Strategy for Series — Question 9

*Question 9 Determine whether $\displaystyle\sum_{n=2}^{\infty}\frac{\log n}{n^2}$ converges. Compare the Integral Test with Cauchy Condensation and identify the shorter calculation.

*Question 9 – Solution

The function $f(x)=\log x/x^2$ is positive and continuous for $x\ge2$. Also, $$f'(x)=\frac{1-2\log x}{x^3}<0\qquad(x\ge2).$$ Integration by parts gives $$\int_2^{\infty}\frac{\log x}{x^2}\,dx=\left[-\frac{\log x+1}{x}\right]_2^{\infty}=\frac{\log2+1}{2}<\infty.$$ Thus the series converges.

Because the terms decrease, $$2^k a_{2^k}=2^k\frac{\log(2^k)}{(2^k)^2}=\frac{k\log2}{2^k},$$ whose series converges by the Ratio Test. Condensation also works, but the integral is slightly shorter and matches the logarithm-over-power form directly.