{N}$ such that $$|a_m-a_n|=|\frac{1}{n+1}+....+\frac{1}{m}|<\frac{1}{2},\forall m>n \geq n_0$$
For $m=2n>n\geq n_0$ we have 
$$\frac{1}{2}=n\frac{1}{2n} =\frac{1}{2n}+...+\frac{1}{2n} \leq\frac{1}{n+1}+...+\frac{1}{2n}<\frac{1}{2}$$
Thus $1/2<1/2$ which clearly is a contradiction
