Shouldn't the harmonic series converge?

In the definition of a Cauchy sequence, you have not gotten the order of the quantifiers correct. The partial sums of a series are Cauchy iff for all $\epsilon > 0$ there exists an $N$ such that for all $m,n \ge N$, $|\sum_{k=m}^{n} a_k| < \epsilon$. Notice the "for all $m,n$." You cannot just let $m=n$ and verify the condition for that case.

It is true that a sequence in $\mathbb{R}$ converges if and only if it is Cauchy, and as such a series in $\mathbb{R}$ converges if and only if the sequence of partials sums $\{S_k\}_{k=0}^\infty$ is Cauchy. The problem is that you are not using the definition of Cauchy correctly. For a sequence $\{x_k\}_{k=0}^\infty$ it reads that for every $\epsilon >0$ there exists $K \ge 0$ such that if $m,n \ge K$, then $|x_n - x_m | < \epsilon$. Here one does not get to restrict to $n = m+1$, one must check all possible values of $m,n \ge K$. Note, though, that it is possible to restrict to $m \ge n \ge K$ in the definition.

Let's look at this for the harmonic series. In this case $S_k = \sum_{m=1}^k 1/m$. You are looking at $S_{k+1} - S_k = 1/(m+1)$, which indeed can be made arbitrarily small. In fact, to use the Cauchy definition we must consider arbitrary $k \ge \ell$ to get $$ S_k - S_\ell = \sum_{m=\ell+1}^k \frac{1}{m}, $$
and it is these partial sums of arbitrary length that cannot be made small.

If $\epsilon>0$ and $m\in\mathbb{N}$ then there is an $n\in\mathbb{N}$, $n>m$ such that

$$ \sum_{k=m}^n\frac{1}{k}>\epsilon $$