The Harmonic Series Diverges

Group the terms: $\frac{1}{3}+\frac{1}{4} > \frac{1}{4}+\frac{1}{4} = \frac{1}{2}$, $\frac{1}{5}+\cdots+\frac{1}{8} > 4\cdot\frac{1}{8} = \frac{1}{2}$. In general, for $m \ge 2$, $\frac{1}{2^{m-1}+1}+\cdots+\frac{1}{2^m} > 2^{m-1}\cdot\frac{1}{2^m} = \frac{1}{2}$ (the $m = 1$ block is the single term $\frac{1}{2}$, computed separately; the strict inequality does not hold there).

Hence $S_{2^m} = 1 + \frac{1}{2} + \left(\frac{1}{3}+\frac{1}{4}\right) + \cdots \ge 1 + m\cdot\frac{1}{2} = 1 + \frac{m}{2}$.

As $m \to \infty$ the bound is unbounded, so $(S_n)$ is unbounded. An increasing unbounded sequence diverges, so the harmonic series diverges.

Alternating series test: if $b_n \downarrow 0$ then $\sum_{n=1}^{\infty}(-1)^{n+1}b_n$ converges. Here $b_n = \frac{1}{n}$ decreases to $0$.

Directly: the even partial sums $T_{2m} = \left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots$ group into positive pairs, so they increase and are bounded by $1$, hence converge. The odd partial sums $T_{2m+1} = T_{2m}+\frac{1}{2m+1}$ tend to the same limit.

Hence the alternating harmonic series converges. (That its sum is $\ln 2$ is not proved here.)

Separate $p \le 0$ first: then $\frac{1}{n^p} = n^{-p} \ge 1$, so the general term does not tend to $0$ and the series diverges by the term test (on this range $f(x) = x^{-p}$ is not decreasing, so the integral comparison below is not applied there).

Now assume $p > 0$; then $f(x) = x^{-p}$ is positive and decreasing.

For $n \ge 2$, $\frac{1}{n^p} \le \int_{n-1}^{n}\frac{dx}{x^p}$, so $S_N \le 1 + \int_{1}^{N}\frac{dx}{x^p}$.

If $p > 1$, $\int_{1}^{N}\frac{dx}{x^p} = \frac{1 - N^{1-p}}{p-1} < \frac{1}{p-1}$ is bounded, so the increasing partial sums are bounded and converge.

If $0 < p \le 1$, $\frac{1}{n^p} \ge \frac{1}{n}$, so comparison with the harmonic series gives divergence (the case $p \le 0$ was already handled by the term test in step 1).