# binomial limit (1 Viewer)

#### juantheron

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

#### s97127

##### Active Member
Evaluation of $\bg_white \displaystyle \lim_{n\rightarrow \infty}\bigg[\binom{n}{0}\cdot \binom{n}{1}\cdot \binom{n}{2}\cdots \cdots \binom{n}{n}\bigg]^{\frac{1}{n(n+1)}}$
what does (m n) stand for?

#### synthesisFR

##### Well-Known Member
Evaluation of $\bg_white \displaystyle \lim_{n\rightarrow \infty}\bigg[\binom{n}{0}\cdot \binom{n}{1}\cdot \binom{n}{2}\cdots \cdots \binom{n}{n}\bigg]^{\frac{1}{n(n+1)}}$
is it 0

#### Average Boreduser

##### Well-Known Member
actually if we observe newton's laws using SMH and mechanics we can assume the answer is actually 3

#### tywebb

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

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#### tywebb

##### dangerman
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)

\bg_white \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}
\bg_white \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}

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

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

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#### synthesisFR

##### Well-Known Member
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

##### Active Member
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

##### dangerman
This problem is too hard for me.
If I can do it then anyone can do it.

#### synthesisFR

##### Well-Known Member
If I can do it then anyone can do it.
yeah except we dont know half the shi u used

#### s97127

##### Active Member
If I can do it then anyone can do it.
but i have not learnt trigamma, digamma, Taylor series...etc yet

#### tywebb

##### dangerman
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

##### Well-Known Member
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

##### Well-Known Member
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

##### Well-Known Member
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

##### Well-Known Member
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

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

So here is Method 2:

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

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

\bg_white \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}

\bg_white \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}

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#### Average Boreduser

##### Well-Known Member
There is another way to do it using Stirling's formula

So here is Method 2:

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

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

\bg_white \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}

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

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

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

#### tywebb

##### dangerman
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.

\bg_white \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}

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

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