Sum of the square roots of the first n natural numbers

AI Thread Summary
The discussion centers on calculating the sum of the square roots of the first n natural numbers, highlighting that an exact closed-form solution is not feasible. Instead, participants suggest using integral approximations, specifically the integral of √x, to estimate the sum with a manageable error margin. A basic integer approximation is provided, though it is noted to be less accurate. The integral approximation of (2/3)n^(3/2) is recommended for better accuracy, assuming the ability to compute n^(3/2). Overall, the conversation emphasizes the challenges and methods for approximating this mathematical sum.
Amith2006
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Is there a way to find the,"Sum of the square roots of the first n natural numbers"?
 
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I don't think you could do it exactly. You could approximate it by the integral of \sqrt{x}, and get a bound on the error.
 
As StatusX says I'm pretty sure there's no way to do in exactly in closed-form. If you don't have a way to calculate square roots at all (ie. you're doing it without a calculator and don't want to go through an approximation method), then a simple integer approximation would be

\frac{2}{3}\lfloor \sqrt{n} \rfloor^3 - \frac{1}{2}\lfloor \sqrt{n} \rfloor^2 - \frac{1}{6} \lfloor \sqrt{n} \rfloor + \lfloor \sqrt{n} \rfloor(n-\lfloor \sqrt{n} \rfloor^2),

but it's not very good. The integral approximation \frac{2}{3} n^{\frac{3}{2}} is much better, but you have to be able to compute n^{3/2}
(2/3 \lfloor n^{3/2}\rfloor is also better than the one I gave above though).
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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