<s>[INST] Show that the sequence $a_n=1+\frac1{\sqrt{2}} +\frac1{\sqrt{3}} + \dots + \frac1{\sqrt{n}}$ is not Cauchy 
Show directly (from the definition) that if
$$a_n=1+\frac{1}{\sqrt{2}} +\frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}}\;,$$
then $(a_n)$ is not a Cauchy sequence.

Attempt
Using $a_n$ is not Cauchy if
$$\exists\, \varepsilon > 0 : \forall N\, \exists\, n, m > N | a_{n} - a_{m}| \geq \varepsilon$$
taking $n>m$.
$$\begin{align}|a_n - b_m| = \left|\sum_{k=1}^{n} \frac{1}{\sqrt{k}} - \sum_{l=1}^{m} \frac{1}{\sqrt{l}}\right| &= \left|\sum_{k=m+1}^{n} \frac{1}{\sqrt{k}}\right| \ge \left|\sum_{k=m+1}^{n} \frac{1}{n}\right| = \frac{n-m}{n}\\  \end{align} $$
Letting $n = 2m$
$$\frac{n-m}{n} = \frac{2m-m}{2m} = \frac{m}{2m} = \frac{1}{2}$$
Thus
$$\exists\, \varepsilon = \frac{1}{2} : \forall N\, \exists\, n=2m > m > N :  \left|\sum_{k=m+1}^{n} \frac{1}{\sqrt{k}}\right| \geq \frac{1}{2}$$
so $a_n$ is not Cauchy.
 [/INST] Another proof:
If
$a_n
=\sum_{k=1}^n \frac1{\sqrt{k}}
$,
$a_{n^2}-a_n
=\sum_{k=n+1}^{n^2} \frac1{\sqrt{k}}
>\sum_{k=n+1}^{n^2} \frac1{\sqrt{n^2}}
=(n^2-n)\frac1{n}
=n-1
$.
</s>