# Deduce

• August 7th 2011, 05:52 AM
Punch
Deduce
Deduce that $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{ 1}{\sqrt{n}}<2\sqrt{n}$
• August 7th 2011, 06:16 AM
alexmahone
Re: 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: $2\sqrt{n}+\frac{1}{\sqrt{n+1}}<2\sqrt{n+1}$

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

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

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

$\Leftrightarrow 0<1$

which is true.
• August 7th 2011, 10:19 AM
CaptainBlack
Re: Deduce
Quote:

Originally Posted by Punch
Deduce that $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{ 1}{\sqrt{n}}<2\sqrt{n}$

You can to use:

$\sum_{k=1}^n \frac{1}{\sqrt{n}} < \int_1^{n+1}\frac{1}{\sqrt{x}}\; dx$

(but that is a calculus derivation)

CB