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Question Number 177835 by infinityaction last updated on 09/Oct/22

Answered by mr W last updated on 09/Oct/22

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{x}^{n-1}=\frac{1}{1-x}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{\int }_0^x{x}^{n-1}dx={\int }_0^x\frac{1}{1-x}dx$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n}=\log \frac{1}{1-x}$$$$withx=\frac{1}{2}$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2^{n}n}=\log 2$$$$$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2^{n}n}=\underset{n=1}{\sum ^{100}}\frac{1}{2^{n}n}+...>\underset{n=1}{\sum ^{100}}\frac{1}{2^{n}n}$$$$\log 2-\underset{n=1}{\sum ^{100}}\frac{1}{2^{n}n}>0$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2^{n}n}=\underset{n=1}{\sum ^{100}}\frac{1}{2^{n}n}+\frac{1}{2^{100}}(\frac{1}{2\times 101}+\frac{1}{2^2\times 102}+\frac{1}{2^3\times 103}+...)$$$$<\underset{n=1}{\sum ^{100}}\frac{1}{2^{n}n}+\frac{1}{2^{100}}(\frac{1}{2\times 101}+\frac{1}{2^2\times 101}+\frac{1}{2^3\times 101}+...)$$$$=\underset{n=1}{\sum ^{100}}\frac{1}{2^{n}n}+\frac{1}{2^{100}\times 101}(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...)$$$$=\underset{n=1}{\sum ^{100}}\frac{1}{2^{n}n}+\frac{1}{2^{100}\times 101}\times \frac{\frac{1}{2}}{1-\frac{1}{2}}$$$$=\underset{n=1}{\sum ^{100}}\frac{1}{2^{n}n}+\frac{1}{2^{100}\times 101}$$$$\Rightarrow \log 2<\underset{n=1}{\sum ^{100}}\frac{1}{2^{n}n}+\frac{1}{2^{100}\times 101}$$$$\Rightarrow 0<\log 2-\underset{n=1}{\sum ^{100}}\frac{1}{2^{n}n}<\frac{1}{2^{100}\times 101}$$$$\Rightarrow \left(c\right)$$

Commented by Tawa11 last updated on 10/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

Commented by infinityaction last updated on 10/Oct/22

$$thankssir$$

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