Find the Pascal's Limit Let $P_{n}$ be the product of the numbers in row  of Pascal's Triangle. Then evaluate $$ \lim_{n\rightarrow \infty} \dfrac{P_{n-1}\cdot P_{n+1}}{P_{n}^{2}}$$
 A: Nice question :)
$P_{n} = \prod_{k=0}^{n}\binom{n}{k} $ 
$=  \prod_{k=0}^{n} \dfrac{n!}{(n-k)!\cdot k!}$
$ = n!^{n+1} \prod_{k=0}^{n} \dfrac{1}{k!^{2}}$
$ \therefore P_{n+1} = (n+1)!^{n+2} \prod_{k=0}^{n+1} \dfrac{1}{k!^{2}}$
$ \Rightarrow \dfrac{P_{n+1}}{P_{n}}=\dfrac{(n+1)^{n}}{n!} ,\dfrac{P_{n}}{P_{n-1}}=\dfrac{(n)^{n-1}}{(n-1)!} $
Now the question asks for, 
$\lim_{n\rightarrow \infty} \dfrac{P_{n-1}P_{n+1}}{P_{n}^{2}} $
So we have , 
$
\lim_{n\rightarrow \infty} \dfrac{P_{n-1}P_{n+1}}{P_{n}^{2}}  = \lim_{n\rightarrow \infty} \dfrac{(n-1)!(n+1)^{n}}{n!\times n^{n-1}}$
$ = \lim_{n\rightarrow \infty} \dfrac{(n+1)^{n}}{n\times n^{n-1}}$
$ = \lim_{n\rightarrow \infty} \left ( \dfrac{n+1}{n}  \right )^{n} $
$ = \lim_{n\rightarrow \infty} \left ( 1 + \frac{1}{n} \right )^{n}$
$ = e $
Btw is where did you get this question from ?
A: The product of terms in the nth line of a Pascal triangle is given by the product of binomial
$$ P_n = \prod\limits_{k=0}^n \binom{n}{k}$$
So your expression evaluates to
$$\lim_{n\rightarrow\infty} \prod_{k=0}^n \prod_{k'=0}^{n+1} \prod_{k''=0}^{n-1}\frac{(n-1)! (n+1)!((n-k)!)^2(k!)^2}{(n!)^2(n-k'-1)!  (n-k''+1)!} = \lim_{n\rightarrow\infty}(n+1)\left(\frac{n+1}{n}\right )^n\frac{n!}{(n+1)!} = e$$
Does this makes sense? 
A: We have that $P_n = \prod_{i=0}^n \binom n i = \prod_{i=0}^n \frac {n!}{i! (n-i)!}$.
So $\frac { P_{n-1} P_{n+1} } {P_n} = (n+1) \prod_{i=1}^{n-1} \frac{(n-1)! (n+1)! i!^2 (n-i)!^2} {n!^2 i!^2 (n-i-1)! (n-i+1)!} = (n+1) \prod_{i=1}^{n-1} \frac{n+1}{n} \frac{n-i}{n-i+1} = \frac {(n+1)^n} {n^{n-1}} \frac {1} {n}$.
Thus, we get that $\frac { P_{n-1} P_{n+1} } {P_n} = (1+1/n)^n$ which is equal to $e$ when $n \rightarrow \infty$
