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Show that $1998 < \sum_1^{10^6}\frac{1}{\sqrt{n}}<1999$ by first recalling the inequality: $$2(\sqrt{n+1}-\sqrt{n})<\frac{1}{\sqrt{n}}<2(\sqrt{n}-\sqrt{n-1})$$ when $n=1, 2, 3...$

What I have tried to show using the inequality:

By the first step of induction to see if the inequality holds : $2(\sqrt{1+1}-\sqrt{1})<\frac{1}{\sqrt{1}}<2(\sqrt{1}-\sqrt{1-1})$ is true

Then by induction we have:

$2(\sqrt{n+1+1}-\sqrt{n+1})<\frac{1}{\sqrt{n+1}}<2(\sqrt{n+1}-\sqrt{n+1-1})$

Then by using the fact that if $a>b$ and $c>0$ then $ac>bc$

The following should be true if the induction is right.

$\frac{2(\sqrt{n+1}-\sqrt{n})}{\sqrt{n+1}}>2(\sqrt{n+1+1}-\sqrt{n+1})$ by taking the square of both sides we have after expanding: $$\frac{8n-8\sqrt{n^2+n}+4}{n+1}>8n+12-8\sqrt{n^2+3n+2}$$

Then placing everything onto one side: $$\frac{-8n^2+8n\sqrt{n^2+3n+2}-12n+8\sqrt{n^2+3n+2}-8\sqrt{n^2+n}-8}{n+1}>0$$

Thouigh I'm not sure whether I'm going down the right direction. I had assumed that the LHS would cancel to provide something more exact. Any detailed hint son how to proceed?

Stackbeans
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2 Answers2

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You don't need induction here. We have $$n(n+1)=n^2+n<n^2+n+\frac14=\left(n+\frac12\right)^2$$ Take square roots: $$\sqrt n\sqrt{n+1}<n+\frac12$$ Multiply by $2/\sqrt n$ and rearrange: $$2(\sqrt{n+1}-\sqrt n)<\frac{1}{\sqrt n}$$

Do the same with $n(n-1)$ to get the right-hand side of the inequality.

TonyK
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More directly, multiply and divide by the conjugate:

$2 ( \sqrt{n+1} - \sqrt{n} ) = 2 \frac{ ( \sqrt{n+1} - \sqrt{n} ) ( \sqrt{n+1} + \sqrt{n} ) }{ ( \sqrt{n+1} + \sqrt{n} )} = \frac{2}{ ( \sqrt{n+1} + \sqrt{n} )} < \frac{2}{2\sqrt{n}} = \frac{1}{ \sqrt{n}}. $

Same for the RHS, can you complete it?

Calvin Lin
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