1. ## Proof By Inducion.

So I need to proof the following inequality

$\sum_{k\ =\ 0}^{n-1} k^{p} < \frac{n^{p+1}}{p+1} < \sum_{k\ =\ 0}^n k^{p}$

If i start it by induction, I get to a point where I have to proove that

$k^p (k+p+1) < (k+1)^{p+1}$ (with the first part)

And I have no idea where to go from here. Any suggestions?

2. Hello,

Originally Posted by ml_lulu
So I need to proof the following inequality

$\sum_{k\ =\ 0}^{n-1} k^{p} < \frac{n^{p+1}}{p+1} < \sum_{k\ =\ 0}^n k^{p}$

If i start it by induction, I get to a point where I have to proove that

$k^p (k+p+1) < (k+1)^{p+1}$ (with the first part)

And I have no idea where to go from here. Any suggestions?
$(k+1)^{p+1}=k^{p+1}+{\color{red}C_{p+1}^p k^p}+{\color{blue}C_{p+1}^{p-1} k^{p-1}+\dots+C_{p+1}^1 k+1}$

${\color{red}C_{p+1}^p}=\frac{(p+1)!}{p!(p+1-p)!}={\color{red}p+1}$

All the blue part is positive (and different to 0 unless k is somehow negative), do you agree ?

Therefore, we get :

$(k+1)^{p+1}={\color{magenta}k^{p+1}+(p+1)k^p}+{\co lor{blue}\text{blue thing}} > {\color{magenta} k^{p+1}+(p+1)k^p}=k^p(k+p+1) \ \square$

Does it make sense ?

3. Hi

Another possible approach, without induction :

Let $k\in\mathbb{N}$.

Using the properties of the integral, $\int_k^{k+1} t^p\,\mathrm{d}t \geq \int_k^{k+1} k^p\,\mathrm{d}t=(k+1-k)k^p=k^p$

Summing these inequalities for $k\in\{0,\,1\ldots,\,n-1\}$ gives $\sum_{k=0}^{n-1} k^p\leq \sum_{k=0}^{n-1} \int_k^{k+1} k^p\,\mathrm{d}t$

The right hand side can be rewritten as $\sum_{k=0}^{n-1} \int_k^{k+1} k^p\,\mathrm{d}t=\int_0^{1} k^p\,\mathrm{d}t+\int_1^{2} k^p\,\mathrm{d}t+\int_2^{3} k^p\,\mathrm{d}t+\ldots+\int_{n-1}^{n} k^p\,\mathrm{d}t=\int_0^n k^p\,\mathrm{d}t$ hence $\sum_{k=0}^{n-1} k^p\leq\int_0^n k^p\,\mathrm{d}t$.

As $\int_0^{n} k^p\,\mathrm{d}t=\left[ \frac{t^{p+1}}{p+1}\right]_0^n=\frac{n^{p+1}}{p+1}$, here is the first inequality : $\boxed{\sum_{k=0}^{n-1} k^p\leq\frac{n^{p+1}}{p+1}}$.

Using the same method, one can show the other one.