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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.

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    $\begingroup$ Real or integer square roots ? As a first approximation, $\frac23n^{3/2}$. For better, use the Euler-MacLaurin summation formula en.wikipedia.org/wiki/… $\endgroup$ – Yves Daoust Apr 19 '15 at 14:16
  • $\begingroup$ Real or integer square roots ? $\endgroup$ – Yves Daoust Apr 19 '15 at 14:19
  • $\begingroup$ Integer roots. Thanks @YvesDaoust $\endgroup$ – lk42392 Apr 19 '15 at 18:31
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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 $m=\lfloor\sqrt n\rfloor$,

$$S_n=\frac{4(m-1)^3+9(m-1)^2+5(m-1)}{6}+m\left(n-m^2+1\right)\\ =m\left(n-\frac{2m^2+3m-5}6\right).$$

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http://ramanujan.sirinudi.org/Volumes/published/ram09.pdf http://arxiv.org/pdf/1204.0877.pdf
Refer to the docs. Simple as that!

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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

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  • $\begingroup$ Could you elaborate, please ? I do not see how to sum over $k$. I have the feeling that I am in a loop. Thanks. $\endgroup$ – Claude Leibovici Apr 19 '15 at 14:55
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    $\begingroup$ I believe you meant $\sum\sqrt k$ in the middle... $\endgroup$ – Simply Beautiful Art Apr 12 '16 at 20:15
  • $\begingroup$ These bounds give you a per-term bound. If you sum all three "sides" to this inequality over $\sum_{k=1}^n$, you'll get that $$\int_0^n \! \sqrt{x} \, \mathrm{d}x < \sum_{k=1}^n \sqrt{k} < \int_1^{n+1} \sqrt{x} \, \mathrm{d}x,$$ as desired. $\endgroup$ – Nicolás Kim Oct 5 '17 at 16:36
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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 $a$

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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,

$$\sum_{n=1}^xn^{1/2}\approx\frac23\left(x+\frac12\right)^{3/2}\underbrace{-0.22474487139}_{\zeta(-1/2)}$$

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  • $\begingroup$ Your exponents that are "2/3" need to be changed to "3/2" in order for your formulas to work/make sense. $\endgroup$ – Olive Stemforn Jul 8 '16 at 19:09
  • $\begingroup$ @OliveStemforn Whoops. $\endgroup$ – Simply Beautiful Art Jul 8 '16 at 19:53

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