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Question Number 114533 by 675480065 last updated on 19/Sep/20

Answered by abdomsup last updated on 19/Sep/20

$$u_n=\prod _{k=1}^n(1+\frac{k}{n^2})$$$$a)let\phi \left(x\right)=x-ln(1+x)$$$$withx⩾0wehave\phi \left(0\right)=0$$$$and{\phi }^{{}^{\prime }}\left(x\right)=1-\frac{1}{1+x}=\frac{x}{1+x}⩾0$$$$\Rightarrow \phi isincreazingon[0,+\mathrm{\infty }[$$$$\Rightarrow x-ln(1+x)⩾0let$$$$g\left(x\right)=ln(1+x)-x+\frac{x^2}{2}$$$$g\left(0\right)=0and{g}^{{}^{\prime }}\left(x\right)=\frac{1}{1+x}-1+x$$$$=\frac{1}{x+1}+x-1=\frac{1+x^2-1}{x+1}=\frac{x^2}{x+1}⩾0$$$$\Rightarrow gisinc.on[0,+\mathrm{\infty }[\Rightarrow$$$$g⩾0\Rightarrow ln)(1+x)⩾x-\frac{x^2}{2}$$$$finallywegetx-\frac{x^2}{2}⩽ln(1+x)$$$$⩽x$$$$wehaveln\left(u_n\right)=\sum _{k=1}^{n}ln(1+\frac{k}{n^2})$$$$and\frac{k}{n^2}-\frac{k^2}{2n^4}⩽ln(1+\frac{k}{n^2})⩽$$$$\frac{k}{n^2}\Rightarrow \frac{1}{n^2}\sum _{k=1}^{n}k-\frac{1}{2n^4}\sum _{k=1}^n{k}^2$$$$⩽\sum _{k=1}^{n}ln(1+\frac{k}{n^2})⩽\frac{1}{n^2}\sum _{k=1}^{n}k\Rightarrow$$$$\frac{n(n+1)}{2n^2}-\frac{n(n+1)(2n+1)}{12n^4}$$$$⩽ln\left(u_n\right)⩽\frac{n(n+1)}{2n^2}$$$$wepassetolimit)n\to +\mathrm{\infty }$$$$\frac{1}{2}⩽limln\left(u_n\right)⩽\frac{1}{2}\Rightarrow$$$${lim}_{n\to \mathrm{\infty }}ln\left(u_n\right)=\frac{1}{2}$$$$\Rightarrow {lim}_{n\to \mathrm{\infty }}{u}_n=\sqrt{e}$$

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