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I came across the following inequality in the introduction of Apostol's Calculus, where he discusses the method of exhaustion. I tried to prove it for arbitrary $p$ by induction, but couldn't.

$$ \sum_{k=1}^{n-1} k^p < \frac{n^{p+1}}{p+1} < \sum_{k=1}^{n} k^p \qquad n,p\in\mathbb{N} $$

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  • $\begingroup$ The two sums are left and right Riemann sums for $\int_{0}^n x^p\,dx$. $\endgroup$ – lulu Oct 24 '16 at 1:22
  • $\begingroup$ Hint: integral test $\endgroup$ – Jacky Chong Oct 24 '16 at 1:23
  • $\begingroup$ @lulu: In Apostol's book, it is a pre-integral exercise. $\endgroup$ – Bernard Oct 24 '16 at 1:28
  • $\begingroup$ @Bernard For general $p$? For natural numbers, the middle term is the lead term in the power sum...but for real $p$? $\endgroup$ – lulu Oct 24 '16 at 1:31
  • $\begingroup$ For a natural number, as far as I remember. $\endgroup$ – Bernard Oct 24 '16 at 1:35
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You can do it without the mean value theorem by noting that, if $0 \le a \lt b$, $b^n-a^n =(b-a)\sum_{k=0}^{n-1} a^k b^{n-1-k} $ so $b^n-a^n \lt (b-a)\sum_{k=0}^{n-1} b^k b^{n-1-k} = n(b-a)b^{n-1} $ and $b^n-a^n \gt (b-a)\sum_{k=0}^{n-1} a^k a^{n-1-k} = n(b-a)a^{n-1} $.

If $b = a+1$, this gives $na^{n-1} \lt (a+1)^{n}-a^n \lt n(a+1)^{n-1} $.

Summing for $a=0$ to $m$, $$\sum_{a=0}^m na^{n-1} \lt \sum_{a=0}^m ((a+1)^{n}-a^n) \lt \sum_{a=0}^m n(a+1)^{n-1} $$ or $$n\sum_{a=0}^m a^{n-1} \lt (m+1)^n \lt n\sum_{a=0}^m (a+1)^{n-1} = n\sum_{a=1}^{m+1} a^{n-1} $$ so $$\sum_{a=0}^m a^{n-1} \lt \dfrac{(m+1)^n}{n} \lt \sum_{a=0}^m (a+1)^{n-1} = \sum_{a=1}^{m+1} a^{n-1}. $$

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You can do it by the Mean Value Theorem, if that's "allowed": $$n^{p+1}=\sum_{k=1}^n(k^{p+1}-(k-1)^{p+1})=\sum_{k=1}^n(p+1)c_k^p\ ,$$ where the $c_k$ are certain constants with $$k-1<c_k<k\ .$$ Hence $$n^{p+1}<\sum_{k=1}^n(p+1)k^p\ ,$$ and the other side of the inequality can be done similarly.

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