Generalized Harmonic numbers 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.
 A: 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.
