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Question Number 100785 by 175 last updated on 28/Jun/20

Answered by john santu last updated on 28/Jun/20

$$\mathrm{consider}\mathrm{m}⩽\sqrt[3]{\mathrm{n}}⩽\mathrm{m}+1$$$${\mathrm{m}}^3⩽\mathrm{n}⩽{\mathrm{m}}^3+{3\mathrm{m}}^2+1$$$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}+1}}{\lfloor \sqrt{\mathrm{n}}\rfloor }=\underset{\mathrm{m}=1}{\sum ^{\mathrm{\infty }}}\sum _{\lfloor \sqrt{\mathrm{n}}\rfloor =\mathrm{m}}\frac{{(-1)}^{\mathrm{n}+1}}{\mathrm{m}}$$$$=-\underset{\mathrm{m}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{m}}}{\mathrm{m}}=-{\left[\underset{\mathrm{m}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-\mathrm{x})}^{\mathrm{m}}}{\mathrm{m}}\right]}_{\mathrm{x}=0}^{\mathrm{x}=1}$$$$=\underset{0}{\stackrel{1}{\int }}\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}{(-\mathrm{x})}^{\mathrm{k}}\mathrm{dx}=\underset{0}{\stackrel{1}{\int }}\frac{\mathrm{dx}}{1+\mathrm{x}}$$$$={\left[\ln (\mathrm{x}+1)\right]}_0^1=\ln \left(2\right)$$

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