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Soumik

$\sqrt{1}>\sqrt{\frac{1}{2}}>....\sqrt{\frac{1}{n}}<br/>\text{For 2 sets we have}\sqrt{1},\sqrt{\frac{1}{2}}....\sqrt{\frac{1}{n}};\sqrt{1},\sqrt{\frac{1}{n}}\\ \implies\ (\sqrt{1}+\sqrt{\frac{1}{2}}+...+\sqrt{\frac{1}{n}})^2<n(1+\frac{1}{2}+....\frac{1}{n})$ ............(1)

Again applying Tchebychef's inequality, $\ (1+\frac{1}{2}+....\frac{1}{n})^2<n(1.1+\frac{1}{2}.\frac{1}{2}+....\frac{1}{n^2})\\ \text{Also} -n<0\implies\frac{1}{n(n-1)}<\frac{1}{n^2}$ .......(2)

Putting n=2,3,4.....n,

$\frac{1}{1.2}>\frac{1}{2^2}\\ \frac{1}{2.3}>\frac{1}{3^2}\\ \ .....\\ \.....\\ \frac{1}{(n-1)n}>\frac{1}{n^2}\\ \implies$

Adding all corresponding sides, we have...

$\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{(n-1)n}>\frac{1}{2^2}+\frac{1}{3^2}+...\frac{1}{n^2}\\ \implies\ 1-\frac{1}{2}+\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{n-1}-\frac{1}{n}>\frac{1}{2^2}+\frac{1}{3^2}+...\frac{1}{n^2}$

$\ n(1+\frac{1}{2^2}+...\frac{1}{n^2})<(2n-1).........(3)$

From (2) and (3), $\ (1+\frac{1}{2}+....\frac{1}{n})^2<(2n-1)\\ \implies\ (1+\frac{1}{2}+...\frac{1}{n})<(2n-1)^{\frac{1}{2}}$

From (1) &(4), $\ (\sqrt{1}+\sqrt{\frac{1}{2}}+...\sqrt{\frac{1}{n}})^2<n(2n-1)^{\frac{1}{2}}$

From the above result, we get the expected result....

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