As the title says, I want to prove that $\frac{n^{p+1}}{p+1}$ < $\sum_{k=1}^{n}{k^p}$ <$\frac{(n+1)^{p+1}}{p+1}$ $\forall{p,n\in{N}},p\ge2$
The problem is in a section on proving various inequalities, and a single page before there was another problem on proving the Faulhaber's formula.
The tip at the back of the book is to use the Bernoulli's inequality. I tried:
- Expanding the $(n+1)^{p+1}$ term or using Bernoulli's inequality, neither of which lead anywhere,
- Using the fact that $\sum_{k=1}^{n}{k^p}<(\sum_{k=1}^{n}{k})^p$ and then using the Bernoulli's inequality, but that also seems to not lead anywhere,
- Proving for $n=1$, $k=2$, then by induction proving for $n=1$, $k \geq 2$, then proving for all n and k, but i was only able to prove it for $n=1$, $k \geq 2$ since afterwards the expression becomes too unwieldy.
I am not sure how to proceed, and if necessary I can provide my current attempts at solving this thing, although they probably would not be very useful, since they are overly verbose and not very enlightening.
Most likely the solution is somehow derived from using the Faulhaber's formula, multiplying all sides by $p+1$ and then finding some constraints on the bernoulli numbers.
${n^{p+1}}$ < $\sum_{j=0}^{p}{{{p+1}\choose{j}}B_j{n^{p+1-j}}}$ <${(n+1)^{p+1}}$
In the book it is stated that $\sum_{j=0}^{p}{{{p+1}\choose{j}}{B_{j}}=0}$, but since i can not isolate the powers of n from this sum I can not get my head around on how to use it.
