Prove that the harmonic series diverges

Group the terms into blocks whose lengths are powers of two:
$$1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\cdots +\frac{1}{16}\right)+\cdots$$

The block running from $\frac{1}{2^k+1}$ to $\frac{1}{2^{k+1}}$ contains $2^k$ terms, and every one of them is at least $\frac{1}{2^{k+1}}$. So each block sums to at least
$$2^k\cdot \frac{1}{2^{k+1}}=\frac{1}{2}.$$

There are infinitely many such blocks, so the partial sums pass every bound. Concretely,
$$\sum _{n=1}^{2^m}\frac{1}{n}\ge 1+\frac{m}{2}.$$

The same argument explains the slowness

That inequality is not only a proof, it is a rate. To push the sum up to $1+m/2$ you need about $2^m$ terms — so the partial sums grow like the logarithm of how many terms you have taken.

Which is exactly what the numbers in the problem statement are showing. Reaching $10$ takes around $e^{10}$ terms. Reaching $20$ takes around $e^{20}$. The series does not diverge slowly despite diverging; it gains a fixed amount on every doubling, and doublings get expensive fast.

Nicole Oresme had this in the fourteenth century, some three hundred years before anyone had a definition of convergence to be careful about. It remains the argument I would show someone first.