Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 1 / 10F / iii

Compare with the divergent series from part (i). The ratio of its new summand to $1/(n\log n)$ isFor completeness, $\log n={\int }_1^n{x}^{-1}dx\le {\int }_1^n{x}^{-1/2}dx=2(\sqrt{n}-1)$, so $(\log n)/n\to 0$. The limit comparison test now gives divergence. Although the exponent $1+1/n$ exceeds one for each summand, it approaches one and does not produce a convergent p-series with a fixed exponent greater than one.