Let $x_n = \sum_{ j=1}^n \frac1j$. Show that for every $k$ we have $\lim _{n→∞} |xn+k −xn| = 0$, yet {$x_n$} is not Cauchy. The problem goes 
Let $x_n = \sum_{j=1}^{n} \frac{1}{j}$. Show that for every k we have $\lim_{n \to \infty}|x_{n+k}-x_n|=0$, yet {$x_n$} is not Cauchy.
My proof for this goes: By Cauchy Condensation, $x_n = \sum_{j=1}^{n} \frac{1}{j}$ converges iff $\sum{1}=n$ converges. So {$x_n$} does not converge since {$n$} does not converge, which implies that {$x_n$} is not Cauchy.
Is my wrong?
 A: A direct proof
is quite straightforward.
If
$x_n 
= \sum_{j=1}^{n} \frac{1}{j}
$
then
$x_{n+k}-x_n
= \sum_{j=n+1}^{n+k} \frac{1}{j}
\lt \sum_{j=n+1}^{n+k} \frac{1}{n}
=\dfrac{k}{n}
\to 0
$
as
$n \to \infty$.
To show that
$x_n$ is not Cauchy,
do the traditional thing
and choose
$m \ge 2n$.
Then
$x_m-x_n
=\sum_{j=n+1}^{m} \dfrac1{j}
\ge\sum_{j=n+1}^{m} \dfrac1{m}
=\dfrac{m-n}{m}
=1-\dfrac{n}{m}
\ge 1-\dfrac{m/2}{m}
=\dfrac12
$.
Note that,
if $m \ge kn$,
then
$\begin{array}\\
x_m-x_n
&=\sum_{j=n+1}^{m} \dfrac1{j}\\
&\ge\sum_{j=n+1}^{kn} \dfrac1{j}\\
&=\sum_{i=1}^k \sum_{j=(i-1)m+1}^{im} \dfrac1{j}\\
&\ge\sum_{i=1}^k \sum_{j=(i-1)m+1}^{im} \dfrac1{im}\\
&\ge\sum_{i=1}^k \dfrac{m}{im}\\
&=\sum_{i=1}^k \dfrac{1}{i}\\
&=x_k\\
\end{array}
$
Since
$x_{2n}-x_n \ge \dfrac12$,
$x_k$ is unbounded
as $k \to \infty$
so that
$x_m-x_n$
can be made arbitrarily large
by making $m$ large enough
compared with $n$.
For example,
since
$x_{2^k}
\ge \dfrac{k}{2}
$,
$x_{2^kn}-x_n
\ge x_{2^k}
\ge \dfrac{k}{2}
$.
