Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 19 / c / Solution

For $n>1$, the Taylor series of the natural logarithm converges absolutely at $1/n$:Multiply this power series by $n+1/2$ and subtract $1$. The constant term cancels the subtracted $1$, and the coefficient of $1/n$ is $-1/2+1/2=0$. For $k\ge 2$, the coefficient of $n^{-k}$ isHenceIn particular $u_n=1/(12n^2)-1/(12n^3)+O(n^{-4})$, in Big O notation.