7
$\begingroup$

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

$\endgroup$

closed as off-topic by heropup, TZakrevskiy, user147263, dustin, Newb Mar 11 '15 at 0:19

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, TZakrevskiy, Community, dustin, Newb
If this question can be reworded to fit the rules in the help center, please edit the question.

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

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 ?

$\endgroup$
  • 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
1
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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 $n \rightarrow \infty$

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.