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Question Number 195693 by deleteduser4 last updated on 08/Aug/23

$$hello$$$$\left[\underset{n=1}{\sum ^{10000}}\frac{1}{\sqrt{n}}\right]=?$$$$\left[\right]:isbracket$$$$thankyou$$$$$$

Commented by York12 last updated on 08/Aug/23

$$youmeangreatestintegerfunction$$

Answered by MM42 last updated on 08/Aug/23

$$198$$

Commented by deleteduser4 last updated on 09/Aug/23

$$thankyou$$$$pleasegiveyoursolution$$

Answered by CrispyXYZ last updated on 09/Aug/23

$$\sqrt{n-1}<\sqrt{n}<\sqrt{n+1}$$$$\Rightarrow \sqrt{n}+\sqrt{n-1}<2\sqrt{n}<\sqrt{n}+\sqrt{n+1}$$$$\Rightarrow \frac{1}{\sqrt{n}+\sqrt{n+1}}<\frac{1}{2\sqrt{n}}<\frac{1}{\sqrt{n-1}+\sqrt{n}}$$$$\Rightarrow 2(\sqrt{n+1}-\sqrt{n})<\frac{1}{\sqrt{n}}<2(\sqrt{n}-\sqrt{n-1})$$$$\mathrm{Therefore},$$$$198=2(100-1)<2(\sqrt{10001}-1)<\underset{n=1}{\sum ^{10000}}\frac{1}{\sqrt{n}}=1$$$$+\underset{n=2}{\sum ^{10000}}\frac{1}{\sqrt{n}}<1+2(100-1)=199$$$$\left[\underset{n=1}{\sum ^{10000}}\frac{1}{\sqrt{n}}\right]=198$$

Commented by MM42 last updated on 09/Aug/23

$$$$$$good$$

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