Asymptotic or Exact formula for Harmonic Number

Question

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.

Answer 1

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

$$\sum_{k=1}^n\frac1k=\ln(n)+\gamma+\frac1{2n}-\sum_{k=1}^p\frac{B_{2k}}{2kn^{2k}}+R_{p,n}$$

$$|R_{p,n}|\le\frac{2\zeta(2p)(2p-1)!}{(2\pi n)^{2p}}$$

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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

Answer 2

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