# How to prove these inequalities without using Faulhaber´s formula and without using the method of proof by mathematical induction?

In a book about calculus, I have seen these inequalities:

$$1^k+2^k+ ... + (n-1)^k< \dfrac {n^{k+1}}{k+1}<1^k+2^k+ ... + n^k$$ and it is stated that they are valid for every integer $$n \geq1$$ and every integer $$k\geq1$$.

I think that they can be proven in at least two different ways, first would be by using Faulhaber´s formula and, possibly, taking into consideration some properties of Bernoulli numbers.

Second way would be to use method of mathematical induction, first by fixing $$k$$ and proving this for all natural $$n$$ and then by fixing $$n$$ and proving this for all natural $$k$$.

However, I would like to know is there any other approach besides these two that I mentioned?

How to prove these inequalities?

• You can approximate $\int_1^n x^k\mathrm{dx}$ as a Riemann sum. (I may not have the limits of integration exactly right.) – saulspatz May 27 at 14:37
• @saulspatz These inequalities are introduced before the chapter on integration, so, hopefully, could there be some other way? – user677585 May 27 at 14:40
• If I were teaching this material, I would just say that I'll prove after we've done integration. It seems to me that a proof using Bernoulli numbers, or Stirling numbers is way too far afield from a calculus course. Perhaps you should give a little more context to your question. Why do you want to prove this? Is it used in the integration chapter? – saulspatz May 27 at 14:48
• @saulspatz I just want to see ideas about with what methods we could prove this, your idea is nice, and it should work, but I would like to know can we find some other method? I do not think that these inequalities are used in the chapter on integration. – user677585 May 27 at 14:52

Although the inequalities $$\ m^k<\frac{(m+1)^{k+1}-\,m^{k+1}}{k+1}< (m+1)^k\$$ can be obtained from the fact that the middle expression is $$\int_\limits{m}^{m+1} x^kdx\$$, they are also easily provable by appealing to the binomial expansions of $$\ (m+1)^{k+1}\$$ and $$\ (m+1)^k\$$. Summing them from $$\ m=0\$$ to $$\ m=n-1\$$ gives $$\ 1^k+2^k+\dots +(n-1)^k <\frac{n^{k+1}}{k+1} < 1^k+2^k+\dots +n^k\$$.
Use the Mean value theorem for $$f(x)=x^{k+1}$$: $$\frac{(n+1)^{k+1}-n^{k+1}}{n+1-n}=(k+1)\theta_n^k,$$ where $$n<\theta_n and the telescoping sum.