Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 336 / 2 / b

Solution

Write the summand as

a_{n} = x^{n} / [Γ (n) n^{n}]

a_{n} \sim \frac{n ^{1/2}}{2 π} exp [n (lo g x + 1 - 2 lo g n)] .

N = \frac{x}{e}, n = Nt .

Then the exponential phase is

n (lo g x + 1 - 2 lo g n) = N f (t), f (t) = 2 t (1 - lo g t) .

It has a unique maximum at

t = 1

, where

f (1) = 2

and

f^{''} (1) = - 2

. The contributing indices satisfy

n - N = O (N)

, so their width tends to infinity and the lattice sum may be replaced by a Riemann sum. The Discrete Laplace method therefore gives

S \sim N \frac{N}{2 π} e^{2 N} \frac{2 π}{2 N} = \frac{N}{2} e^{2 N} .

Hence

S \sim \frac{x}{2 e} exp (2 \frac{x}{e}) (x \to \infty) .

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