Let $P_{n}$ be the product of the numbers in row of Pascal's Triangle. Then evaluate $$ \lim_{n\rightarrow \infty} \dfrac{P_{n-1}\cdot P_{n+1}}{P_{n}^{2}}$$


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  • 8
    $\begingroup$ . . . and you tried . . . what? $\endgroup$ – imallett Mar 10 '15 at 16:13

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 ?

  • 3
    $\begingroup$ It's from an old calculus book of my father's $\endgroup$ – Andy Mar 10 '15 at 13:05
  • $\begingroup$ Could you explain how you got from step 2 to 3? $\endgroup$ – Brad Mar 10 '15 at 18:54
  • $\begingroup$ @Brad It is a product of three terms multiplied together, so it splits into $\prod_{k=0}^{n}n!\prod_{k=0}^{n}\frac1{(n-k)!}\prod_{k=0}^{n}\frac1{k!}$; then you evaluate the first product, reverse the second, and put them back together. $\endgroup$ – Mario Carneiro Mar 10 '15 at 20:08
  • $\begingroup$ @Andy: See math.stackexchange.com/a/700946/215011 , which references johncarlosbaez.wordpress.com/2014/02/12/triangular-numbers . So this result was known before 2012? $\endgroup$ – grand_chat Mar 10 '15 at 22:13

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?


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 $n \rightarrow \infty$

  • $\begingroup$ ...which tends to $e$, not 'is equal' to $e$. $\endgroup$ – CiaPan Mar 10 '15 at 16:44

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