Help with a limit $S_{n}=\frac{1}{n}\sum _{k=1}^{n} \sqrt{\frac{k}{n}}$

Question

How can I calculate the next limit? $S_{n}=\frac{1}{n}\sum _{k=1}^{n} \sqrt{\frac{k}{n}}$ when "n" goes to infinity

Answer 1

As alreasy answered by Winther and copper.hat, the simplest way is to use Riemann sum. From Winther's answer for the infinite sum, you would get $\frac{2}{3}$ and from copper.hat's answer, you would get for the partial sum $$\frac{2}{3} \leq S_n \leq \frac{2}{3} \left(\left(1+\frac{1}{n}\right)^{3/2}-\left(\frac{1}{n}\right)^{3/2}\right)$$ For the rhs of the last expression, an asymptotic expansion for large values of $n$ is $$\frac{2}{3}+\frac{1}{n}-\frac{2}{3} \left(\frac{1}{n}\right)^{3/2}+O\left(\left(\frac{1}{n}\right)^{2}\right)$$

I don't know how much you are familiar with harmonic numbers, so forgive me if I am off topic for your question. $$\sum _{k=1}^{n} \sqrt{\frac{k}{n}}=\sqrt{\frac{1}{n}} H_n^{\left(-\frac{1}{2}\right)}$$ and so $$S_{n}=\frac{1}{n}\sum _{k=1}^{n} \sqrt{\frac{k}{n}}=\left(\frac{1}{n}\right)^{3/2} H_n^{\left(-\frac{1}{2}\right)}$$ and an asymptotic expansion for large values of $n$ gives $$S_n=\frac{2}{3}+\frac{1}{2 n}+\left(\frac{1}{n}\right)^{3/2} \zeta \left(-\frac{1}{2}\right)+O\left(\left(\frac{1}{n}\right)^2\right)$$ (where $\zeta(-1/2) \approx -0.207886$)

Answer 2

Let $f(x) = \sqrt{x}$ and note that $S_n = {1 \over n} \sum_{k=1}^n f({ k \over n})$. $f$ is increasing, so $\int_0^1 f(x) dx \le S_n \le \int_{1 \over n}^{1+{1 \over n}} f(x) dx$.