binomial limit (1 Viewer)

juantheron · Mar 17, 2023

Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\bigg[\binom{n}{0}\cdot \binom{n}{1}\cdot \binom{n}{2}\cdots \cdots \binom{n}{n}\bigg]^{\frac{1}{n(n+1)}}$

s97127 · Mar 17, 2023

juantheron said:
Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\bigg[\binom{n}{0}\cdot \binom{n}{1}\cdot \binom{n}{2}\cdots \cdots \binom{n}{n}\bigg]^{\frac{1}{n(n+1)}}$

what does (m n) stand for?

synthesisFR · Mar 17, 2023

juantheron said:
Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\bigg[\binom{n}{0}\cdot \binom{n}{1}\cdot \binom{n}{2}\cdots \cdots \binom{n}{n}\bigg]^{\frac{1}{n(n+1)}}$

is it 0

Average Boreduser · Mar 17, 2023

actually if we observe newton's laws using SMH and mechanics we can assume the answer is actually 3

tywebb · Mar 18, 2023

I put it into wolframalpha and got $\sqrt{e}$

synthesisFR · Mar 18, 2023

tywebb said:
I put it into wolframalpha and got $\sqrt{e}$
View attachment 37999

canu explain how to do it

tywebb · Mar 19, 2023

This is how I did it without wolframalpha. First use a logarithm to turn the product into a sum so it is easier to deal with. Thereafter it is a matter of simplifying the limit as much as possible.

Method 1 (see more methods in subsequent posts)

$\begin{aligned}\ln\lim_{n\rightarrow\infty}\left(\prod_{r=0}^n{n\choose r}\right)^{\frac{1}{n(n+1)}}&=\ln\left(\lim_{n\rightarrow\infty}\prod_{r=1}^nr^{2r-n-1}\right)^{\frac{1}{n(n+1)}}\text{ algebraic simplification }\\&=\lim_{n\rightarrow\infty}\sum_{r=1}^n\frac{(2r-n-1)\ln r}{n(n+1)}\text{ by log laws}\\&=\lim_{x\rightarrow\infty}\frac{2\ln H(x)-(x+1)\ln\Gamma(x+1)}{x^2+x}\text{ where }H=\text{the hyperfactorial function and }\Gamma=\text{the gamma function}\\&=\lim_{x\rightarrow\infty}\frac{\frac{d}{dx}(2\ln H(x)-(x+1)\ln\Gamma(x+1))}{\frac{d}{dx}(x^2+x)}\text{ using l'H\^opital's rule}\\&=\lim_{x\rightarrow\infty}\frac{\ln\Gamma(x+1)+2x-(x+1)\psi(x+1)-\ln(2\pi)+1}{2x+1}\text{ where }\psi=\text{the digamma function}\end{aligned}$
$\begin{aligned}\color{white}{\ln\lim_{n\rightarrow\infty}\left(\prod_{r=0}^n{n\choose r}\right)^{\frac{1}{n(n+1)}}}&=\lim_{x\rightarrow\infty}\frac{\frac{d}{dx}(\ln\Gamma(x+1)+2x-(x+1)\psi(x+1)-\ln(2\pi)+1)}{\frac{d}{dx}(2x+1)}\text{ by l'H\^opital again}\\&=\lim_{x\rightarrow\infty}\textstyle(1-\frac{1}{2}(x+1)\psi^{(1)}(x+1))\text{ where }\psi^{(1)}=\text{the trigamma function}\\&=\lim_{x\rightarrow\infty}\textstyle(1-\frac{1}{2}(1+\frac{1}{2x}-\frac{1}{3x^2}+\frac{1}{6x^3}-\frac{1}{30x^4}+\cdots))\text{ by Laurent series expansion}\color{white}{\text{on and }\Gamma=\text{the gamma function}}\\&=\textstyle1-\frac{1}{2}\\&=\textstyle\frac{1}{2}\end{aligned}$

$\text{whereupon exponentiation with base }e\text{ we find that$

$\lim_{n\rightarrow\infty}\left(\prod_{r=0}^n{n\choose r}\right)^\frac{1}{n(n+1)}=e^\frac{1}{2}=\sqrt e$

synthesisFR · Mar 19, 2023

Ok so its not a 4u question
WHY TF WOULD this person put it in the extension 2 thread i almost was gonna drop 4U looking at this working out

s97127 · Mar 19, 2023

synthesisFR said:
Ok so its not a 4u question
WHY TF WOULD this person put it in the extension 2 thread i almost was gonna drop 4U looking at this working out

This problem is too hard for me.

tywebb · Mar 19, 2023

s97127 said:
This problem is too hard for me.

