# binomial limit (1 Viewer)

#### s97127

##### Active Member
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}

$\bg_white \text{Hence }\lim_{n\rightarrow\infty}\left(\prod_{r=0}^n{n\choose r}\right)^\frac{1}{n(n+1)}=e^\frac{1}{2}=\sqrt e$
Someone gives me this solution using Stolz Cesaro as well but i don't really get it. Can you please write down a detailed solution for me? Thanks

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#### tywebb

##### dangerman
I might look at that later but if we call your way Method 4 then there is yet another method using integration.

Method 5

$\bg_white \text{If }\textstyle p_n=\ln\prod_{r=0}^n{n \choose r}=\ln\prod_{r=1}^nr^{2r-n-1}=\sum_{r=1}^n(2r-n)\ln\frac{r}{n}+n\ln n-\ln n!$

$\bg_white \text{noting that }\lim_{n\rightarrow\infty}\frac{n\ln n}{n^2}=\lim_{n\rightarrow\infty}\frac{\ln n!}{n^2}=0\text{ then}$

\bg_white \begin{aligned}\ln\lim_{n\rightarrow\infty}\left(\prod_{r=0}^n{n\choose r}\right)^{\frac{1}{n(n+1)}}&=\lim_{n\rightarrow\infty}\frac{p_n}{n(n+1)}\\&=\lim_{n\rightarrow\infty}\frac{p_n}{n^2}\\&=\lim_{n\rightarrow\infty}\frac{1}{n^2}\sum_{r=1}^n(2r-n)\ln\frac{r}{n}\\&=\lim_{n\rightarrow\infty}\sum_{r=1}^n\left(\frac{2r}{n}-1\right)\left(\ln\frac{r}{n}\right)\cdot\frac{1}{n}\\&=\int_0^1(2x-1)\ln x\ \!dx\\&=\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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#### tywebb

##### dangerman
$\bg_white \text{By the way did you know that for any real number }\alpha,$
$\bg_white \lim_{n\rightarrow\infty}\left(\prod_{r=0}^n{n\choose r}\right)^\frac{1}{n(n+\alpha)}=\sqrt e\ ?$

#### tywebb

##### dangerman
$\bg_white \text{You can generalise it more to }\forall\alpha,\beta\in\Bbb R,\lim\limits_{n\rightarrow\infty}\left(\prod_{r=0}^n{n\choose r}\right)^\frac{1}{(n+\alpha)(n+\beta)}=\sqrt e$

I rewrote the proofs for this more general formula and it is now at https://www.angelfire.com/ab7/fourunit/e-binomial-product.pdf

$\bg_white \text{With }\alpha=\beta=0\text{ we get a formula for }e:$

$\bg_white \textstyle\lim\limits_{n\rightarrow\infty}\sqrt[n]{\prod_{r=0}^n\sqrt[n]{{n \choose r}^2}}=e$

#### juantheron

##### Active Member
Thanks Tywebb for different Methods.