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 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} $.