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For an integer $r\ge1$, consider the sequence $(a_n)_{n\ge2}$ defined by $$a_n=\frac{1}{n^{r+1}}\sum_{k=1}^{n-1}k^r$$ It is easy to prove, for $r=1,2,3$, that this sequence in increasing using the known closed form of the sum.

But is it true in general that for any integer $r\ge1$ this sequence is increasing?

In fact, it can be proved that this is true starting from a certain index $n_0$ that depends on $r$, and numerical evidence suggests that the answer is yes, but I could not find a proof, Any help?

Omran Kouba
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  • Proving $a_n$ is increasing is equivalent to prove that $$a_n\le \frac{n^r}{(n+1)^{r+1}-n^{r+1}}$$ I don't know if it helps. Very happy to see you @Omran Kouba. – Olivier Oloa Dec 20 '16 at 08:41

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A general result holds (see Theorem 4 in Monotonicity of Certain Riemann Sums by D. Borwein, J. Borwein and B. Sims).

If the function $f: [0,1]\to \mathbb{R}$ is concave on the interval $[0,c]$, convex on $[c,1]$ and increasing on $[0,1]$ then $$\sigma_n(f)=\frac{1}{n}\sum_{k=0}^{n-1}f(k/n)$$ is an increasing sequence, and $$\tau_n(f)=\frac{1}{n}\sum_{k=1}^{n}f(k/n)$$ is a decreasing sequence.

Robert Z
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