real analysis - Does $1/n$ satisfy the Cauchy criterion?

I have been told that if a series satisfies the Cauchy criterion, it converges. I was not quite convinced on this, however, since some of the posts on this website seemed to imply otherwise. I came up with what I thought was a counterexample: I argued that $1/n$ satisfies the Cauchy criterion but diverges. However, I've been told afterwards that $1/n$ doesn't satisfy the Cauchy criterion. Now, I've tried to show that the $1/n$ does not satisfy the Cauchy criterion but couldn't come up with a proof. Can someone provide me a proof that shows that the series $1/n$ does not satisfy the Cauchy criterion?

Answer

The sequence $\{1/n : n=1,2,3,\ldots\}$ satisfies the Cauchy criterion and converges.

The series $\displaystyle\sum_{n=1}^\infty \frac1n$ diverges. That is the same as saying that its sequence of partial sums
$$
\left\{ \sum_{n=1}^m \frac1n : m = 1,2,3,\ldots \right\}
$$
diverges. The latter sequence is not a Cauchy sequence. To see that it's not a Cauchy sequence without first considering whether it converges, consider the difference between the $m_1$th term and the $m_2$th term, whree $m_2>m_1$:
\begin{align}
& \phantom{={}} \left| \sum_{n=1}^{m_1} \frac1n - \sum_{n=1}^{m_2} \frac1n \right| \\[6pt]

& =\left| \sum_{n=m_1+1}^{m_2} \frac1n \right| \ge \left| \int_{m_1+1}^{m_2} \frac1x\,dx \right| \\[6pt]
& = \left|\log\frac{m_2}{m_1+1}\right|\to\infty\text{ as }m_2\to\infty.
\end{align}

Don't confuse sequences with series.