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Question Number 167335 by Jamshidbek last updated on 13/Mar/22

$${\mathrm{x}}_{\mathrm{n}}=\mathrm{n}\cdot \sin (2\pi \mathrm{en}!)$$$$\mathrm{Calculate}$$$$\text{Double subscripts: use braces to clarify}$$

Commented by MJS_new last updated on 13/Mar/22

$$\mathrm{limit}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}$$$$-1⩽\sin (2\pi \mathrm{e}n!)⩽1$$

Commented by Jamshidbek last updated on 13/Mar/22

$$\mathrm{You}\mathrm{are}\mathrm{wrong}\mathrm{thinking}$$

Commented by MJS_new last updated on 13/Mar/22

$$\mathrm{then}\mathrm{explain}$$

Answered by abdomsup last updated on 14/Mar/22

$$\mathrm{n}!\sim {\mathrm{n}}^{\mathrm{n}}.{\mathrm{e}}^{-\mathrm{n}}\sqrt{2\pi \mathrm{n}}\Rightarrow$$$$2\pi \mathrm{en}!\sim 2\pi {\mathrm{n}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{n}+1+\frac{1}{2}}\sqrt{2\pi }$$$$={\left(2\pi \right)}^{\frac{3}{2}}{\mathrm{n}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{n}+\frac{3}{2}}\Rightarrow$$$$\mathrm{nsin}(2\pi \mathrm{en}!)\sim \mathrm{nsin}\left\{{\left(2\pi \right)}^{\frac{3}{2}}{\mathrm{n}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{n}+\frac{3}{2}}\right\}\to 0$$$$$$

Commented by MJS_new last updated on 14/Mar/22

$$2\pi \mathrm{e}n!\approx 2\pi n^{n+\frac{1}{2}}{\mathrm{e}}^{-n+1}\sqrt{2\pi }$$$$={\left(2\pi \right)}^{\frac{3}{2}}{n}^{n+\frac{1}{2}}{\mathrm{e}}^{-n+1}$$$$\mathrm{why}\mathrm{should}\sin \left({\left(2\pi \right)}^{\frac{3}{2}}{n}^{n+\frac{1}{2}}{\mathrm{e}}^{-n+1}\right)\to 0?$$$$\mathrm{explanation}\mathrm{please}$$

Commented by Mathspace last updated on 14/Mar/22

$$e^{-n}defeatallthefunction$$

Commented by MJS_new last updated on 14/Mar/22

$$\mathrm{if}{a}_n\to +\mathrm{\infty }\mathrm{then}\mathrm{\forall }r\in \mathbb{R}\wedge r>1:r\times a_n\to +\mathrm{\infty }$$$$a_n=\frac{n^n}{{\mathrm{e}}^n}\sqrt{2\pi n}\to +\mathrm{\infty }\wedge r=2\pi \mathrm{e}>1\Rightarrow r\times a_n\to +\mathrm{\infty }$$

Answered by qaz last updated on 15/Mar/22

$$\mathrm{en}!=\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\frac{\mathrm{n}!}{\mathrm{k}!}+\underset{\mathrm{k}=\mathrm{n}+1}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{n}!}{\mathrm{k}!}$$$$={\mathbb{N}}^{+}+\frac{1}{\mathrm{n}+1}+\frac{1}{(\mathrm{n}+1)(\mathrm{n}+2)}+...$$$$={\mathrm{N}}^{+}+\frac{1}{\mathrm{n}}(1-\frac{1}{\mathrm{n}}+\frac{1}{{\mathrm{n}}^2}-...)+\frac{1}{{\mathrm{n}}^2}(1-\frac{1}{\mathrm{n}}+\frac{1}{{\mathrm{n}}^2}-...)(1-\frac{2}{\mathrm{n}}+\frac{4}{{\mathrm{n}}^2}-...)+...$$$$={\mathrm{N}}^{+}+\frac{1}{\mathrm{n}}-\frac{1}{{\mathrm{n}}^3}+...$$$$\Rightarrow \underset{\mathrm{n}\to \mathrm{\infty }}{\lim n}\cdot \sin (2\pi \mathrm{en}!)=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim n}\cdot \sin \left[2\pi ({\mathrm{N}}^{+}+\frac{1}{\mathrm{n}}-\frac{1}{{\mathrm{n}}^3}+...)\right]$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim n}\cdot \sin \left[2\pi (\frac{1}{\mathrm{n}}-\frac{1}{{\mathrm{n}}^3}+...)\right]$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim n}\cdot [2\pi (\frac{1}{\mathrm{n}}-\frac{1}{{\mathrm{n}}^3}+...)-\frac{{\left(2\pi \right)}^3}{3!}\cdot {(\frac{1}{\mathrm{n}}-\frac{1}{{\mathrm{n}}^3}+...)}^3+...]$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }[2\pi -\frac{6\pi +4{\pi }^3}{{3\mathrm{n}}^2}+\mathcal{O}\left(\frac{1}{{\mathrm{n}}^3}\right)]$$$$=2\pi$$

Commented by MJS_new last updated on 15/Mar/22

$$\mathrm{great}!$$$$\mathrm{continuous}\mathrm{limit}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}$$$$\mathrm{discrete}\mathrm{limit}=2\pi$$$$\mathrm{I}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{of}\mathrm{this}$$

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