Show that for any sequence of positive reals, a_{n}, we have lim sup_{n→∞} n( (a_{n+1} + 1)/a_{n} - 1) ≥ 1. Show that we can find a sequence where equality holds.

**Solution**

Suppose lim sup < 1. Then we cannot find an infinite subsequence with n( (a_{n+1} + 1)/a_{n} - 1) ≥ 1, so we must have n( (a_{n+1} + 1)/a_{n} - 1) < 1 for all sufficiently large n. Suppose it is true for all n ≥ N. Then (a_{N+1} + 1)/a_{N} < (N + 1)/N and hence a_{N}/N > (a_{N+1} + 1)/(N + 1) = a_{N+1}/(N+1) + 1/(N+1). But similarly, a_{N+1}/(N+1) > a_{N+2}/(N+2) + 1/(N+2) and so on. Hence a_{N} ≥ 1/(N+1) + 1/(N+2) + ... , which is impossible since the rhs diverges. So we have established that lim sup ≥ 1.

We can experiment with various series. a_{n} = n gives 2. a_{n} = n^{2} also gives 2. Higher powers give higher values. So we try looking at exponents between 1 and 2. a_{n} = n^{1+k} gives 1+k. So by taking k arbitrarily small we can get close to 1, but we cannot reach it. It is often helpful to think of log n as n^{k} with k infinitesimally small. So we try a_{n} = n log n. That gives n( (a_{n+1} + 1)/a_{n} - 1) = (1 + n log (1 + 1/n) + log(n + 1) )/log n = (1 + n log (1 + 1/n) + (log(1 + 1/n) )/log n + 1. But 1 + n log(1 + 1/n) + log (1 + 1/n) is bounded and so n( (a_{n+1} + 1)/a_{n} - 1) has limit 1 (and hence also lim sup = 1).

© John Scholes

jscholes@kalva.demon.co.uk

5 Feb 2002