Proof that the harmonic series diverges by grouping terms

Proof

We bound the harmonic numbers at powers of $2$ from below, by induction.

1. For $N \ge 1$, $H_N = \sum_{n=1}^{N} 1/n$. Since $H_{N+1} = H_N + 1/(N+1) > H_N$, the sequence $(H_N)$ is increasing.
2. Let $j \ge 1$. The block of terms $1/n$ with $2^{j-1} < n \le 2^j$ has $2^j - 2^{j-1} = 2^{j-1}$ terms.
3. Each term in that block satisfies $1/n \ge 1/2^j$. So the block has sum at least $2^{j-1}/2^j = 1/2$.
4. We claim $H_{2^k} \ge 1 + k/2$ for every natural number $k$. For $k = 0$, $H_1 = 1$.
5. Assume the claim for $k$. Then $H_{2^{k+1}} = H_{2^k} + \sum_{2^k < n \le 2^{k+1}} 1/n$. By step 3 with $j = k + 1$, this is at least $1 + k/2 + 1/2$.
6. By induction, steps 4 and 5 prove the claim for every $k$.
7. Let $M$ be a real number. By the Archimedean property there is a natural number $k$ with $2(M - 1) < k$, that is, $1 + k/2 > M$.
8. Let $N_0 = 2^k$. For $N \ge N_0$, steps 1, 4 and 7 give $H_N \ge H_{N_0} \ge 1 + k/2 > M$.
9. So $H_N \to +\infty$, in the sense of a sequence tending to infinity.
10. In particular $(H_N)$ is not bounded. Convergent sequences are bounded, so $(H_N)$ does not converge.
11. The $H_N$ are the partial sums of the harmonic series. So the harmonic series is a divergent series.