Inequalities to give bounds on generalised harmonic numbers? Let $H_n$ be the $n$th harmonic number and $H_n^{(k)}$ be the $n$th harmonic number of order $k$ as follows:
$$H_n=\sum_{m=1}^{n}\frac{1}{m}$$
$$H_n^{(k)}=\sum_{m=1}^{n}\frac{1}{m^k}$$
There are several inequalities giving upper and lower bounds on $H_n$, such as this one found on MathWorld (eqn 14):
$$\frac{1}{2(n+1)}<H_n-\ln n-\gamma<\frac{1}{2n}$$
where $\gamma$ is the Euler-Mascheroni constant:

Are there any equivalent inequalities for $H_n^{(k)}$? And how does one arrive at them?
Heuristically, the following seems to hold, and offer nice tight bounds:
$$n^{-k}
   \left(-\frac{n}{k-1}+\gamma-\frac{k}{12
   n}-\frac{1}{n^3}\right)
   +\zeta (k)<H_n^{(k)}<n^{-k}
   \left(-\frac{n}{k-1}+\gamma-\frac{k}{12
   n}+\frac{1}{n^3}\right)
   +\zeta (k)$$
For example, this is a plot with $k=1.8$:

Is this inequality valid? And how do I prove it?
NOTE: This is a substantial revision of the original question, which was unclear - and since which, I have found the above potential bounds on my own. The bounty is for validation and proof.
 A: Let $n\geq1$ and $k\geq 2$. By the result of this paper, it holds that
\begin{align*}
H_n^{(k)}  = \zeta (k) & + n^{ - k} \left(  - \frac{n}{{k - 1}} + \frac{1}{2} - \sum\limits_{m = 1}^{M - 1} \frac{{B_{2m} }}{{(2m)!}}\frac{{\Gamma (k + 2m - 1)}}{{\Gamma (k)}}\frac{1}{{n^{2m - 1} }}  \right. \\ & -\left. \theta _M (n,k)\frac{{B_{2M} }}{{(2M)!}}\frac{{\Gamma (k + 2M - 1)}}{{\Gamma (k)}}\frac{1}{{n^{2M - 1} }}  \right),
\end{align*}
where $M\geq 1$, and $0<\theta _M (n,k)<1$ is an appropriate number. The $B_m$ are the Bernoulli numbers. In particular, with $M=2$,
$$
H_n^{(k)}  < \zeta (k) + n^{ - k} \left( { - \frac{n}{{k - 1}} + \frac{1}{2} - \frac{k}{{12}}\frac{1}{n} + \frac{{k(k + 1)(k + 2)}}{{720}}\frac{1}{{n^3 }}} \right)
$$
and
$$
H_n^{(k)}  > \zeta (k) + n^{ - k} \left( { - \frac{n}{{k - 1}} + \frac{1}{2} - \frac{k}{{12}}\frac{1}{n}} \right).
$$
Note that the constant must be $1/2$ and not $\gamma$. It is also seen that for sufficiently large values of $k$, your upper bound is not valid.
A: We can obtain bounds from the asymptotic expansions of $H_{n}^{(k)}$ which can be derived from this exact relation valid for $k\ge 2$
$$H_{n}^{(k)} = \zeta(k) + \frac{1}{(k-1)!} \left(-\frac{\partial }{\partial n}\right)^{k-1} H_{n}\tag{1}$$
where $\zeta(k)=\sum_{i=1}^{\infty}\frac{1}{i^k}$ is the Riemann zeta function.
$(1)$ can be easily derived from the well known representation, valid for $k \ge 1$
$$H_{n}^{(k)}=\sum_{m=1}^{\infty}\left(\frac{1}{m^k}-\frac{1}{(n+m)^k}\right)\tag{2}$$
which, for $k=1$ reads
$$H_{n}^{(1)}=H_{n} = \sum_{m=1}^{\infty}\left(\frac{1}{m}-\frac{1}{(n+m)}\right)\tag{3}$$
Inserting the asymptotic expansion of $H_{n}$
$$H_{n} \underset{n\to\infty}\simeq \log(n) +\gamma +\frac{1}{2n} -\frac{1}{12 n^2}+\frac{1}{120 n^4} \mp\ldots\tag{4}$$
we get
$$H_{n}^{(k)} \underset{n\to\infty}\simeq \zeta (k)+\frac{1}{n^k}\left(\frac{1}{2}-\frac{n}{k-1}-\frac{k}{12 n}+\frac{\binom{k+2}{3}}{120 n^3}\mp \ldots\right)\tag{5}$$
Taking more an more terms of the asymptotic expansion into account we can easily derive a chain of inequalities starting like this (notice that they are valid even for $n \ge 1$, and, of course, $k\ge 2$)
$$H_n^{(k)}-\zeta (k)>-\frac{1}{n^k}\frac{n}{(k-1)}\tag{6a}$$
$$H_n^{(k)}-\zeta (k)<\frac{1}{n^k}\left(\frac{1}{2}-\frac{n}{k-1}\right)\tag{6b}$$
$$H_n^{(k)}-\zeta (k)>\frac{1}{n^k}\left(\frac{1}{2}-\frac{n}{k-1}-\frac{k}{12n}\right)\tag{6c}$$
$$H_n^{(k)}-\zeta (k)<\frac{1}{n^k}\left(\frac{1}{2}-\frac{n}{k-1}-\frac{k}{12n}+\frac{k (k+1) (k+2)}{720 n^3}\right)\tag{6d}$$
A: LEMMA (see [1]) 
Let $f(x)$ be a function with Taylor series in $(-a,a)$, $a\geq 1$. Let also its Taylor series converges absolutely in $1$. Then exists a constant $c=c(f)$ such that
$$
\sum^{M}_{k=1}f\left(\frac{1}{k}\right)=\int^{M}_{1}f\left(\frac{1}{t}\right)dt+c(f)+O\left(\frac{1}{M}\right)\tag 1
$$
Moreover 
$$
c(f)=f(0)+f'(0)\gamma+\sum^{\infty}_{s=2}\frac{f^{(s)}(0)}{s!}\left(\zeta(s)-\frac{1}{s-1}\right).
