# Approximation to an average of integer square roots

1. Nov 11, 2013

### phasic

I have stumbled upon an approximation to the average of integer square roots.

$\sum^{n}_{k=1}{\sqrt{k}/n} \approx \sqrt{median(1,2,...,n)}$

Sorry I am not very good at LaTeX, but I hope this comes across okay. Could anyone explain why this might be happening?

In fact, I just discovered that the error increases in the shape of the sqrt function and is only off by about 4 at n = 10000, about a 5% error. Interestingly, this relative error is asymptotic as n tends towards infinity and may converge? If that were the case, this approximation will always have around 5% relative error...?

This had an interesting consequence on finding where Ʃ(tanh(x-sqrt(k))) = 0. It turns out if x is the average of the square roots up to n, since tanh is an odd function, the sum will be close to 0. Indeed, the approximate minimum at the average is close to the numerically calculated minimum of Ʃ log(cosh(x-sqrt(k)))

I hope this makes sense! Any thoughts are welcome.

Last edited: Nov 11, 2013
2. Nov 11, 2013

### lurflurf

Do you know any calculus? That is related to something called a Riemann sum.
In calculus one learns that
$$\lim_{n \rightarrow \infty} \sum_{k=1}^n \sqrt{\frac{k}{n^3}}=\frac{2}{3}$$
So we can see that when n is large
$$\frac{1}{n} \sum_{k=1}^n \sqrt{k} \sim \frac{2}{3}\sqrt{n}$$
A more accurate approximation if you are interested is
$$\frac{1}{n} \sum_{k=1}^n \sqrt{k} \sim \frac{2}{3} \sqrt{n} - \frac{1}{2 \sqrt{n}}+\frac{1}{n} \zeta \left( -\frac{1}{2} \right) + \frac{1}{24n \sqrt{n}}$$
where
$$\zeta \left( -\frac{1}{2}\right) \sim -0.20788622497735456601730672539704930222626853128767253761011...$$
http://www.wolframalpha.com/input/?....**Integral.rangestart-.*Integral.rangeend---