Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 19 / d

The function $g(x)=1/(x\ln x)$ is positive and decreasing for $x\ge 2$, since $g^{\prime }(x)=-(\ln x+1)/(x^2{\ln }^{2}x)<0$. The integral test for convergence therefore applies. Its comparison improper integral isConsequently $\sum _{n=2}^{\infty }1/(n\ln n)$ diverges.

The function $g(x)=1/(x{\ln }^{2}x)$ is likewise positive and decreasing on $[2,\infty )$, with $g^{\prime }(x)=-(\ln x+2)/(x^2{\ln }^{3}x)<0$. NowThe integral test for convergence shows that $\sum _{n=2}^{\infty }1/(n{\ln }^{2}n)$ converges. This comparison integral determines convergence, not the numerical value of the series itself.