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Question Number 74041 by FCB last updated on 18/Nov/19

Commented by mathmax by abdo last updated on 18/Nov/19

$$let{A}_n={(1+\frac{{(-1)}^n}{n})}^{\frac{1}{sin\left(\pi \sqrt{1+n^2}\right)}}\Rightarrow$$$$ln\left(A_n\right))=\frac{1}{sin\left(\pi \sqrt{n^2+1}\right)}ln(1+\frac{{(-1)}^n}{n})wehave$$$$ln(1+\frac{{(-1)}^n}{n})\sim \frac{{(-1)}^n}{n}and\pi \sqrt{n^2+1}=\pi n\sqrt{1+\frac{1}{n^2}}$$$$\sim n\pi (1+\frac{1}{2n^2})=n\pi +\frac{\pi }{2n}\Rightarrow sin\left(\pi \sqrt{n^2+1}\right)\sim {(-1)}^{n}sin\left(\frac{\pi }{2n}\right)\Rightarrow$$$$ln\left(A_n\right)\sim \frac{{(-1)}^n}{sin\left(\frac{\pi }{2n}\right)}\times \frac{{(-1)}^n}{n}=\frac{1}{nsin\left(\frac{\pi }{2n}\right)}\sim \frac{1}{n\times \frac{\pi }{2n}}=\frac{2}{\pi }$$$${lim}_{n\to +\mathrm{\infty }}{lnA}_n=\frac{2}{\pi }\Rightarrow {lim}_{n\to +\mathrm{\infty }}{A}_n=e^{\frac{2}{\pi }}$$$$anyanswergivenisnotcorrect...!$$

Commented by FCB last updated on 18/Nov/19

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by abdomathmax last updated on 18/Nov/19

$$youarewelcome.$$

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