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

The obvious comparison is the wrong one

The instinct is to compare with $\frac{1}{n^{2}}$, which converges. Check the direction before writing it down: for $n \ge 3$ the logarithm is bigger than $1$.

So the terms sit above a convergent series, which permits no conclusion at all. Direct comparison only decides when the inequality faces the useful way, and here it faces the other way.

Give away a power of n

A logarithm grows more slowly than any positive power of $n$, so $\ln n$ can be traded for $\sqrt{n}$ and the inequality still holds for every $n \ge 1$.

The p-series with $p = \frac{3}{2}$ converges, and now the comparison points downhill onto a convergent series, so the original converges. Half a power of $n$ was more than enough slack.

The integral test settles it directly

Take $f(x) = \frac{\ln x}{x^{2}}$. It is positive and continuous on $[2, \infty)$, and $f'(x) = \frac{1 - 2\ln x}{x^{3}}$ is negative once $\ln x > \frac{1}{2}$, so $f$ decreases from $x = 2$ onwards. Integration by parts gives the antiderivative.

The integral converges, so the series does too. Starting at $n = 2$ costs nothing, because the $n = 1$ term is $\ln 1 = 0$.

The mistakes students make

Compare this page with $\sum \frac{\ln n}{n}$, which diverges. The logarithm is the same; the power of $n$ decides everything.

• Comparing with $\frac{1}{n^{2}}$ and stopping. From $n = 3$ the terms are larger than $\frac{1}{n^{2}}$, and sitting above a convergent series proves nothing.
• Assuming $\ln n$ eventually overpowers a power of $n$. It never does: $\frac{\ln n}{n^{p}} \to 0$ for every $p > 0$, which is why the trade for $\sqrt{n}$ is safe.
• Using the integral test without checking that $f$ decreases. It rises until $x = \sqrt{e}$, so state the test on $[2, \infty)$.