Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics II / 2 / b / ii

The implication is false. SetThen $n{a}_n=1/\log (n+1)\to 0$. For $n\ge 2$, $\log (n+1)\le \log (2n)\le 2\log n$, so $a_n\ge 1/(2n\log n)$. The integral test givesThus the comparison test shows that $\sum a_n$ diverges.