AP Calculus BC

Does the Sum of (-1)^n ln(n)/n Converge? Yes

The sum of negative 1 to the n times ln n over n converges CONDITIONALLY. The alternating series test applies because ln n over n decreases once n passes 3 and tends to 0. Without the signs the terms sit above 1 over n from n = 3 on, so the absolute series diverges.

∑n=2∞(−1)nln⁡nn\sum_{n=2}^{\infty}\frac{(-1)^{n}\ln n}{n}

Converges

Settled by the alternating series test, and only conditionally.

Decreasing eventually is enough

The magnitudes here are bn=ln⁡nnb_{n} = \frac{\ln n}{n}, and they do not decrease from the start. The first two are b2≈0.347b_{2} \approx 0.347 and b3≈0.366b_{3} \approx 0.366, so the sequence goes up before it comes down.

Differentiating shows exactly where the turn happens.

ddx(ln⁡xx)=1−ln⁡xx2\frac{d}{dx}\left(\frac{\ln x}{x}\right) = \frac{1 - \ln x}{x^{2}}

That derivative is negative once ln⁡x>1\ln x > 1, so ln⁡xx\frac{\ln x}{x} falls for every x>ex > e, which covers n=3n = 3 onwards. The limit is 00, by L'Hopital's rule or by the fact that a logarithm loses to any positive power of nn. Both hypotheses of the alternating series test hold from n=3n = 3, and the test asks for nothing more.

Finitely many terms never change a verdict

Deleting or altering the first few terms shifts the value of a convergent series but cannot make it diverge, and cannot rescue a divergent one. That is why every convergence test only needs its hypothesis to hold for large n.

Conditional, because the absolute series is too big

Stripping the signs leaves ∑n=2∞ln⁡nn\sum_{n=2}^{\infty}\frac{\ln n}{n}. For n≥3n \ge 3 the numerator is bigger than 11, which puts every term above the matching harmonic term.

ln⁡nn>1n(n≥3)\frac{\ln n}{n} > \frac{1}{n} \quad (n \ge 3)

The harmonic series diverges, so direct comparison sends ∑ln⁡nn\sum \frac{\ln n}{n} the same way. Converging with the signs and diverging without them is the definition of conditional convergence.

The mistakes students make

Two of these come from reading the alternating series test as stricter than it is.

  • Rejecting the test because b2<b3b_{2} < b_{3}. The magnitudes have to decrease eventually, not from the first term, and here they do so from n=3n = 3.
  • Checking only that bn→0b_{n} \to 0 and calling the series convergent on that basis. A limit of 00 is a hypothesis of the alternating series test, not a test on its own.
  • Recording the convergence as absolute. The absolute series beats the harmonic series from n=3n = 3 onwards, so it diverges and the convergence is conditional.

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 sum of (-1)^n ln(n)/n converge?

Yes, conditionally. The alternating series test applies from n=3n = 3 onwards, and the absolute series diverges.

Do the terms have to decrease from the very first one?

No. ln⁡nn\frac{\ln n}{n} rises until n=en = e and falls afterwards, and eventual decrease is all the test requires.

Is this absolutely or conditionally convergent?

Conditionally. ∑ln⁡nn\sum \frac{\ln n}{n} diverges by comparison with the harmonic series.