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

All $a_n$ are positive, and $a_1=e^{-1}$. The logarithm turns consecutive ratios into differences, producing the finite telescoping sumThus, already for every finite $n$,If the partial sums converge to $S$, the continuity of the exponential function givesThe shift from $a_{n+1}$ to $a_n$ preserves the limit of a sequence. Convergence of the partial sums is established in part (f), so no convergence assumption is left unsupported.

Cancellation of the factorials in the consecutive ratio givesSet $t=1/n$. The derivative of the natural logarithm at $1$ gives $\ln (1+t)/t\to 1$. ThereforeThus the necessary condition from the term test for divergence is satisfied. On its own it does not prove that $\sum u_n$ converges.

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.

A rational lower bound for the natural logarithm proves positivity for every $n$, including $n=1$. DefineThen $h(0)=0$, while direct differentiation givesThus $h(t)>0$ for $t>0$. Setting $t=1/n$ in the expression from part (b) givesTherefore $\boxed{u_n>0\text{ for every }n\ge 1}$, and $(a_n)$ is a strictly increasing sequence.

For $k\ge 2$, writeUse positivity from part (d) and bound the absolute terms of the power series in part (c) by a geometric series. For $n\ge 2$,For $n=1$, $u_1=\frac{3}{2}\ln 2-1<\frac{1}{2}<1$, since $\ln 2={\int }_1^{2}dt/t<1$. Thus a bound valid for every positive integer $n$ is

By part (e) and the comparison test for series, $\sum u_n$ converges, since the P-series $\sum n^{-2}$ converges. Put $S={\sum }_{n=1}^{\infty }{u}_n$. Part (a) now provesPart (d) also shows that $(a_n)$ increases towards this finite positive limit. This proves convergence of the Stirling sequence without presupposing its limiting constant.

For identification of that constant, the independently known Stirling's formula $n!\sim \sqrt{2\pi n}(n/e{)}^n$ givesThe comparison argument above supplies the convergence proof requested by the preceding parts; the Stirling's formula identification is an additional evaluation.