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Question Number 161745 by HongKing last updated on 21/Dec/21

$$\mathrm{Prove}\mathrm{that}:$$$$3\sqrt{\mathrm{e}}=\frac{1}{3}\underset{\mathbf{k}=1}{\sum ^{\mathrm{\infty }}}\left[\underset{\mathbf{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{n}!}\right]{\mathrm{k}2}^{-\mathbf{k}}$$

Commented by mr W last updated on 22/Dec/21

$$questionseemsfalse.$$

Answered by mr W last updated on 22/Dec/21

$$\underset{k=1}{\sum ^{\mathrm{\infty }}}{x}^k=\frac{x}{1-x}for\mid x\mid <1$$$$\underset{k=1}{\sum ^{\mathrm{\infty }}}{kx}^{k-1}=\frac{1}{1-x}+\frac{x}{{(1-x)}^2}$$$$\underset{k=1}{\sum ^{\mathrm{\infty }}}{kx}^k=\frac{x}{1-x}+\frac{x^2}{{(1-x)}^2}$$$$withx=\frac{1}{2}$$$$\Rightarrow \underset{k=1}{\sum ^{\mathrm{\infty }}}k{2}^{-k}=\frac{\frac{1}{2}}{1-\frac{1}{2}}+\frac{{\left(\frac{1}{2}\right)}^2}{{(1-\frac{1}{2})}^2}=2$$$$e^x=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}$$$$withx=1$$$$\Rightarrow \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}=e$$$$$$$$\frac{1}{3}\underset{\mathbf{k}=1}{\sum ^{\mathrm{\infty }}}\left[\underset{\mathbf{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{n}!}\right]{\mathrm{k}2}^{-\mathbf{k}}$$$$=\frac{1}{3}\underset{\mathbf{k}=1}{\sum ^{\mathrm{\infty }}}\left(e\right){\mathrm{k}2}^{-\mathbf{k}}$$$$=\frac{e}{3}\underset{\mathbf{k}=1}{\sum ^{\mathrm{\infty }}}{\mathrm{k}2}^{-\mathbf{k}}$$$$=\frac{e}{3}\times 2$$$$=\frac{2e}{3}\ne 3\sqrt{e}$$

Commented by HongKing last updated on 22/Dec/21

$$\mathrm{SORRY}\mathrm{my}\mathrm{dear}\mathrm{Sir},[\underset{\mathbf{n}=0}{\sum ^{\mathbf{k}}}...$$$$\mathrm{no}\mathrm{\infty },\mathbf{k}$$

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