MHB Prove Inequality: $18<\sum\limits_{i=2}^{99}\dfrac 1{\sqrt i} <19$

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The discussion focuses on proving the inequality $18 < \sum_{i=2}^{99} \frac{1}{\sqrt{i}} < 19$. Participants explore the implications of the inequality, particularly how it relates to the expressions $\sqrt{n+1} + \sqrt{n} > 2\sqrt{n}$ and $\sqrt{n-1} + \sqrt{n} < 2\sqrt{n}$. The reciprocal forms of these inequalities are also examined, leading to the conclusion that $\frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{2\sqrt{n}} < \frac{1}{\sqrt{n-1} + \sqrt{n}}$. The thread emphasizes the validity and effectiveness of the proposed solutions. Overall, the discussion provides a thorough mathematical exploration of the inequality.
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prove :
$18<1+\dfrac {1}{\sqrt 2}+\dfrac {1}{\sqrt 3}+----+\dfrac{1}{\sqrt {99}}<19$
 
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we have
$n+ 1 \gt n\gt n-1$

hence$\sqrt{n+1} + \sqrt{n}> 2 \sqrt{n} > \sqrt{n-1}+ \sqrt{n}$

hence taking reciprocal
$\dfrac{1}{\sqrt{n+1} + \sqrt{n}}<\dfrac{1}{ 2 \sqrt{n}} < \dfrac{1}{\sqrt{n-1}+ \sqrt{n}} $

or
$\sqrt{n+1} - \sqrt{n}< \dfrac{1}{ 2 \sqrt{n}} < \sqrt{n}- \sqrt{n-1}$

using the above we get the telescopic sum we get
the result
 
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hence$\sqrt{n+1} + \sqrt{n}> 2 \sqrt{n} > \sqrt{n-1}+ \sqrt{n}$

taking reciprocal it should be:
$\dfrac{1}{\sqrt{n+1} + \sqrt{n}}<\dfrac{1}{ 2 \sqrt{n}} < \dfrac{1}{\sqrt{n-1}+ \sqrt{n}} $

or
$\sqrt{n+1} - \sqrt{n}< \dfrac{1}{ 2 \sqrt{n}} < \sqrt{n}- \sqrt{n-1}$

a very good solution !
 
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