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.
Best Answer
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.