When sums blow up
The harmonic series $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots$ has terms that shrink to zero. Does it converge?
At $p = 1.00$, five hundred terms total $S_{500} = 6.7928$, and the last stretch of the trace looks almost flat — each new term adds about $0.002$. It is a bluff. The verdict reads DIVERGES: this sum passes $7$, then $10$, then any number you name. At $p = 2.00$ the same flatness is honest: $S_{500} = 1.6429$, pinned under a ceiling at about $1.6449$ — which is $\pi^2/6$, the famous Basel value.
Shrinking terms are necessary for convergence — a series whose terms do not die can never settle — but they are not sufficient. The harmonic terms shrink too slowly. Group them: $\frac{1}{2}$, then $\frac{1}{3} + \frac{1}{4}$, then $\frac{1}{5} + \cdots + \frac{1}{8}$, and so on, doubling each time. Every group totals more than $\frac{1}{2}$, and there are infinitely many groups — so no ceiling exists. Raise the exponent and the tail thins fast enough for the total to stay finite. The dividing line is exact:
At $p = 1$ it diverges; at $p = 1.05$ it already converges, though to a distant ceiling. The instrument cannot see that line in the pictures — only the grouping argument draws it.
About this model
What it represents: exact partial sums $S_N$ of the p-series up to $N = 500$, on a log-scaled $N$ axis. The ceiling for $p > 1$ is $S_{500}$ plus an integral estimate of the missing tail, accurate to the digits shown.
What it leaves out: the tail itself — and with it, the truth about divergence. No finite window can display unbounded growth: at $p = 1$ the plotted curve is indistinguishable from a convergent one, because $S_N$ grows like $\ln N + 0.577$, reaching only $6.7928$ at $N = 500$ and $9.0945$ at $N = 5000$. The verdict stat comes from the $p > 1$ test, never from the picture.
Source: Oresme's grouping proof and the $\ln N + \gamma$ growth rate — Harmonic series (mathematics).
At $p = 1$ the trace looks flat by $N = 500$. What does $S_N$ actually do as $N \to \infty$?
Both $1/n$ and $1/n^2$ shrink to zero. Why does only $\sum 1/n^2$ converge?
Transfer: what does $\sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n}}$ do? (That is $p = 0.5$.)
Terms shrinking to zero is the entry fee, not the ticket: convergence depends on how fast the tail thins. For $\sum 1/n^p$ the answer is exact — finite precisely when $p > 1$ — and no finite plot can tell you; the test can.