Harmonic Series and p-Series BC only

$\frac{1}{x}$ is positive, continuous and decreasing on $[1, \infty)$, and $\int_1^{\infty} \frac{dx}{x} = \lim_{b\to\infty} \ln b = \infty$, so the harmonic series diverges, which settles the counterexample 10.1 and 10.3 both relied on. Its terms still tend to zero; what grows is $\ln b$, and it grows so slowly that the partial sums need over $12{,}000$ terms to pass $10$ and more than $272$ million to pass $20$. Running the same argument on $x^{-p}$ gives $\int_1^{b} x^{-p}\,dx = \frac{b^{1-p}-1}{1-p}$ for $p \neq 1$, which settles exactly when $p > 1$, while $p = 1$ produces $\ln x$ instead. So $\sum \frac{1}{n^{p}}$ converges when $p > 1$ and diverges when $p \le 1$, with the logarithm at the boundary being the reason that case behaves unlike its neighbours.

The errors cluster at the line: placing $p = 1$ on the convergent side, remembering the inequality backwards (a sanity check settles it, since larger $p$ means smaller terms and so convergence must be the upper region), and reading $p$ off the wrong part of an expression, so that $\frac{1}{\sqrt{n}}$ is $p = \frac{1}{2}$ and diverges while $\frac{1}{n\sqrt{n}}$ is $p = \frac{3}{2}$ and converges, and $\frac{1}{2^{n}}$ is not a $p$-series at all. Drawn on one set of axes, $\frac{1}{x}$ and $\frac{1}{x^{2}}$ start at the same height and both fall to zero, and only one encloses a finite area, which is why the threshold has to be known rather than seen. From 10.6 onward $p$-series are the benchmark other series are compared against, and the integral's value at $p > 1$ is $\frac{1}{p-1}$, a bound and not the sum.