Taking reciprocals

Why take reciprocals

u(n) tends to 0: the terms pile up on one another and comparing them becomes awkward. Setting

   v(n) = 1 / u(n)

turns a sequence that collapses into one that blows up. Above all, the recurrence relation becomes almost an arithmetic progression.

The heart of the problem

   v(n+1) − v(n) = 1/ln(1 + u(n)) − 1/u(n)

Now part I showed that 1/ln(1 + x) − 1/x equals 1/2 up to a small gap, a gap that vanishes with x. Since u(n) tends to 0, the difference v(n+1) − v(n) tends to 1/2: the terms of v advance by one half at each step.

From the average step to the limit

A sequence whose successive increments tend to 1/2 behaves like n/2 — that is Cesàro's lemma. The assignment avoids invoking it and does better: it bounds the increment at every rank,

   1/2 − (3/16)·u(n)  ≤  v(n+1) − v(n)  ≤  1/2

then sums these inequalities from 0 to n−1. The sum telescopes:

   v(n) − v(0) = [v(1) − v(0)] + [v(2) − v(1)] + … + [v(n) − v(n−1)]

This bounds v(n), hence u(n), hence n·u(n). Every question in part II is one step of that staircase.