The line is at $p = 1$, and it falls on the divergent side BC only

$f(x) = \frac{1}{x}$ is positive, continuous and decreasing on $[1, \infty)$, so the integral test applies. Then

$$\int_{1}^{\infty} \frac{dx}{x} = \lim_{b \to \infty} \big[\ln x\big]_{1}^{b} = \lim_{b \to \infty} \ln b = \infty.$$

The integral diverges, so $\sum \frac{1}{n}$ diverges. That is the whole proof, and it retires the loose end 10.1 and 10.3 both left open.

What the proof does not say is that the terms misbehave. They tend to zero perfectly well. What diverges is $\ln b$, and it does so at a crawl: the partial sums track $\ln n$ closely, which is why reaching $10$ takes over $12{,}000$ terms and reaching $20$ takes more than $272$ million. Slow divergence is still divergence, and no amount of numerical evidence would have revealed it.

Run the same argument on $f(x) = \frac{1}{x^{p}}$ for $p > 0$. The antiderivative splits into two cases, and the split is the threshold:

1. $p \neq 1$: $\displaystyle\int_{1}^{b} x^{-p}\,dx = \frac{b^{1-p} - 1}{1-p}$. As $b \to \infty$ this settles exactly when the exponent $1 - p$ is negative, that is when $p > 1$, and then the integral is $\frac{1}{p-1}$.
2. $p = 1$: the antiderivative is $\ln x$ instead, and it grows without bound.

So $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{p}}$ converges if $p > 1$ and diverges if $p \le 1$. The logarithm appearing only at $p = 1$ is not an accident of technique; it is the reason the boundary behaves differently from everything on either side of it.

For $p \le 0$ the series diverges for a cruder reason: the terms do not tend to zero, so 10.3's test finishes it without any of this.

Nobody misjudges $p = 5$. The errors cluster at the line, and they take three forms:

1. Putting $p = 1$ on the convergent side. It is the harmonic series, and it diverges. This is the single most common error the code records.
2. Remembering the inequality backwards, so that large $p$ diverges. A sanity check kills this instantly: bigger $p$ means smaller terms, and smaller terms are more likely to converge, so convergence must be the upper region.
3. Reading $p$ off the wrong part of the expression. $\sum \frac{1}{\sqrt{n}}$ has $p = \frac{1}{2}$ and diverges; $\sum \frac{1}{n\sqrt{n}}$ has $p = \frac{3}{2}$ and converges. Radicals are exponents, and $\sum \frac{1}{2^{n}}$ is not a $p$-series at all, since its $n$ is in the exponent rather than the base.

The two curves in the figure make the threshold visible. $\frac{1}{x}$ and $\frac{1}{x^{2}}$ both start at height $1$ and both fall to zero, and they look much alike. Only one encloses a finite area all the way out: $\int_1^{\infty} x^{-2}\,dx = 1$ while $\int_1^{\infty} x^{-1}\,dx$ is infinite. The eye cannot separate these two cases, which is exactly why the rule has to be memorised rather than intuited.

Beyond their own verdicts, $p$-series are the standard against which other series get measured. From 10.6 onward the most common argument in this unit has the shape "this series behaves like $\sum \frac{1}{n^{p}}$, and I know what that does".

That makes fluency with the threshold worth more than the individual results. Two habits:

1. Read off $p$ by comparing degrees. $\frac{2n+1}{n^{3}+5}$ behaves like $\frac{2n}{n^{3}} = \frac{2}{n^{2}}$, so it behaves like a $p$-series with $p = 2$ and converges.
2. Watch the boundary. $\frac{n+1}{n^{2}+3}$ behaves like $\frac{1}{n}$, landing exactly on $p = 1$, and therefore diverges. Terms that look like they are shrinking respectably are the ones that catch people out.

Note also what the integral gives you and what it does not. When $p > 1$ the integral evaluates to $\frac{1}{p-1}$, and that is a bound on the sum, never the sum: at $p = 2$ it gives $1$ while the series sums to $\frac{\pi^{2}}{6} \approx 1.6449$. That is 10.4's distinction, and it does not soften here.