Limit problem in book: $\lim_{n\to\infty}\frac{1+\sqrt{2}+...+\sqrt{n}}{n^{3/2}}$ My book states that the
$$
\lim_{n\to\infty}\frac{1+\sqrt{2}+...+\sqrt{n}}{n^{3/2}}=
\lim_{n\to\infty}\frac{1}{n}\left[\frac{1+\sqrt{2}+...+\sqrt{n}}{n^{1/2}}\right]=\frac{2}{3}.
$$
But I just don't see it. I was thinking that if the $$\lim_{n\to\infty}\frac{1}{n}=0$$ then the whole thing would equal 0 but I don't see how to get $\frac{2}{3}$. Thanks
 A: Draw the graph of $y=\sqrt{x}$. In red, draw the rectangle with base $[0,1]$ and height $\sqrt{1}$, the rectangle with base $[1,2]$ and height $\sqrt{2}$, and so on up to the rectangle with base $[n-1,n]$ and height $\sqrt{n}$.
In blue, draw the rectangle with base $[1,2]$ and height $\sqrt{1}$, the rectangle with base $[2,3]$ and height $\sqrt{2}$, and so on up to the rectangle with base $[n,n+1]$ and height $\sqrt{n}$.
Then $\sqrt{1}+\sqrt{2}+\cdots +\sqrt{n}$ is the sum of the areas of the red rectangles, and it is also the sum of the areas of the blue rectangles. Call this sum $S_n$. 
Then $S_n$ is greater than the area under $y=\sqrt{x}$ from $x=0$ to $x=n$, and less than the area under $y=\sqrt{x}$ from $x=1$ to $x=n$. Thus
$$\int_0^n \sqrt{x}\,dx\lt S_n\lt \int_1^{n+1} \sqrt{x}\,dx.$$
Calculate. We get 
$$\frac{2}{3}n^{3/2} \lt S_n\lt \frac{2}{3} ((n+1)^{3/2}-1).$$
Divide through by $n^{3/2}$. We get
$$\frac{2}{3}\lt \frac{S_n}{n^{3/2}}\lt \frac{2}{3}\frac{(n+1)^{3/2}-1}{n^{3/2}}.$$ 
Now let $n\to\infty$. The expression $\frac{(n+1)^{3/2}-1}{n^{3/2}}$ has limit $1$, so by Squeezing we get $\lim_{n\to\infty} \frac{S_n}{n^{3/2}}=\frac{2}{3}$. 
A: As 
 $\displaystyle\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$
$$\lim_{n\to\infty}\frac1n\sum_{1\le r\le n}\sqrt{\frac rn}=\int_0^1\sqrt xdx$$
A: Another variant is to use the theorem of Cesaro-Stolz. There
\begin{align}
\lim_{n\to\infty}\frac{1+\sqrt2+...+\sqrt n}{n^{3/2}}
&=\lim_{n\to\infty}\frac{\sqrt{n}}{n^{3/2}-(n-1)^{3/2}}
\\
&=\lim_{n\to\infty}\frac{\sqrt{n}(n^{3/2}+(n-1)^{3/2})}{n^3-(n-1)^3}
\\
&=\lim_{n\to\infty}\frac{n^2(1+(1-\tfrac1n)^{3/2})}{3n^2-3n+1}=\frac23
\end{align}
