Binomial limit

[man111]

[tex]\displaystyle \lim_{n\rightarrow \infty}\prod^{n}_{k=0}\frac{2n+2k+1}{2n+2k}[/tex]

[Петър Евгениев]

Is it 1 by any chance? (If you have the answer, ofc.)

[drago]

$\sqrt{2}$

[man111]

Thanks friends Got it .

Using A. M>= G.M

[tex]\displaystyle \frac{(n+k)+(n+k+1)}{2}>\sqrt{(n+k)(n+k+1)}\Rightarrow \frac{2n+2k+1}{2n+2k}>\sqrt{\frac{n+k+1}{n+k}}\cdots (1)[/tex]

Also [tex]\displaystyle \frac{(2n+2k)(2n+2k-1)}{2}>\sqrt{(2n+2k+1)(2n+2k-1)}\Rightarrow \frac{2n+2k}{2n+2k+1}>\sqrt{\frac{2n+2k-1}{2n+2k+1}}[/tex]

So we have [tex]\displaystyle \frac{2n+2k+1}{2n+2k}<\sqrt{\frac{2n+2k+1}{2n+2k-1}}\cdots (2)[/tex]

So [tex]\displaystyle \prod^{n}_{k=0}\sqrt{\frac{n+k+1}{n+k}}<\prod^{n}_{k=0}\frac{2n+2k+1}{2n+2k}<\prod^{n}_{k=0}\sqrt{\frac{2n+2k+1}{2n+2k-1}}[/tex]

So [tex]\displaystyle \lim_{n\rightarrow \infty}\sqrt{\frac{2n+1}{n}}<\lim_{n\rightarrow \infty}\prod^{n}_{k=0}\frac{2n+2k+1}{2n+2k}<\lim_{n\rightarrow \infty}\sqrt{\frac{4n+1}{2n-1}}[/tex]

Using squeeze Theorem, We have [tex]\displaystyle \lim_{n\rightarrow \infty}\prod^{n}_{k=0}\frac{2n+2k+1}{2n+2k}=\sqrt{2}.[/tex]