1. ## Acceptable proof?

Hi,

Is this an acceptable proof of induction?

show that, for all n >= 2

$\frac{1}{2}n^{3/2} \textless \sum_{k=1}^n\sqrt{k}~\textless~n^{3/2}$

multiply by 2 and verify the statement holds for n=2

$\sqrt{2^3} = 2\sqrt2 ~\textless~ 2\sum_{k=1}^n 1 + \sqrt{2} ~\textless~ 2\sqrt{8}$

The inductive step:

$n^{3/2}+(n+1)^{1/2} \rightarrow n^2 +1 \textless \sum_{k=1}^{n+1}\sqrt{k}~\textless~2n^{3/2} + (n+1)^{1/2} \rightarrow 2n^2 +1$

Acceptable?

2. 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 ( $\sqrt{2^3} \neq \sqrt{2}$...) or what you did in the final step...

for n=2 it should be:

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

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

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

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

Now please explain in more detail the final step

3. 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 ( $\sqrt{2^3} \neq \sqrt{2}$...) or what you did in the final step...

for n=2 it should be:

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

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

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

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

Now please explain in more detail the final step
Hi,
Sorry it was a typo
I think you figured it out, was i meant was that $2^{3/2} = \sqrt{2^3} = 2\sqrt{2}$