Sequences — Question 9

*Question 9 Let $a_n=\dfrac{(n!)^{1/n}}{n}$ for $n\ge1$.

1. Apply Stirling’s formula $n!\sim\sqrt{2\pi n}(n/e)^n$ to rewrite $a_n$ as a dominant constant times a factor tending to $1$.

2. Find $\lim_{n\to\infty}a_n$ and give its decimal value.

3. Explain why the first dozen terms alone do not reveal the limit accurately.

*Question 9 – Solution

Stirling’s formula states $$n!=\sqrt{2\pi n}\left(\frac ne\right)^n(1+o(1)).$$

All quantities are positive, so taking $n$th roots and dividing by $n$ gives $$a_n=\frac1n(2\pi n)^{1/(2n)}\frac ne\,(1+o(1))^{1/n} =\frac1e(2\pi n)^{1/(2n)}(1+o(1))^{1/n}.$$

The extra factors tend to $1$. For the first factor, take logarithms: $$\ln\left((2\pi n)^{1/(2n)}\right)=\frac{\ln(2\pi n)}{2n}\longrightarrow0,$$ so exponentiating gives $(2\pi n)^{1/(2n)}\to1$. Also, if $u_n=o(1)$, then $(1+u_n)^{1/n}\to1$ because its logarithm is $\ln(1+u_n)/n\to0$. Consequently, $$\boxed{\lim_{n\to\infty}a_n=\frac1e\approx0.367879.}$$

At $n=12$, $a_{12}\approx0.4407$, still about $0.0729$ above $1/e$. The slowly decaying correction factor $(2\pi n)^{1/(2n)}$ explains why a short numerical table is suggestive but cannot identify the hidden constant reliably.