Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 37 / 2 / a / Solution

This is an asymmetrically almost-subadditive sequence. The limit must be allowed to take the value $-\infty$; the stated hypotheses alone do not imply a finite real limit. DefineFix $k$, and write $N=qk+r$ with $1\le r\le k$. Repeatedly apply the given inequality with the first summand equal to $k$. Induction givesFor fixed $k$, there are only finitely many possible residual terms $x_r$, so division by $N$ and passage to the upper limit yieldThis holds for every $k$, and therefore the upper limit is at most $I$. If $I$ is finite, its definition also gives $x_N/N\ge I-{\alpha }_N/N$, so the lower limit is at least $I$ because ${\alpha }_N/N\to 0$. If $I=-\infty$, the upper bound by every $(x_k+{\alpha }_k)/k$ makes the upper limit $-\infty$ directly. Thus in both casesIn particular,When $\lambda =-\infty$ this inequality is interpreted in the extended-real sense. For example, $x_n=-n^2$ and ${\alpha }_n=0$ satisfy the assumptions but have normalized limit $-\infty$. A finite-limit assertion would need an additional lower linear bound.