AP Calculus BC

Does the Sum of 1/n Converge? No

The sum of 1 over n diverges. It is the p-series with p = 1, and a p-series converges only when p is greater than 1, so this is the boundary case that fails. Its terms do go to 0, which is exactly why the nth term test cannot settle it.

∑n=1∞1n\sum_{n=1}^{\infty}\frac{1}{n}

Diverges

Settled by the p-series test.

Why p = 1 is the boundary

The p-series rule is a single comparison: ∑1np\sum \frac{1}{n^{p}} converges when p>1p > 1 and diverges when p≤1p \le 1. Here p=1p = 1 exactly, which falls on the diverging side.

∑n=1∞1np converges  ⟺  p>1\sum_{n=1}^{\infty}\frac{1}{n^{p}} \text{ converges} \iff p > 1

The integral test is where that rule comes from. ∫1∞dxx=lim⁡b→∞ln⁡b=∞\int_{1}^{\infty}\frac{dx}{x} = \lim_{b \to \infty}\ln b = \infty, and since the integral diverges so does the series.

It diverges astonishingly slowly

The partial sums grow like ln n, so reaching a total of 20 takes about 272 million terms. Slow divergence is still divergence, and no amount of numerical evidence can distinguish it from convergence.

Why the nth term test says nothing

The terms 1n\frac{1}{n} do go to 00, so the nth term test is silent. That test can only ever prove DIVERGENCE, by finding terms that fail to vanish; it never proves convergence.

The harmonic series is the standard counterexample to the belief that vanishing terms are enough. They are necessary and not sufficient, and this is the series that proves it.

lim⁡n→∞1n=0yet∑n=1∞1n=∞\lim_{n \to \infty}\frac{1}{n} = 0 \quad \text{yet} \quad \sum_{n=1}^{\infty}\frac{1}{n} = \infty

The mistakes students make

  • Concluding convergence because the terms go to 00. That is the single most common error in Unit 10, and this series exists to refute it.
  • Citing the nth term test as proof of convergence. It has no such power in either direction when the limit is 00.
  • Trying the ratio test. It gives L=1L = 1, which is inconclusive, so it wastes a step.
  • Confusing it with the ALTERNATING harmonic series, which does converge, to ln⁡2\ln 2, conditionally.

Not sure which test a series wants?

The Convergence Test Chooser walks the decision in order: nth term first, then geometric and p-series pattern matching, then alternating structure, then the ratio test, and finally the comparison family.

Frequently asked questions

Does the harmonic series converge?

No. It is the pp-series with p=1p = 1, and convergence needs p>1p > 1.

Its terms go to zero, so why does it diverge?

Vanishing terms are necessary for convergence but not sufficient. The harmonic series is the classic proof that the converse fails.

What does the ratio test give?

L=1L = 1, which is inconclusive. When the ratio test returns 11 you must switch tests, and here the pp-series rule settles it immediately.