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Question Number 104604 by ~blr237~ last updated on 22/Jul/20

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}(1+\frac{1}{2}+...+\frac{1}{n})\frac{1}{2^n}=ln4$$

Answered by OlafThorendsen last updated on 22/Jul/20

$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}(1+\frac{1}{2}+...+\frac{1}{n})\frac{1}{2^n}$$$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2^n}\underset{k=1}{\sum ^n}\frac{1}{k}$$$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}\underset{k=n}{\sum ^{\mathrm{\infty }}}\frac{1}{2^k}$$$$\mathrm{S}=\underset{\mathrm{N}\to \mathrm{\infty }}{\lim }\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}.\frac{1}{2^n}.\frac{1-{\left(\frac{1}{2}\right)}^{\mathrm{N}}}{1-\frac{1}{2}}$$$$\mathrm{S}=2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}.\frac{1}{2^n}$$$$\mathrm{S}=2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2^{-n}}{n}$$$$\frac{1}{1-x}=\underset{n=0}{\sum ^{\mathrm{\infty }}}{x}^k,\mid x\mid <1$$$$-\ln \mid 1-x\mid =\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^k}{k}$$$$\mathrm{S}=2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2^{-n}}{n}=-2\ln \mid 1-\frac{1}{2}\mid$$$$\mathrm{S}=2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2^{-n}}{n}=-2\ln \frac{1}{2}=\ln 4$$

Commented by ~blr237~ last updated on 23/Jul/20

$$NiceworkSir$$

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