AP Calculus BC

Does the Sum of ln(n)/n^3 Converge? Yes

Q: Does the logarithm ever change the verdict?

Only at the boundary. For p > 1 a logarithm in the numerator is harmless; at p = 1 it turns the harmonic series into something that diverges even faster.

Q: Which test is cleanest here?

Direct comparison with 1/n^{2} is fastest. The integral test also works, since \int \frac{\ln x}{x^{3}}dx converges by parts, but it costs more effort for the same answer.

The series converges. A logarithm grows more slowly than any positive power of n, so the terms are eventually smaller than one over n squared, and comparison with that convergent p-series settles it.

\sum_{n=1}^{\infty}\frac{\ln n}{n^{3}}

Converges

Settled by the limit comparison test.

Spending part of the exponent on the logarithm

For every $\varepsilon > 0$ there is a point past which $\ln n < n^{\varepsilon}$ . Taking $\varepsilon = 1$ gives $\ln n < n$ for $n \ge 1$ , so the terms are eventually below $\frac{n}{n^{3}} = \frac{1}{n^{2}}$ .

That is a p-series with $p = 2 > 1$ , which converges, so direct comparison gives convergence. The logarithm cost one unit of the exponent and there were three to spend.

How much room there is

The same argument works for $\sum \frac{\ln n}{n^{p}}$ whenever $p > 1$ : choose $\varepsilon$ small enough that $p - \varepsilon$ is still above 1. So $\sum \frac{\ln n}{n^{1.01}}$ converges too, though comparison with $1/n^{2}$ is far too crude there.

At $p = 1$ the room runs out. The series $\sum \frac{\ln n}{n}$ diverges, because its terms are LARGER than the harmonic terms once $\ln n > 1$ . A logarithm is slow, but it is not free.

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 logarithm ever change the verdict?

Only at the boundary. For $p > 1$ a logarithm in the numerator is harmless; at $p = 1$ it turns the harmonic series into something that diverges even faster.

Which test is cleanest here?

Direct comparison with $1/n^{2}$ is fastest. The integral test also works, since $\int \frac{\ln x}{x^{3}}dx$ converges by parts, but it costs more effort for the same answer.