1. Integral Proof,

Hi,

I am trying to prove:

$\displaystyle \sum_{k=1}^{n-1} k^p < \frac{n^{p+1}}{p+1} < \sum_{k=1}^{n} k^p$

I have tried a few things, my problem is that I cannot seem to find a general formula for $\displaystyle \sum_{k=1}^{n-1} k^p$

I know that $\displaystyle \sum_{k=1}^{n-1} k^2 = \frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}$

But that is only order 2. I don't know if there is a general method for order p.

Thanks
Regards
Craig.

2. Originally Posted by craigmain
Hi,

I am trying to prove:

$\displaystyle \sum_{k=1}^{n-1} k^p < \frac{n^{p+1}}{p+1} < \sum_{k=1}^{n} k^p$

I have tried a few things, my problem is that I cannot seem to find a general formula for $\displaystyle \sum_{k=1}^{n-1} k^p$

I know that $\displaystyle \sum_{k=1}^{n-1} k^2 = \frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}$

But that is only order 2. I don't know if there is a general method for order p.

Thanks
Regards
Craig.
Do you know to convert summations to integrals? Consider the curve $\displaystyle x^p$. If we try to approximate the area by dividing the x-axis into pieces chosen in two different ways, you can solve the problem. Also compare your approximations to the integrals to get the inequalities..

For starters, observe:
$\displaystyle \frac{n^{p+1}}{p+1} < \sum_{k=1}^{n} k^p \Leftrightarrow \frac{1}{p+1} < \frac1{n}\sum_{k=1}^{n} \left(\frac{k}{n}\right)^p$ and also observe that $\displaystyle \int_0^1 x^p \, dx = \frac{1}{p+1}$