Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 1 / 9F (Analysis I)

The ratio test says that ifthen $\sum a_n$ converges absolutely. If the ratio has a limit $L>1$, the series diverges.

For the first assertion, choose $q$ with $L<q<1$. For all sufficiently large $n$,Iteration bounds the tail by a constant multiple of the convergent geometric series $\sum q^n$. If $L>1$, choose $q$ with $1<q<L$. Eventually $|a_{n+1}|\ge q|a_n|$, so $a_n$ cannot tend to zero; the series therefore diverges.

First suppose $L>0$. Taking logarithms,Divide by $n$. Since the summands tend to $\log L$, their Cesàro means also tend to $\log L$. Henceand exponentiation gives $a_n^{1/n}\to L$.

If $L=0$, then for every $\epsilon >0$ the ratios are eventually at most $\epsilon$, so $a_n\le C{\epsilon }^n$ for a suitable constant $C$. ThusLetting $\epsilon \downarrow 0$ proves the ratio limit implies root limit in all cases.

SetThenThe ratio test therefore proves

For the sequence in part (iii), part (ii) givesTaking reciprocals yields the factorial-over-power series limit