Pascal's Limit

$\displaystyle P_n = \prod _{k=0}^n {\begin{pmatrix} n \\ k \end{pmatrix} } = \prod _{k=0}^n {\frac {n!}{k!(n-k)!}} = (n!)^{n+1} \prod _{k=0}^n {\frac {1}{(k!)^2}}$

Therefore, $\displaystyle \quad P_{n+1} = (n+1)!^{n+2} \prod _{k=0}^{n+1} {\frac {1}{(k!)^2}}$

$\Rightarrow \dfrac {P_{n+1}}{P_n} = \dfrac {(n+1)^n} {n!} \quad \Rightarrow \dfrac {P_{n}}{P_{n-1}} = \dfrac {n^{n-1}} {(n-1)!}$

Therefore,

$\displaystyle \lim _{n\rightarrow \infty} {\frac {P_{n-1}P_{n+1}} {P_n^2}} = \lim _{n\rightarrow \infty} {\frac {(n-1)!\space (n+1)^n} {n!\space n^{n-1}}} = \lim _{n\rightarrow \infty} {\frac {(n+1)^n}{n\dot{} n^{n-1}}}$

$\displaystyle = \lim _{n\rightarrow \infty} { \left( \frac {n+1}{n}\right)^n} = \lim _{n\rightarrow \infty} { \left( 1 + \frac {1}{n}\right)^n} = \boxed{e}$

Using brute force method, for n=2, the quotient is 2; n = 3, the quotient is 2.25; n=4, the quotient is 2.37; n=5, the quotient is 2.44 and so on increasing and approaching 2.71828..or e