Sequences — Question 9

Question 9 – Solution

Step 1: Insert Stirling’s approximation.

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

Step 2: Apply the nth root and normalize.

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

Step 3: Evaluate the correction factors.

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, $$\boxed{\lim_{n\to\infty}a_n=\frac1e\approx0.367879.}$$

Step 4: Interpret the numerical evidence.

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.