I'd like to be able to prove the following inequality: $\frac{{{H_{n, - r}}}}{{{n^r}\left( {n + 1} \right)}} \le \frac{{{H_{n - 1, - r}}}}{{n{{\left( {n - 1} \right)}^r}}}$.
It's clear that as $n \to \infty$ we get equality, the limit on each side is $1/(r+1)$, and it also seems clear that this is a lower limit. But I haven't been able to find an argument for the inequality for finite $n$.
Any advice or hints greatly appreciated-- Number theory was not my strong suit when I was a mathematics student.
We want to show $\frac{H_{n, - r}}{n^r\left( n + 1 \right) } \le \frac{H_{n - 1, - r}}{n\left( n - 1 \right)^r} $ where $H_{n, a} =\sum_{k=1}^n \frac1{k^a} $, so $H_{n, -r} =\sum_{k=1}^n k^r $.
From this, $H_{n - 1, - r} =H_{n , - r}-n^r $, so we want $\frac{H_{n, - r}}{n^r\left( n + 1 \right) } \le \frac{H_{n , - r}-n^r}{n\left( n - 1 \right)^r} $.
Bringing the $H$ term together, this becomes $H_{n , - r}(\frac1{n\left( n - 1 \right)^r}-\frac1{n^r\left( n + 1 \right) }) \ge\frac{n^r}{n\left( n - 1 \right)^r} $ or $H_{n , - r}(\frac1{( n - 1)^r}-\frac1{n^{r-1}\left( n + 1 \right) }) \ge\frac{n^r}{\left( n - 1 \right)^r} $ or $H_{n , - r}(1-\frac{(n-1)^r}{n^{r-1}\left( n + 1 \right) }) \ge n^r $ or $H_{n , - r}\frac{n^{r-1}(n+1)-(n-1)^r}{n^{r-1}( n + 1) } \ge n^r $ or $H_{n , - r} \ge \frac{n^{2r-1}( n + 1) }{n^{r-1}(n+1)-(n-1)^r} $ or $H_{n , - r} \ge \frac{n^{r-1}( n + 1) }{(1+1/n)-(1-1/n)^r} $.
Since $H_{n , - r} > \frac{n^{r+1}}{r+1}+\frac12 n^r $, this is true if $\frac{n^{r+1}}{r+1}+\frac12 n^r \ge \frac{n^{r-1}( n + 1) }{(1+1/n)-(1-1/n)^r} $ or $\frac{n^2}{r+1}+\frac{n}{2} \ge \frac{( n + 1) }{(1+1/n)-(1-1/n)^r} $ or $\frac{n}{r+1}+\frac{1}{2} \ge \frac{( 1 + 1/n) }{(1+1/n)-(1-1/n)^r} $ or $\frac{2n+r+1}{2(r+1)} \ge \frac{( 1 + 1/n) }{(1+1/n)-(1-1/n)^r} $ or $(1+1/n)-(1-1/n)^r \ge \frac{2(r+1)(1+1/n)}{2n+r+1} $.
If $n >> r$, $(1-1/n)^r \approx 1-r/n $, so $(1+1/n)-(1-1/n)^r \approx (1+1/n)-(1-r/n) =(r+1)/n $, so this becomes $(r+1)/n \ge \frac{2(r+1)(1+1/n)}{2n+r+1} $ or $1 \ge \frac{2(n+1)}{2n+r+1} $ or $2n+r+1 \ge 2n+2 $ and this is true.
(Whew!)
Therefore, for $n$ large compared to $r$, your inequality is true.