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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/… $\endgroup$
    – user65203
    Apr 19, 2015 at 14:16
  • $\begingroup$ Integer roots. Thanks @YvesDaoust $\endgroup$
    – lk42392
    Apr 19, 2015 at 18:31
  • $\begingroup$ If you want to sum only integers, that is, numbers of the form floor(√n), then that needs to be stated clearly in the question. $\endgroup$
    – Dan Asimov
    Jan 16 at 18:10

6 Answers 6

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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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  • $\begingroup$ Claude, do you know of reading material on this ? The Wikipedia article on this seems just to be sums of $1/n$ $\endgroup$
    – Will Jagy
    Aug 6, 2021 at 4:10
  • $\begingroup$ @WillJagy. Have a look at the excellent (as always !) answer from robjohn in math.stackexchange.com/questions/1583452/… . Cheers :-) $\endgroup$ Aug 6, 2021 at 5:51
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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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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. $\endgroup$ Apr 19, 2015 at 14:55
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    $\begingroup$ I believe you meant $\sum\sqrt k$ in the middle... $\endgroup$ Apr 12, 2016 at 20:15
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    $\begingroup$ 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. $\endgroup$ Oct 5, 2017 at 16:36
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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 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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  • $\begingroup$ This is very beautiful. Thank you! $\endgroup$ Sep 2, 2021 at 4:45
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Not sure if there's a closed form, but some helpful inequalities are possible without calculus.

$S(N)=\sum_{k=1}^N \sqrt{k} $

$\implies S(N)^2=1+2+...+N +2\sqrt{1}(\sqrt{2}+\sqrt{3}+...+\sqrt{N})+2\sqrt{2}(\sqrt{3}+\sqrt{4}+...+\sqrt{N})+...+2\sqrt{(N-1)}\sqrt{N}$

$S(N)^2=\frac{N(N+1)}{2}+2\sum_{k=1}^{N-1}\sqrt{k}[S(N)-S(k)]$

$S(N)^2=\frac{N(N+1)}{2}+2S(N-1)S(N)-2\sum_{k=1}^{N-1}\sqrt{k}S(k)$

$S(N)^2=\frac{N(N+1)}{2}+2[S(N)-\sqrt{N}]S(N)-2\sum_{k=1}^{N-1}\sqrt{k}S(k)$

$S(N)^2=\frac{-N(N+1)}{2}+2\sum_{k=1}^{N}\sqrt{k}S(k)$

Now you can bind it with quadratic expression in S(N).

$\frac{-N(N+1)}{2}+2S(N)\le S(N)^2\le \frac{-N(N+1)}{2}+2S(N)^2$

$\implies S(N)\ge \sqrt{\frac{N(N+1)}{2}}$

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