Sum of Square roots formula. I would like to know if there is formula to calculate sum of series of square roots $\sqrt{1} + \sqrt{2}+\dotsb+ \sqrt{n}$ like the one for the series $1 + 2 +\ldots+ n = \frac{n(n+1)}{2}$.
Thanks in advance.
 A: http://ramanujan.sirinudi.org/Volumes/published/ram09.pdf
http://arxiv.org/pdf/1204.0877.pdf
 Refer to the docs.
Simple as that!
A: For a better upper bound than Alex's answer,
$$\sum_{n=1}^xn^{1/2}\le\frac23\left(x+\frac12\right)^{3/2}$$
And if you want to improve upon that,
$$\sum_{n=1}^xn^{1/2}\approx\frac23\left(x+\frac12\right)^{3/2}\underbrace{-0.22474487139}_{\zeta(-1/2)}$$
A: The definition of harmonic numbers is $$H_p^{(-a)}=\sum_{i=1}^p i^a $$ When $a$ is not a positive integer, there is no closed form but, as Yves Daoust commented, there are quite nice expansions.
For example, if $n=\frac 12$ as in the post, you have $$H_p^{\left(-\frac{1}{2}\right)}=\frac{2 p^{3/2}}{3}+\frac{\sqrt{p}}{2}+\zeta
   \left(-\frac{1}{2}\right)+\frac{1}{24\sqrt p}+O\left(\left(\frac{1
   }{p}\right)^{5/2}\right)$$ where $\zeta
   \left(-\frac{1}{2}\right)\approx -0.2078862250$.
For example, for $p=10$, the exact value is $\approx 22.46827819$ while the above approximation gives $\approx 22.46827983$. By itself, the first term already gives $21.0819$; the sum of first and second term gives $\approx 22.6629$. For $p=100$, the approximation leads to $12$ exact significant figures.
There are similar expansions for any value of the exponent $a$
A: For integer square roots, one should note that there are runs of equal values and increasing lengths
$$1,1,1,2,2,2,2,2,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4\dots$$
For every integer $i$ there are $(i+1)^2-i^2=2i+1$ replicas, and by the Faulhaber formulas
$$\sum_{i=1}^m i(2i+1)=2\frac{2m^3+3m^2+m}6+\frac{m^2+m}2=\frac{4m^3+9m^2+5m}{6}.$$
When $n$ is a perfect square minus $1$, all runs are complete and the above formula applies, with $m=\sqrt{n+1}-1$.
Otherwise, the last run is incomplete and has $n-\left(\lfloor\sqrt n\rfloor\right)^2+1$ elements.
Hence, with $m=\lfloor\sqrt n\rfloor$,
$$S_n=\frac{4(m-1)^3+9(m-1)^2+5(m-1)}{6}+m\left(n-m^2+1\right)\\
=m\left(n-\frac{2m^2+3m-5}6\right).$$
A: For an easier solution notice that $f(x) = \sqrt{x}$ is a monotone increasing function, hence for every $[k ,k+1]$
$$
\int_{k-1}^{k} \sqrt{x} dx<\sqrt{k}<\int_{k}^{k+1} \sqrt{x} dx
$$
Now sum over k, you'll get a sharp approximation
A: 
Hopefully you can figure it out using this sketch.
