I encountered inequalities $$\sum_{i=1}^{n-1}i^k < \dfrac {n^{k+1}}{k+1} < \sum_{i=1}^{n}i^k$$ valid for every integer $n \geq1$ and every integer $k \geq 1$ (if $n=1$ then we interpret $\sum_{i=1}^{n-1}i^k$ as $0$).

I think that these inequalities could be proved with the help of Faulhaber´s formula or some similar formula and it looks like they could be proved by using method of mathematical induction.

But I would like to know how to prove these inequalities without using Faulhaber´s formula or some similar formula and without using method of mathematical induction?

  • 1
    $\begingroup$ Hint: Use integral to estimate. $\endgroup$ – ntt Apr 10 '17 at 17:51
  • $\begingroup$ What springs to mind is to write the middle term as $\int_0^n x^k\,dx$ and view the first and last terms as Riemann sums. See for instance this older answer: math.stackexchange.com/a/1240555/137524 $\endgroup$ – Semiclassical Apr 10 '17 at 17:52

$\forall x\in (m,m+1)$

$$m^{k}<x^{k}<(m+1)^{k} \tag{$m\in \mathbb{N}$, $k\ge 1$}$$

$$m^{k}<\int_{m}^{m+1} x^{k} \, dx<(m+1)^{k}$$

$$m^{k}< \left[ \frac{x^{k+1}}{k+1} \right]_{m}^{m+1} <(m+1)^{k}$$

$$m^{k}< \frac{(m+1)^{k+1}}{k+1}-\frac{m^{k+1}}{k+1} <(m+1)^{k}$$

$$\sum_{m=1}^{n-1} m^{k}< \frac{n^{k+1}}{k+1}-\frac{1}{k+1} < \sum_{m=1}^{n-1} (m+1)^{k}$$

$$\sum_{m=1}^{n-1} m^{k}< \frac{n^{k+1}}{k+1}-\frac{1}{k+1} < \sum_{m=2}^{n} m^{k}$$

$$\frac{1}{k+1}+\sum_{m=1}^{n-1} m^{k}< \frac{n^{k+1}}{k+1} < \frac{1}{k+1}+ \sum_{m=2}^{n} m^{k}$$

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


Because of the identity $$b^p-a^p=(b-a)(b^{p-1}+b^{p-2}a+...+ba^{p-2}+a^{p-1})$$ we have $$ n^p < \dfrac {(n+1)^{p+1}-n^{p+1}}{p+1} < (n+1)^p$$ and if we now take the sum from $n=1$ to $n=m-1$ we obtain $$\sum_{n=1}^{m-1}n^p<\sum_{n=1}^{m-1} \dfrac {(n+1)^{p+1}}{p+1} - \dfrac {n^{p+1}}{p+1}<\sum_{n=1}^{m-1}(n+1)^p$$ and from here we easily obtain $$\sum_{n=1}^{m-1}n^p< \dfrac {m^{p+1}}{p+1} <\sum_{n=1}^{m}n^p$$

This looks similar to the answer by the user Ng Chung Tak but here the concept of integral is not used so this proof is accessible to those who do not know integral calculus.


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