If I can do it then anyone can do it.

synthesisFR · Mar 19, 2023

tywebb said:
If I can do it then anyone can do it.

yeah except we dont know half the shi u used

s97127 · Mar 19, 2023

tywebb said:
If I can do it then anyone can do it.

but i have not learnt trigamma, digamma, Taylor series...etc yet

tywebb · Mar 19, 2023

The gamma function is just a continuous version of factorial and the digamma and trigamma just come from the derivatives.

The hardest function that appeared was the hyperfactorial but I got rid of it pretty quickly by using l'Hôpital's rule.

That's a nifty rule to use in limits - even for much simpler questions.

synthesisFR · Mar 19, 2023

tywebb said:
The gamma function is just a continuous version of factorial and the digamma and trigamma just come from the derivatives.

The hardest function that appeared was the hyperfactorial but I got rid of it pretty quickly by using l'Hospital rule.

That's a nifty rule to use - even for much simpler questions.

do u do maths at uni?

Average Boreduser · Mar 19, 2023

idk bruh. It's pretty damn simple

just limit properties and gamma function is just a continuous version of factorial and the digamma and trigamma just come from the derivatives. hardest function that appeared was the hyperfactorial but I got rid of it pretty quickly by using l'Hospital rule.
That's a nifty rule to use - even for much simpler questions. But realistically speaking if ur doing 4U and u cant even do smn as simple as that bruh

good luck bud.

synthesisFR · Mar 19, 2023

Pethmin said:
But realistically speaking if ur doing 4U and u cant even do smn as simple as that bruh good luck bud.

yeah man...
im considering to drop it tbh, even 3u for that matter. im just not good enough ...

Average Boreduser · Mar 19, 2023

synthesisFR said:
yeah man...
im considering to drop it tbh, even 3u for that matter. im just not good enough ...

honestly just dont do math at all for that matter, Ik kindergaetners who could do this qn (they go to ruse)

tywebb · Mar 20, 2023

There is another way to do it using Stirling's formula

So here is Method 2:

$\text{If }p(n)=\left(\prod_{r=0}^n{n \choose r}\right)^\frac{1}{n(n+1)}\text{ then }p(n)^{n(n+1)}=p(n-1)^{n(n-1)}\cdot\frac{n^n}{n!}.$

$\text{ If }L=\lim_{n\rightarrow\infty}p(n)\text{ then }$

$\begin{aligned}\lim_{n\rightarrow\infty}\frac{L^{n(n+1)}}{L^{n(n-1)}\cdot\frac{n^n}{n!}}&=\lim_{n\rightarrow\infty}\frac{L^{2n}n!}{n^n}\\&=\lim_{n\rightarrow\infty}\left(L^{2n}\cdot\frac{\sqrt{2\pi n}}{e^n}\right)\cdot\lim_{n\rightarrow\infty}\left(\frac{n!}{n^n}\cdot\frac{e^n}{\sqrt{2\pi n}}\right)\\&=\lim_{n\rightarrow\infty}\left(L^{2n}\cdot\frac{\sqrt{2\pi n}}{e^n}\right)\cdot1\text{ by Stirling's formula}\\&=1\end{aligned}$

$\begin{aligned}\therefore L&=\lim_{n\rightarrow\infty}\left(\frac{e^n}{\sqrt{2\pi n}}\right)^\frac{1}{2n}\\&=\sqrt{e}\cdot\lim_{n\rightarrow\infty}\left(\frac{1}{\sqrt{2\pi}}\right)^\frac{1}{2n}\cdot\sqrt[4]{\lim_{x\rightarrow 0^+}x^x}\text{ if }x=\frac{1}{n}\\&=\sqrt{e}\cdot1\cdot1\\&=\sqrt{e}\end{aligned}$

Average Boreduser · Mar 20, 2023

tywebb said:
There is another way to do it using Stirling's formula

So here is Method 2:

$\text{If }p(n)=\left(\prod_{r=0}^n{n \choose r}\right)^\frac{1}{n(n+1)}\text{ then }p(n)^{n(n+1)}=p(n-1)^{n(n-1)}\cdot\frac{n^n}{n!}.$

