The question is: There is a collection of numbers: $\sqrt{1},\sqrt{2},\sqrt{3},\dotsc \sqrt{n}$. $\mu$ is its arithmetic mean or average; $\sigma$ is its standard deviation. Calculate $$\lim_{n\to\infty}\frac{\mu}{\sigma} = \ ?$$
I think of the Squeeze Theorem, but I don't know how to do it, and maybe there is a more proper way to solve it. Can you help me, please?
Since $\sigma^2=\frac{1}{n}\sum_{k=1}^n(\sqrt k)^2-\mu^2=\frac{n+1}{2}-\mu^2$, we have $$\frac{\mu^2}{\sigma^2}=\frac{1}{\frac{n+1}{2\mu^2}-1}.$$ Therefore, we need to compute $\mu/\sqrt n$. We write it explicitly $$\frac{\mu}{\sqrt n}=\frac{1}{n}\sum_{k=1}^n\sqrt\frac{k}{n}\to\int_0^1\sqrt x\mathrm dx=\frac{2}{3},$$ by the definition of Riemann integral.
Hence, we have $\mu^2/\sigma^2\to 1/((9/8)-1)=8$, and the limit is thus $2\sqrt2$.