Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 1 / 9E (Analysis)

The arithmetic--geometric mean inequality gives $x_n\ge 1$ from $n=2$ onward. For $x>1$,so the tail decreases to a limit $L\ge 1$. Passing to the recurrence gives $2L=L+L^{-1}$, hence $L=1$.

For the subadditive sequence, $0\le x_n/n\le x_1$, so it is bounded. The Fekete lemma can be proved directly here as follows. Let $\alpha ={\inf }_k{x}_k/k$. Fix $k$ and write $n=qk+r$, $0\le r<k$. ThenThus $\mathrm{limsup}{x}_n/n\le \alpha$, while the definition gives $\mathrm{liminf}{x}_n/n\ge \alpha$. Hence $x_n/n\to \alpha$.

Let $S_n={\sum }_{i\le n}{a}_i$ and $T_n={\sum }_{i\le n}|a_i|$. Conditional convergence gives $S_n\to S$ and $T_n\to \infty$. Since $P_n=T_n+S_n$ and $N_n=T_n-S_n$,

The alternating-series test says that $b_n\downarrow 0$ implies convergence of $\sum (-1{)}^n{b}_n$. The hypothesis implies $b_n/b_{n+1}>1$ eventually, so $b_n$ is eventually decreasing. For some $c>0$, eventually $b_n/b_{n+1}\ge 1+c/n$; the divergent product of these factors forces $b_n\to 0$. The test now applies.