TITLE: Prove   $1^a+2^a+\cdots+n^a < \frac{(n+1)^{(a+1)}-1}{a+1} $ for any $a >0$ and $n \in \mathbb{Z^+}$
QUESTION [6 upvotes]: 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.
$$1^a+2^a+\cdots+m^a < \frac{(m+1)^{(a+1)}-1}{a+1}$$
Now
$$1^a+2^a+\cdots+m^a + (m+1)^a < \frac{(m+1)^{(a+1)}-1}{a+1} + (m+1)^a $$
$$ < \frac{(m+1)^a (m+1+a) + (m+1)^a -1}{a+1} $$
I think I am making progress so far, now what?

REPLY [9 votes]: $$\sum_{k=1}^{n} k^a \lt \int_{1}^{n+1} x^a \text{ d}x = \frac{(n+1)^{a+1} - 1}{a+1}$$