A calculus problem by Aaron Jerry Ninan

$\begin{aligned} L & = \lim_{n \to \infty} \left(\frac {n!}{n^n}\right)^{1/n} \\ & = \exp \left(\lim_{n \to \infty} \frac 1n \ln \left(\frac {n!}{n^n} \right) \right) \\ & = \exp \left({\color{#3D99F6} \lim_{n \to \infty} \frac 1n \sum_{k=1}^n \ln \left(\frac kn \right)} \right) & \small \color{#3D99F6} \text{By Riemann's sum} \\ & = \exp \left({\color{#3D99F6}\int_0^1 \ln(x) \ dx} \right) & \small \color{#3D99F6} \text{See Note.} \\ & = e^{\color{#3D99F6}-1} \end{aligned}$

$\implies k = \boxed{-1}$

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Note: $\ln x$ has a discontinuity at $x=0$ , which produces an improper bound. We can solve the integral by integration by parts with $f = \ln x$ and $dg = 1$ .

$\begin{aligned} I & = \int_0^1 \ln x \ dx \\ & = x \ln x \big|_0^1 - \int_0^1 1 \ dx \\ & = 1 \ln 1 - {\color{#3D99F6}\lim_{x \to 0^+} x \ln x} - x \big|_0^1 & \small \color{#3D99F6} \text{By L'Hôpital's rule: } \lim_{x \to 0^+} x \ln x = 0 \\ & = 0 - {\color{#3D99F6}0} - 1 + 0 \\ & = \boxed {-1} \end{aligned}$

Stirling's approximation:(n!/n^n)^(1/n)=((√(2πn))^(1/n))/e and hence the ans is 1/e