AP Calculus BC

Does the Sum of ln(n)/n Converge? No

The sum of ln n over n DIVERGES. For n of 3 or more, ln n is greater than 1, so ln n over n is greater than 1 over n. The harmonic series already diverges, and a series lying above a divergent series of positive terms must diverge as well.

∑n=2∞ln⁡nn\sum_{n=2}^{\infty}\frac{\ln n}{n}

Diverges

Settled by the direct comparison test.

One line of direct comparison

Direct comparison needs a known divergent series sitting underneath. The harmonic series is the obvious candidate, and the logarithm hands you the inequality.

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

Since ∑1n\sum \frac{1}{n} diverges and these terms are larger from n=3n = 3 onwards, the series diverges. The n=2n = 2 term is a single finite number and cannot affect a verdict.

The integral test agrees

For x>ex > e the function f(x)=ln⁡xxf(x) = \frac{\ln x}{x} is positive, continuous and decreasing, since f′(x)=1−ln⁡xx2f'(x) = \frac{1 - \ln x}{x^{2}} is negative there. The substitution u=ln⁡xu = \ln x handles the integral.

∫3∞ln⁡xx dx=lim⁡b→∞[(ln⁡x)22]3b=∞\int_{3}^{\infty}\frac{\ln x}{x}\,dx = \lim_{b \to \infty}\left[\frac{(\ln x)^{2}}{2}\right]_{3}^{b} = \infty

The antiderivative (ln⁡x)22\frac{(\ln x)^{2}}{2} grows without bound, so the improper integral diverges and the series does the same. Two tests, one verdict.

Slow growth is still growth

Adding the first thousand terms gives about 23.8, and the first million gives about 95.4. Partial sums that look tame settle nothing either way. The comparison is what decides this, not the arithmetic.

The mistakes students make

The first one is the single most common error in the whole series unit.

  • Arguing that ln⁡nn→0\frac{\ln n}{n} \to 0, so the series converges. The nth term test can only prove divergence. A limit of 00 leaves the question open, as the harmonic series itself shows.
  • Comparing with 1n2\frac{1}{n^{2}}. Direct comparison works in one direction only: above a divergent series proves divergence, below a convergent one proves convergence, and anything else concludes nothing.
  • Starting the comparison at n=2n = 2, where ln⁡2<1\ln 2 < 1 and the inequality is false. State it for n≥3n \ge 3 and note that the first term is irrelevant.

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 ln(n)/n converge?

No. It diverges by direct comparison with the harmonic series, since ln⁡nn>1n\frac{\ln n}{n} > \frac{1}{n} for n≥3n \ge 3.

Why does the terms going to zero not help?

Because an→0a_{n} \to 0 is necessary for convergence, not sufficient. The harmonic series has terms tending to 00 and still diverges.

Can I use the integral test instead?

Yes. The antiderivative is (ln⁡x)22\frac{(\ln x)^{2}}{2}, which is unbounded, so the improper integral and the series both diverge.