# $\lim_{n \to \infty} \frac {\sqrt{n+1}+\sqrt{n+2}+…+\sqrt{2n}}{n^{3/2}}$

Evaluate the limit

$$\lim_{n \to \infty} \frac {\sqrt{n+1}+\sqrt{n+2}+...+\sqrt{2n}}{n^{3/2}}.$$

Rearranging I can get

$$\lim_{n \to \infty} \frac {\sqrt{\frac{n+1}{n}}+\sqrt{\frac{n+2}{n}}+...+\sqrt{\frac{2n}{n}}}{n}.$$

but I do not see how that helps me. Perhaps I use L'Hopital's rule?

• Do you know the Riemann sum? – user63181 Jan 9 '15 at 21:54

$$\sum_{k=1}^n\frac{\sqrt{n+k}}{n^{3/2}}=\frac1n\sum_{k=1}^n\sqrt{1+\frac kn}\xrightarrow[n\to\infty]{}\int_0^1\sqrt{1+x}\;dx$$
• This is the perfect opportunity for $\sim$ – Mohamad Ali Baydoun Jan 9 '15 at 22:10
• For $\sim$ rather than an arrow with $n \to \infty$. It's just really satisfying – Mohamad Ali Baydoun Jan 9 '15 at 22:10
Using Stolz-Cesaro: \eqalign{ \lim_{n\to\infty}\frac{\sqrt{n+1}+\cdots+\sqrt{2n}}{\sqrt{n^3}} & = \lim_{n\to\infty}\frac{(\sqrt{(n+1)+1}+\cdots+\sqrt{2(n+1))}-(\sqrt{n+1}+\cdots+\sqrt{2n})}{\sqrt{(n+1)^3}-\sqrt{n^3}}\cr & = \lim_{n\to\infty}\frac{\sqrt{2n+1}+\sqrt{2(n+1)}-\sqrt{n+1}}{\sqrt{(n+1)^3}-\sqrt{n^3}}\cr &=\lim_{n\to\infty}\frac{\sqrt{2n+1}+\sqrt{2(n+1)}-\sqrt{n+1}}{\sqrt{(n+1)^3}-\sqrt{n^3}}\frac{\sqrt{(n+1)^3}+\sqrt{n^3}}{\sqrt{(n+1)^3}+\sqrt{n^3}}=\cdots }.