2

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?

  • 1
    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 '19 at 14:37
  • @saulspatz These inequalities are introduced before the chapter on integration, so, hopefully, could there be some other way? –  May 27 '19 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 '19 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. –  May 27 '19 at 14:52

2 Answers2

2

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\ $.

lonza leggiera
  • 28,646
  • 2
  • 12
  • 33
0

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 $n<\theta_n<n+1$ and the telescoping sum.

Michael Rozenberg
  • 194,933