AP Calculus BC

Does the Sum of 1/(n(ln n)^2) Converge? Yes

The sum of 1 over n times ln n squared CONVERGES. The integral test decides it: substituting u equals ln x turns the improper integral into the integral of 1 over u squared, which equals 1 over ln 2. Squaring the logarithm is what separates this from 1 over n ln n, which diverges.

∑n=2∞1n(ln⁡n)2\sum_{n=2}^{\infty}\frac{1}{n\left(\ln n\right)^{2}}

Converges

Settled by the integral test.

One substitution decides it

The function f(x)=1x(ln⁡x)2f(x) = \frac{1}{x(\ln x)^{2}} is positive, continuous and decreasing on [2,∞)[2, \infty), so the integral test applies. Setting u=ln⁡xu = \ln x, so that du=dxxdu = \frac{dx}{x}, turns the integral into one of the easiest there is.

∫2∞dxx(ln⁡x)2=∫ln⁡2∞duu2=lim⁡b→∞[−1u]ln⁡2b=1ln⁡2\int_{2}^{\infty}\frac{dx}{x(\ln x)^{2}} = \int_{\ln 2}^{\infty}\frac{du}{u^{2}} = \lim_{b \to \infty}\left[-\frac{1}{u}\right]_{\ln 2}^{b} = \frac{1}{\ln 2}

The integral is finite, so the series converges. Note what the test returns: a verdict, not a total. The number 1ln⁡2\frac{1}{\ln 2} is the value of the integral and not the sum of the series.

Why the twin series diverges

Run the same substitution on ∑1nln⁡n\sum \frac{1}{n \ln n}, one power of the logarithm lighter.

∫2∞dxxln⁡x=∫ln⁡2∞duu=lim⁡b→∞[ln⁡u]ln⁡2b=∞\int_{2}^{\infty}\frac{dx}{x\ln x} = \int_{\ln 2}^{\infty}\frac{du}{u} = \lim_{b \to \infty}\Big[\ln u\Big]_{\ln 2}^{b} = \infty

So one series converges and the other diverges, and the only difference between them is the exponent on the logarithm. More generally ∑1n(ln⁡n)p\sum \frac{1}{n(\ln n)^{p}} converges exactly when p>1p > 1, which mirrors the p-series rule one level down.

Nothing else on the syllabus tells them apart

The nth term test gives 0 for both. The ratio test gives 1 for both, so it is inconclusive twice over. Comparison with 1 over n says both sets of terms are smaller than a divergent series from n = 3 on, which proves nothing. Only the integral test separates them.

The mistakes students make

This pair punishes guessing from the look of the terms.

  • Assuming that terms visibly smaller than 1n\frac{1}{n} must give a convergent series. 1nln⁡n\frac{1}{n \ln n} is smaller than 1n\frac{1}{n} from n=3n = 3 onwards, by a factor of 1ln⁡n\frac{1}{\ln n}, and still diverges.
  • Integrating 1u2\frac{1}{u^{2}} as ln⁡u\ln u after the substitution. The two antiderivatives, −1u-\frac{1}{u} and ln⁡u\ln u, are exactly what produce the two different verdicts.
  • Applying the integral test from n=1n = 1. The term is undefined there, since ln⁡1=0\ln 1 = 0, which is why the series starts at n=2n = 2.

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)^2) converge?

Yes, by the integral test. The improper integral equals 1ln⁡2\frac{1}{\ln 2}, which is finite.

Why does 1/(n ln n) diverge when this one converges?

Its integral becomes ∫duu=ln⁡u\int \frac{du}{u} = \ln u, which is unbounded. One extra power of the logarithm is the whole difference.

Is the sum equal to 1/ln 2?

No. The integral test returns convergence or divergence, never a value. The sum is a different number and has no elementary closed form.