AP Calculus BC

Does the Sum of 1/ln(n) Converge? No

The sum of 1 over ln n, starting at n equals 2, diverges. Since ln n is less than n, the terms are LARGER than the harmonic terms, and the harmonic series already diverges, so direct comparison proves divergence.

∑n=2∞1ln⁡n\sum_{n=2}^{\infty}\frac{1}{\ln n}

Diverges

Settled by the direct comparison test.

Comparing upward with the harmonic series

The logarithm grows more slowly than its input, so its reciprocal is larger.

ln⁡n<n  ⟹  1ln⁡n>1nfor n≥2\ln n < n \implies \frac{1}{\ln n} > \frac{1}{n} \quad \text{for } n \ge 2

The harmonic series diverges, and these terms are bigger term by term, so the comparison test forces divergence.

Direction is everything in comparison

To prove DIVERGENCE your terms must be at least as large as a known divergent series. To prove CONVERGENCE they must be no larger than a known convergent one. Getting this backwards is the most common comparison error.

The nth term test is silent here

The terms do go to 00, since ln⁡n→∞\ln n \to \infty, so the nth term test says nothing. That is the usual situation: it settles very few series and this is not one of them.

lim⁡n→∞1ln⁡n=0\lim_{n \to \infty}\frac{1}{\ln n} = 0

The mistakes students make

  • Concluding convergence because 1ln⁡n→0\frac{1}{\ln n} \to 0. Vanishing terms are never sufficient.
  • Comparing with 1n\frac{1}{n} in the wrong direction, which would require these terms to be smaller.
  • Starting the sum at n=1n = 1, where ln⁡1=0\ln 1 = 0 makes the term undefined.

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/ln(n) converge?

No. Its terms exceed 1n\frac{1}{n}, and the harmonic series already diverges.

Why does it start at n = 2?

ln⁡1=0\ln 1 = 0, so the n=1n = 1 term is undefined. The starting index does not affect convergence.