The series ∑∞n=11n\sum_{n=1}^\infty\frac1n diverges!

The proof can be made a bit more rigorous by setting
$$\begin{align} a_n=\frac1n:&\,\quad1,\,\frac12,\frac13,\frac14,\frac15,\frac16,\dots\\b_n=\frac1{2\lfloor(n+1)/2\rfloor}:&\quad\frac12,\frac12,\frac14,\frac14,\frac16,\frac16,\dots \end{align}\tag{1}$$
Note that $a_n\ge b_n$, $a_n\gt b_n$ when $n$ is odd, and $a_n=b_{2n-1}+b_{2n}$.

Assuming that
$$\sum_{n=1}^\infty a_n\tag{2}$$
converges, then
$$\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty(b_{2n-1}+b_{2n})=\sum_{n=1}^\infty a_n\tag{3}$$
also converges. However,
$$\sum_{n=1}^\infty(a_n-b_n)\gt0\tag{4}$$
Since $a_n\ge b_n$ and $a_n\gt b_n$ when $n$ is odd.

Now, $(3)$ says that
$$\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty a_n\tag{5}$$
and $(4)$ says that
$$\sum_{n=1}^\infty b_n\lt\sum_{n=1}^\infty a_n\tag{6}$$
These last two statements are contradictory, so the assumption that $(2)$ converges must be false.