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Prove for any $a >0$ and $n \in \mathbb{Z^+} $

$$1^a+2^a+\cdots+n^a < \frac{(n+1)^{(a+1)}-1}{a+1}$$

Also for $a \in (-1,0)$ the above inequality is reversed.

For $n=1, 2^{(a+1)}-1 > (a+1)$ is true

Let us assume the result is true for $n=m$, i.e.

$$1^a+2^a+\cdots+m^a < \frac{(m+1)^{(a+1)}-1}{a+1}$$


$$1^a+2^a+\cdots+m^a + (m+1)^a < \frac{(m+1)^{(a+1)}-1}{a+1} + (m+1)^a $$

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

I think I am making progress so far, now what?

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Have you tried induction? – Pedro Tamaroff Mar 27 '12 at 1:36
For $n=1, 2^{(a+1)}-1 > a+1$ and assume for $n=m$ and prove for $n=m$ right? – Kirthi Raman Mar 27 '12 at 1:41
I mean prove for $n=m+1$ – Kirthi Raman Mar 27 '12 at 1:50
up vote 9 down vote accepted

$$\sum_{k=1}^{n} k^a \lt \int_{1}^{n+1} x^a \text{ d}x = \frac{(n+1)^{a+1} - 1}{a+1}$$

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Wow!, one liner proof (Genius). How about $a \in (-1,0)$, any suggestions ? – Kirthi Raman Mar 27 '12 at 1:52
@AlexSmith: Your question clearly states $a \gt 0$, but I believe the inequality is reversed when $a \lt 0$ ($x^a$ becomes a decreasing function). Try drawing the graphs and one should be able to see. – Aryabhata Mar 27 '12 at 1:54
I got it same kind with inequality reversed (I think I understand how to get it) – Kirthi Raman Mar 27 '12 at 1:55

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