$ \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?
 A: $$\sum_{k=1}^n\frac{\sqrt{n+k}}{n^{3/2}}=\frac1n\sum_{k=1}^n\sqrt{1+\frac kn}\xrightarrow[n\to\infty]{}\int_0^1\sqrt{1+x}\;dx$$
If you like this answer please do not upvote it . Apparently someone in this site's administration believes some people(s) upvoted some of my posts unduely and some 695 (!) points from my reputation have been deleted, which seems to me exaggerated, but also there is at least once some 15 points ("accepted answer") that were deleted, as the number of points is not a multiple of ten. So someone  who posted a question to which my answers were the accepted ones have their points given to me deleted.
I don't care about the points, and if you need to take them all off please do it, but leave alone trying to do some mathematics. 
A: Using Stolz-Cesaro:
$$\eqalign{
\lim_{n\to\infty}\frac{\sqrt{n+1}+\cdots+\sqrt{2n}}{\sqrt{n^3}}
& = 
\lim_{n\to\infty}\frac{(\sqrt{(n+1)+1}+\cdots+\sqrt{2(n+1))}-(\sqrt{n+1}+\cdots+\sqrt{2n})}{\sqrt{(n+1)^3}-\sqrt{n^3}}\cr
& =
\lim_{n\to\infty}\frac{\sqrt{2n+1}+\sqrt{2(n+1)}-\sqrt{n+1}}{\sqrt{(n+1)^3}-\sqrt{n^3}}\cr
&=\lim_{n\to\infty}\frac{\sqrt{2n+1}+\sqrt{2(n+1)}-\sqrt{n+1}}{\sqrt{(n+1)^3}-\sqrt{n^3}}\frac{\sqrt{(n+1)^3}+\sqrt{n^3}}{\sqrt{(n+1)^3}+\sqrt{n^3}}=\cdots
}.$$
