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Question Number 205716 by universe last updated on 28/Mar/24

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{(2n+1)(2n+3)...(4n+1)}{\left(2n\right)(2n+2)...\left(4n\right)}=?$$

Answered by MM42 last updated on 28/Mar/24

$$\frac{\left(2n\right)(2n+2)...\left(4n\right)}{(2n-1)(2n+1)...(4n-1)}<A<\frac{(2n+2)(2n+4)...(4n+2)}{(2n+1)(2n+3)...(4n+1)}$$$$\Rightarrow \frac{4n+1}{(2n-1)A}<A<\frac{4n+2}{\left(2n\right)A}$$$$\Rightarrow \frac{4n+1}{2n-1}<A^2<\frac{4n+2}{2n}$$$$\Rightarrow {lim}_{n\to \mathrm{\infty }}{A}^2=2$$$$\Rightarrow {lim}_{n\to \mathrm{\infty }}A=\sqrt{2}✓$$$$$$

Answered by Berbere last updated on 28/Mar/24

$$=\underset{n\to \mathrm{\infty }}{\lim }\underset{k=0}{\prod ^n}\frac{(2n+2k+1)}{(2n+2k)}=\underset{n\to \mathrm{\infty }}{\lim }{U}_n$$$$U_n=e^{\underset{k=0}{\sum ^n}ln\left(\frac{2k+2n+1}{2n+2k}\right)}=e^{\underset{k=0}{\sum ^n}ln(1+\frac{1}{2n+2k})}=e^{V_n}$$$$\mathrm{\forall }x\in [0,1[$$$$x-\frac{x^2}{2}⩽ln(1+x)⩽x;$$$$proof(1-x^2)⩽1\Rightarrow 1-x⩽\frac{1}{1+x}\Rightarrow {\int }_0^{t}1-x⩽{\int }_0^t\frac{dx}{1+x}$$$$\iff t-\frac{t^2}{2}⩽ln(1+x);\frac{1}{1+x}⩽1\Rightarrow ln(1+x)⩽x$$$$\Rightarrow \frac{1}{2(n+k)}-\frac{1}{8{(n+k)}^2}⩽ln(1+\frac{1}{2(n+k)})⩽\frac{1}{2(n+k)}$$$$\Rightarrow \frac{1}{2n}\underset{k=0}{\sum ^n}\frac{1}{1+\frac{k}{n}}-\frac{1}{8}{T}_n⩽V_n⩽\frac{1}{2n}\underset{k=0}{\sum ^n}\frac{1}{1+\frac{k}{n}}$$$$0⩽T_n=\underset{k=0}{\sum ^n}\frac{1}{{(n+k)}^2}⩽\frac{n+1}{n^2}\to 0$$$$\underset{n\to \mathrm{\infty }}{\lim }{V}_n=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{2n}\underset{k=0}{\sum ^n}\frac{1}{1+\frac{k}{n}}RiemannSum=\frac{1}{2}{\int }_0^1\frac{1}{1+x}dx=\frac{ln\left(2\right)}{2}$$$$U_n=\underset{n\to \mathrm{\infty }}{\lim }{e}^{V_n}=e^{\frac{1}{2}ln\left(2\right)}=\sqrt{2}$$$$$$

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