Hot topic: Limits

$\begin{aligned} \alpha & =\lim_{n \to \infty} \left [ n \left(\frac{e^n n!} {\sqrt{2\pi} n^{n+1/2}}-1\right) \right] \\ & =\lim_{n \to \infty} \left [ n \left(\frac{e^n n \color{#3D99F6}{\Gamma (n)}}{\sqrt{2\pi} n^{n+1/2}}-1\right) \right] \quad \quad \small \color{#3D99F6}{\text{By Stirling series}} \\ & =\lim_{n \to \infty}\left[ n \left(\frac{e^n n}{\sqrt{2\pi} n^{n+1/2}}\cdot \frac{\sqrt{2\pi} n^{n-1/2}} {e^n} \left(1 +\frac {1} {12n} +O\left ( \frac {1} {n^2} \right) \right)-1 \right) \right] \\ & = \lim_{n \to \infty} \left [\frac {1} {12} +O\left ( \frac {1} {n} \right)\right] \\ & = \frac {1} {12} \end{aligned}$

$\implies 84\alpha =\boxed {7}$

We use stirling's approximation, particularly the upper bound for factorials. (Why the upper bound, I have no idea why, dont ask me!)

$\displaystyle \lim_{n \to \infty} (n!) =\sqrt{2\pi n} \cdot( \frac{n}{e})^n \cdot (1+ \frac{1}{12n-1})$ .

Therefore,

$\large \displaystyle \lim_{n \to \infty} n( \frac{e^n n!}{n^{n+1/2} (\sqrt{2\pi})} -1) = \lim_{n \to \infty} n( \frac{e^n (\sqrt{2\pi n} \cdot( \frac{n}{e})^n \cdot (1+ \frac{1}{12n-1}))}{n^{n} (\sqrt{2\pi n})} -1) = \lim_{n \to \infty} n(1+\frac{1}{12n-1}-1)$

Which arises after multiple cancellations and substitution of stirling's approximation.

Thus,

$\displaystyle \lim_{n \to \infty} n(1+\frac{1}{12n+1}-1) = \lim \frac{n}{12n-1} = \lim_{n \to \infty} \frac{1}{12}+\frac{1}{(12)(12n-1)} = \boxed{\frac{1}{12}}$

So, the limit is $\frac{1}{12}$ . Our required answer is $84 \cdot \frac{1}{12} = \boxed{7}$ .