Deduce that $\displaystyle 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{ 1}{\sqrt{n}}<2\sqrt{n}$

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- Aug 7th 2011, 05:52 AMPunchDeduce
Deduce that $\displaystyle 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{ 1}{\sqrt{n}}<2\sqrt{n}$

- Aug 7th 2011, 06:16 AMalexmahoneRe: Deduce
Since the question says deduce, I'm not sure if a solution that involves induction is acceptable.

Anyway, using induction,

We will need to prove: $\displaystyle 2\sqrt{n}+\frac{1}{\sqrt{n+1}}<2\sqrt{n+1}$

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

$\displaystyle \Leftrightarrow 2\sqrt{n}\sqrt{n+1}<{2n+1}$

$\displaystyle \Leftrightarrow 4n(n+1)<4n^2+4n+1$

$\displaystyle \Leftrightarrow 0<1$

which is true. - Aug 7th 2011, 10:19 AMCaptainBlackRe: Deduce