Is this a valid proof that the harmonic series diverges?
- Assume the series converges to a value, S:
$$S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...$$
- Split the series into two, with alternating even and odd denominators. Since the original series converges, the component series will converge.
$$S_{EVEN}=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...$$ $$S_{ODD}=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...$$ $$S=S_{EVEN}+S_{ODD}$$
- Show that $S_{EVEN}=\frac{1}{2}S$
$$\frac{1}{2}S=\frac{1}{2}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...)=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...=S_{EVEN}$$
Show $S_{ODD}>S_{EVEN}$ because each odd term is greater than its corresponding even term: $$1>\frac{1}{2}\qquad \frac{1}{3}>\frac{1}{4}\qquad \frac{1}{5}>\frac{1}{6}\qquad ...$$
Show $S_{ODD}=S_{EVEN}$ $$S_{ODD}=S-S_{EVEN}=S-\frac{1}{2}S=\frac{1}{2}S=S_{EVEN}$$
The contradiction implies that the original assumption of convergence is false:
$$S_{ODD}>S_{EVEN}$$ $$S_{ODD}=S_{EVEN}$$ $$\therefore S\ne 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...$$
This is almost valid. We need to justify the second step, as mentioned by Ross Millikan, as it is not always valid to split a series into their even and odd terms.
Take, as a simple example, the alternating harmonic series, where you would get it equalling $\infty-\infty$ which is indeterminate, but it does not make sense for the convergence of a series to be indeterminate.
This can be justified by seeing your series is absolutely convergent, assuming it converges.
If one must be pedantic, the same issue occurs showing $S_\mathrm{ODD}>S_\mathrm{EVEN}$, but this can be more easily justified by the fact we are comparing terms in the order that they are summed. If they were not compared in this order and their respective series converged conditionally, this may not be true.
Aside from all that it looks good. If I may provide an alternative proof of similar approach, it would've sufficed to have shown that
$$S=1+\frac12+\frac13+\frac14+\dots>\frac12+\frac12+\frac14+\frac14+\dots=S$$