Asymptotic or Exact formula for Harmonic Number I know Faulhaber's formula for positive integers. However, is there an asymptotic or exact formula for generalized Harmonic number. For example,
how can I calculate
\begin{align*}
\sqrt{1}+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}.
\end{align*}
I am looking forward to an exact formula or asymptotic formula.
Any help or references will be appreciated.
 A: By the Euler-Maclaurin summation formula, if $s\ne-1$, then we have
$$\sum_{k=1}^n k^s=\frac{n^{s+1}}{s+1}+\frac{n^s}2+\zeta(-s)+\sum_{k=1}^p\frac{B_{2k}(s)_{2k-1}}{(2k)!}n^{s+1-2k}+R_{p,n}$$
where $\zeta(s)$ is the Riemann zeta function, $B_k$ are the Bernoulli numbers, $(s)_k$ is the falling factorial, and $R_{p,n}$ is a remainder term given by
$$|R_{p,n}|\le\frac{2\zeta(2p)(s)_{2p-1}}{(2\pi)^{2p}}n^{s+1-2p}$$
If $s=-1$, then we have instead
$$\sum_{k=1}^n\frac1k=\ln(n)+\gamma+\frac1{2n}-\sum_{k=1}^p\frac{B_{2k}}{2kn^{2k}}+R_{p,n}$$
$$|R_{p,n}|\le\frac{2\zeta(2p)(2p-1)!}{(2\pi n)^{2p}}$$

For your example, we have
\begin{align}\sum_{k=1}^n\sqrt k&=\frac23n^{3/2}+\frac12n^{1/2}+\zeta(-1/2)+\sum_{k=1}^p\frac{B_{2k}}{2^{3k-1}k!}n^{-\frac12-2k}+R_{p,n}\\&=\frac23n^{3/2}+\frac12n^{1/2}-0.2078862249773545+\frac1{24}n^{-1/2}-\frac1{1920}n^{-3/2}+\mathcal O(n^{-7/2})\end{align}
As demonstrated by this graph: https://www.desmos.com/calculator/mkffzsvvhs
A: If $k$ is any complex number, a generalized harmonic number is given by:
$$H_k(n)=\sum_{j=1}^{n}\frac{1}{j^k}$$
I will answer your question nonetheless for particular cases:
$$\sum_{j=1}^{n}\sqrt{j}=-2\sum_{i=0}^{\infty} (-1)^i (2\pi)^{2i}H_{-2i}(n)\sum_{j=0}^{i} \frac{(-1)^j (2\pi)^{-2j}}{(2i+1-2j)!}\zeta\left(-\frac{1}{2}+2j\right)$$
where $H_{-2i}(n)$ is given by Faulhaber's formula (sum of the first $n$ integers raised to $2i$) and $\zeta$ is Riemann's zeta function.
For the detailed proof of the above, please see:
https://arxiv.org/abs/1905.09818 (An exact formula for the prime counting function).
The same reasoning explained in this paper can be applied to create exact power series for your all generalized harmonic numbers (where $k$ is any complex), the formula changes very little (only -1/2 is replaced by some other number, such as -1/3 for cubic roots).
Other papers that present formulae for the discrete cases of the generalized harmonic numbers (or generalized harmonic progressions) are:
https://arxiv.org/abs/1810.07877 (Generalized harmonic numbers revisited)
https://arxiv.org/abs/1811.11305 (Generalized harmonic progressions).
I'm in the process of typing out a paper with a formula for all, which is of course an integral, it can't be simpler than that.
A: There is an exact formula, that I just created.
It is pretty similar to the Euler-Mclaurin formula, except it gives you the exact residual terms.
For all complex $k$, except -1:
$$\sum_{j=1}^{n}j^{k}=\frac{n^{k+1}}{k+1}+\frac{n^{k}}{2}+\zeta(-k)+\frac{\pi\,n^{k+2}}{k+1}\int_{0}^{\pi/2}\left(\sec{v}\,\text{csch}{(n\pi\tan{v})}\right)^2\left(1-\frac{\cos{(k+1)\,v}}{(\cos{v})^{k+1}}\right)\,dv$$
Please refer to my RG paper for the detailed proof:
Analytic continuation for various functions
