AP Calculus BC

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

Q: Does the sum of 1/(n ln n) converge?

No. The integral test gives \ln(\ln x), which grows without bound.

Q: Why can it not start at n = 1?

Because \ln 1 = 0, so the first term would divide by zero. Starting index never changes convergence, but the terms must exist.

The sum of 1 over n times ln n, starting at n equals 2, diverges. The integral test settles it: the matching integral evaluates to ln of ln x, which grows without bound. The terms shrink faster than the harmonic terms and it still is not fast enough.

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

Diverges

Settled by the integral test.

The integral test

The function $f(x) = \frac{1}{x\ln x}$ is positive, continuous and decreasing on $[2, \infty)$ , which is what the integral test requires.

\int_{2}^{\infty}\frac{dx}{x\ln x} = \lim_{b \to \infty}\left[\ln\left(\ln x\right)\right]_{2}^{b} = \infty

The substitution $u = \ln x$ turns the integrand into $\frac{1}{u}$ , whose antiderivative is a logarithm, and a logarithm of a logarithm still grows without bound. The integral diverges, so the series does too.

Why it starts at n = 2

The sum cannot start at $n = 1$ , because $\ln 1 = 0$ makes the first term undefined. Where a series starts never affects convergence, but the terms do have to exist.

A boundary case worth remembering

This series sits between the divergent harmonic series and the convergent $p$ -series. Its terms are strictly smaller than $\frac{1}{n}$ , so comparison with the harmonic series proves nothing, and yet it still diverges.

Change the power on the logarithm and the answer flips: $\sum \frac{1}{n\left(\ln n\right)^{2}}$ CONVERGES, by the same integral test. The two differ only in that exponent.

\int_{2}^{\infty}\frac{dx}{x\left(\ln x\right)^{2}} = \frac{1}{\ln 2}

The mistakes students make

Comparing with the harmonic series to prove divergence. The terms here are SMALLER, so that comparison is the wrong direction and proves nothing.
Using the ratio test, which returns $L = 1$ and is inconclusive.
Assuming faster-shrinking terms must converge. These shrink faster than harmonic and diverge anyway.

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) converge?

No. The integral test gives $\ln(\ln x)$ , which grows without bound.

Why can it not start at n = 1?

Because $\ln 1 = 0$ , so the first term would divide by zero. Starting index never changes convergence, but the terms must exist.

What about 1/(n (ln n)^2)?

That one CONVERGES, by the same integral test, which evaluates to $\frac{1}{\ln 2}$ . The exponent on the logarithm is the only difference.