6
$\begingroup$

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?

  • 0
    I mean prove for $n=m+1$2012-03-27

1 Answers 1

9

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

  • 0
    I got it same kind with inequality reversed (I think I understand how to get it)2012-03-27