Limit of biggest share of the pie

Question

A huge pie is divided among $N$ guests. The first guest gets $\frac{1}{N}$ of the pie. Guest number $k$ guest gets $\frac{k}{N}$ of what's left, for all $1\leq k\leq N$. (In particular, the last guest gets all of what is left.)

Let $s(N)$ be the maximum of all shares of pie if we have $N$ guests. What is the value of $$\lim\sup_{N\to\infty}N\cdot s(N)$$ ?

(The term $N\cdot s(N)=\frac{s(N)}{\frac{1}{N}}$ denotes the quotient of how much the luckiest guest does better than the average guest.)

Answer 1

This limit is infinite. Guest number $k$ gets the quotient $x_k=\frac kN(1-\frac1N)(1-\frac2N)\dots (1-\frac{k-1}N)$, for $k$ about $\sqrt{N}$ this behaves like $\frac{1}{\sqrt{N}}$, cince the product is bounded by below (actually it tends to $e^{-1/2}$). It is easy to see that $k$ corresponding to the maximal quotient $x_k$ is a minimal $k$ satisfying $k(k+1)\geqslant N$, for seeing this simplify the inequality $x_k\geqslant x_{k+1}$.