How to solve $\lim\limits_{n\to∞}\dfrac1{n^{3/2}}(\sqrt{2n+1}+\sqrt{2n+2}+\cdots+\sqrt{2n+n})$
This problem was asked by another user, but was deleted when I found the answer below. So I thought I should post it here.
Please check if my use of Riemann Integral was correct, since I have very little knowledge on it.
Thanks.
$$\lim _{n\to \:∞}\left(\frac{\left(\sqrt{2n+1}+\sqrt{2n+2}+\ldots+\sqrt{2n+n}\right)}{n^{\frac{3}{2}}}\right)=\lim _{n\to \:∞}\dfrac{1}{n}\left(\frac{(\sqrt{2n+1}+\sqrt{2n+2}+\ldots+\sqrt{2n+n})}{\sqrt{n}}\right)=\lim _{n\to \:∞}\dfrac{1}{n}\left(\sqrt{2+\frac{1}{n}}+\sqrt{2+\frac{2}{n}}+\ldots+\sqrt{2+\frac{n}{n}}\right) =\lim _{n\to \:∞}\dfrac{1}{n}\sum_{i=1}^n\sqrt{2+\dfrac{i}{n}}\\ =\int_0^1\sqrt{2+x}dx\\ =\frac 23\left[(2+x)^\frac 32\right]_0^1\\ =1.578$$