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Question Number 211919 by Spillover last updated on 24/Sep/24

Answered by Frix last updated on 24/Sep/24

$$x^{1+\frac{3}{2}+\frac{4}{4}+\frac{5}{8}+\frac{6}{16}...}=x^5$$$$1+\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{k+2}{2^k}=1+2\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2^k}+\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{k}{2^k}$$$$2\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2^k}=2\times 1$$$$\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{k}{2^k}=\underset{k\to \mathrm{\infty }}{\lim }(2-\frac{k+2}{2^k})=2$$

Commented by Spillover last updated on 24/Sep/24

$$igot1$$

Commented by Frix last updated on 24/Sep/24

$$\mathrm{I}\mathrm{misread}x\sqrt{x^2\sqrt{x^3\sqrt{x^4...}}}$$$$x\sqrt[2]{x\sqrt[3]{x\sqrt[4]{x...}}}=x^{1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+...}=x^{\mathrm{e}-1}$$$$\underset{0}{\stackrel{1}{\int }}{x}^{\mathrm{e}-1}dx={\left[\frac{x^{\mathrm{e}}}{\mathrm{e}}\right]}_0^1=\frac{1}{\mathrm{e}}$$

Answered by Spillover last updated on 25/Sep/24

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