inequality.

by hemangsarkar, Sep 5, 2012, 2:30 PM

Q) prove that –
$1  <  \frac{1}{1001} + \frac{1}{1002} + \frac{1}{1003} +$ ….. $+\frac{1}{3001} < \frac{4}{3}$



My bashed up solution :
let $S$ be $  \frac{1}{1001} + \frac{1}{1002} + \frac{1}{1003} +$ ….. $+\frac{1}{3001}$

Note that :
$(1001)(3001) < (2000)(2002)$

$(1002)(3000) < (2000)(2002)$

………

$(1999)(2003) < (2000)(2002)$

This is because –

$(2001-n)(2001+n) < (2000)(2002)$

Or, $ n^2 > 1$.


Hence we have,

$\frac{1}{1001*3001} > \frac{1}{2000*2002}$

$\frac{1}{1002*3000} > \frac{1}{2000*2002}$
…..
$\frac{1}{1999*2003} > \frac{1}{2000*2002}$

Also, we have –

$\frac{1}{1001} + \frac{1}{3001} = \frac{4002}{1001*3001}$

$\frac{1}{1002} + \frac{1}{3000} = \frac{4002}{1000*3000}$
and so on

So we have

$S > \frac{4002*1000}{2000*2002} + \frac{1}{2001} > 1$

For the other inequality, we use a similar approach –

$(2001 + n)(2001-n) > (1001)(3001)$

That means, $1000 > n$.

n can take values from $0$ to $999$.

So we must have

$(1002)(3000) > (1001)(3001)$

$(1003)(2999) > (1001)(3001)$
….

$(2000)(2002) > (1001)(3001)$


So we have $S < \frac{1000*4002}{1001*3001} + \frac{1}{2001} < \frac{4}{3}$.

Official solution : use am – hm for $ > 1$ part.

And for the other part,

Grouping together $500$ terms at a time,

$S  < \frac{500}{1000} +\frac{500}{1500} + \frac{500}{2000} +\frac{1}{3001} < \frac{4}{3}$
This post has been edited 2 times. Last edited by hemangsarkar, Sep 5, 2012, 2:33 PM

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