Originally Posted by

**Defunkt** It would be a lot easier (for me, at least) to help you if you gave more detail on your steps...

I have no idea how you ended up with that for n=2 ($\displaystyle \sqrt{2^3} \neq \sqrt{2}$...) or what you did in the final step...

for n=2 it should be:

$\displaystyle \frac{\sqrt{2^3}}{2}=\sqrt{\frac{2^3}{2^2}} = \sqrt{2}$

$\displaystyle \sum_{k=1}^{2}\sqrt{k} = 1 + \sqrt{2}> \sqrt{2}$

$\displaystyle \sqrt{2^3}=\sqrt{8} = 2\sqrt{2} > 1 + \sqrt{2}$

So $\displaystyle \sqrt{2}<1+\sqrt{2}<2\sqrt{2}$ is true...

Now please explain in more detail the final step