need help with Induction

**alpha** · January 18th 2010, 09:35 AM

Sum of n=1 to N of 1/sqrt(n) > sqrt(N) for N>= 2.

Thanks for any help.

**hatsoff** · January 18th 2010, 11:39 AM

Originally Posted by alpha

Show using an inductive or non-inductive argument that $\sum_{n=1}^N\frac{1}{\sqrt{n}} > \sqrt{N}$ for $N\geq 2$ .
Thanks for any help.

Clearly $N<N+1$ for all $N\geq 2$ . Then

$\sqrt{N}<\sqrt{N+1}$

$\implies$ $N<\sqrt{N}\sqrt{N+1}$

$\implies$ $N+1-\sqrt{N}\sqrt{N+1}<1$

$\implies$ $\sqrt{N+1}-\sqrt{N}<\frac{1}{\sqrt{N+1}}$ .

So if $\sqrt{N}<\sum_{n=1}^N\frac{1}{\sqrt{n}}$ , then

$\sqrt{N}+\sqrt{N+1}-\sqrt{N}<\sum_{n=1}^N\frac{1}{\sqrt{n}}+\frac{1}{\ sqrt{N+1}}$

$\implies$ $\sqrt{N+1}<\sum_{n=1}^{N+1}\frac{1}{\sqrt{n}}$ .

Since furthermore $\sqrt{2}<1+\frac{1}{\sqrt{2}}=\sum_{n=1}^2\frac{1} {\sqrt{n}}$ , then $\sum_{n=1}^N\frac{1}{\sqrt{n}} > \sqrt{N}$ for all $N\geq 2$ .

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