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Question Number 97795 by abdomathmax last updated on 09/Jun/20

$$\mathrm{calculate}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\left[\sqrt{\mathrm{n}}\right]}$$$$[..]\mathrm{meant}\mathrm{the}\mathrm{floor}$$

Answered by bobhans last updated on 10/Jun/20

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

Commented by mathmax by abdo last updated on 11/Jun/20

$$\mathrm{thanks}\mathrm{sir}\mathrm{your}\mathrm{answer}\mathrm{is}\mathrm{correct}$$

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