$$
where $\zeta(s)$ is Riemann's zeta function.
PROOF. See [1].
PROPOSITION.
For a function $f$ as in lemma It holds the following usefull generalized expansion:
$$
\frac{1}{x}\sum^{x}_{k=1}f\left(\frac{x}{k}\right)-\int^{1}_{1/x}f\left(\frac{1}{t}\right)dt=\frac{f(0)}{x}+f'(0)\left(\gamma+\frac{1}{2x}-\frac{1}{12 x^2}\right)+
$$
$$
+\frac{c(f,x)}{x}+\frac{f(1)-f(0)-f'(0)}{2x}-\frac{f'(1)-f'(0)}{12x^2}+O\left(x^{-4}\right)\textrm{, as }x\rightarrow+\infty\tag 2
$$
where 
$$
c(f,x)=\sum^{\infty}_{s=2}\frac{f^{(s)}(0)x^s}{s!}\left(\zeta(s)-\frac{1}{s-1}\right)
$$
which is generalization of LEMMA. 
For to prove it one can use 
$$
\sum^{x}_{k=1}\frac{1}{k}=\log(x)+\gamma+\frac{1}{2x}-\frac{1}{12x^2}+O\left(x^{-4}\right)\textrm{, }x\rightarrow\infty\tag 3
$$
$$
\sum^{\infty}_{k=x+1}\frac{1}{k^s}=\frac{1}{(s-1)x^{s-1}}-\frac{1}{2x^s}+\frac{s}{12x^{s+1}}+O\left(x^{-s-3}\right)\textrm{, }x\rightarrow\infty\tag 4
$$
$$
\frac{1}{x}\int^{x}_{1}f\left(\frac{1}{t}\right)dt=\frac{x-1}{x}f(0)+f'(0)\log(x)+\frac{1}{x}\sum^{\infty}_{s=2}\frac{f^{(s)}(0)x^s}{s!(s-1)}-\sum^{\infty}_{s=2}\frac{f^{(s)}(0)}{s!(s-1)}\tag 5
$$
and change of variables $t\rightarrow tx$
$$
\int^{x}_{1}f\left(\frac{x}{t}\right)dt=x\int^{1}_{1/x}f\left(\frac{1}{t}\right)dt.
$$ 
Your relation is Application of Proposition with $f(x)=x^k$.
Also
$$
\frac{1}{x}\int^{x}_{1}f\left(\frac{1}{t}\right)dt=-\int^{1/x}_{0}f\left(\frac{1}{t}\right)dt+\int^{1}_{0}f\left(\frac{1}{t}\right)dt\tag 6
$$
Hence if $E(f,N)$ denotes the error terms of the Riemann approximation of integral 
$$
\int^{1}_{0}f\left(\frac{1}{t}\right)dt\tag 7
$$
with the usual rectangular method, then 
$$
E(f,N)=-\int^{1/N}_{0}f\left(\frac{1}{t}\right)dt+\frac{f(0)}{N}+f'(0)\left(\gamma+\frac{1}{2N}-\frac{1}{12N^2}\right)+
$$
$$
+\frac{c(f,N)}{N}+\frac{f(1)-f(0)-f'(0)}{2N}-\frac{f'(1)-f'(0)}{12N^2}+O\left(N^{-4}\right)\textrm{, }N\rightarrow+\infty.\tag 8
$$
Provited that (7) exists.
REFERENCES
[1]: Nikos Bagis, M.L. Glasser. 'Integrals and Series Resulting from Two Sampling Theorems'. Sampling Theory in Singnal and Image Processing., Sampling Publishing, Vol. 5, No. 1, 2006.