$\text{ If }L=\lim_{n\rightarrow\infty}p(n)\text{ then }$

$\begin{aligned}\lim_{n\rightarrow\infty}\frac{L^{n(n+1)}}{L^{n(n-1)}\cdot\frac{n^n}{n!}}&=\lim_{n\rightarrow\infty}\frac{L^{2n}n!}{n^n}\\&=\lim_{n\rightarrow\infty}\left(L^{2n}\cdot\frac{\sqrt{2\pi n}}{e^n}\right)\cdot\lim_{n\rightarrow\infty}\left(\frac{n!}{n^n}\cdot\frac{e^n}{\sqrt{2\pi n}}\right)\\&=\lim_{n\rightarrow\infty}\left(L^{2n}\cdot\frac{\sqrt{2\pi n}}{e^n}\right)\cdot1\text{ by Stirling's formula}\\&=1\end{aligned}$

$\begin{aligned}\therefore L&=\lim_{n\rightarrow\infty}\left(\frac{e^n}{\sqrt{2\pi n}}\right)^\frac{1}{2n}\\&=\sqrt{e}\cdot\lim_{n\rightarrow\infty}\left(\frac{1}{\sqrt{2\pi}}\right)^\frac{1}{2n}\cdot\sqrt[4]{\lim_{x\rightarrow 0^+}x^x}\text{ if }x=\frac{1}{n}\\&=\sqrt{e}\cdot1\cdot1\\&=\sqrt{e}\end{aligned}$

r u sure its
$\text{If }p(n)=\left(\prod_{r=0}^n{n \choose r}\right)^\frac{1}{n(n+1)}\text{ then }p(n)^{n(n+1)}=p(n-1)^{n(n-1)}\cdot\frac{n^n}{n!}.$

$\text{ If }L=\lim_{n\rightarrow\infty}p(n)\text{ then }$ ? I got
$\text{If }x(n)=\left(\prod_{r=0}^y{n \choose r}\right)^\frac{1}{n(n+1)}\text{ then }a(r)^{b(z+t)}=u(ny-1)^{n(n-1)}\cdot\frac{n^n}{n!}.$

$\text{ If }L=\lim_{n\rightarrow\infty}p(n)\text{ then }$

tywebb · Mar 24, 2023

Here is another method using the Stolz–Cesàro theorem. It starts the same as the first method but then goes in a completely different direction, nevertheless ending up at the same answer.

Method 3.

$\begin{aligned}\ln\lim_{n\rightarrow\infty}\left(\prod_{r=0}^n{n\choose r}\right)^{\frac{1}{n(n+1)}}&=\ln\left(\lim_{n\rightarrow\infty}\prod_{r=1}^nr^{2r-n-1}\right)^{\frac{1}{n(n+1)}}\text{ algebraic simplification }\\&=\lim_{n\rightarrow\infty}\sum_{r=1}^n\frac{(2r-n-1)\ln r}{n(n+1)}\text{ by log laws}\\&=\lim_{n\rightarrow\infty}\frac{2\ln\prod_{r=1}^n r^r-(n+1)\ln n!}{n(n+1)}\\&=\lim_{n\rightarrow\infty}\frac{2\ln\prod_{r=1}^{n+1}r^r-(n+2)\ln(n+1)!-(2\ln\prod_{r=1}^nr^r-(n+1)\ln n!)}{(n+1)(n+2)-n(n+1)}\text{ using the Stolz-Ces\`{a}ro theorem}\\&={\textstyle\frac{1}{2}}\lim_{n\rightarrow\infty}\frac{\ln(n+1)^{n+1}-\ln(n+1)!}{n+1}\\&={\textstyle\frac{1}{2}}\lim_{n\rightarrow\infty}\frac{\ln(n+1)^{n+1}-\ln(n+1)!-(\ln n^n-\ln n!)}{n+1-n}\text{ by Stolz-Ces\`{a}ro again}\\&={\textstyle\frac{1}{2}}\lim_{n\rightarrow\infty}\textstyle{\ln(1+\frac{1}{n})^n}\\&=\textstyle\frac{1}{2}\ln e\\&=\textstyle\frac{1}{2}\end{aligned}$

$\text{Hence }\lim_{n\rightarrow\infty}\left(\prod_{r=0}^n{n\choose r}\right)^\frac{1}{n(n+1)}=e^\frac{1}{2}=\sqrt e$

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