Show that this difference goes to zero, $$\frac{1+\sqrt{2} + ... + \sqrt{N}}{N} - \frac{2}{3}\sqrt{N} \to  0.$$
The hint given in the question is this: choose appropriate Riemann sums and estimate the approximation error.
My current work:
$$\frac{1+\sqrt{2} + ... + \sqrt{N}}{N} - \frac{2}{3}\sqrt{N}$$
$$=: A_n =(\sum_{k=1}^N \sqrt{k}\frac{1}{N}) - \frac{2}{3}\sqrt{N}$$ 
The first term is in the form of a Riemann sum, so letting N go to infinity, we see that mesh(p) goes to zero, for some partition p, which gives the (improper) Riemann integral, over the interval [1,N]:
$$\lim_{N->\infty}\int_1^N \sqrt{x}dx$$
Evaluation of the integral, without evaluating the limit, gives:
$$\frac{2}{3}N^{\frac{3}{2}} - \frac{2}{3}$$ 
Then $$A_n = \frac{2}{3}N^{\frac{3}{2}} - \frac{2}{3} - \frac{2}{3}\sqrt{N}$$
And this is where I am currently stuck.  The above equation is a little suspect, because I let N go to infinity to get the improper integral, while I did nothing with the $\frac{2}{3}\sqrt{N}$ term -- and just included this term into the equation, since I feel it gets me a little closer to do some kind of approximation.
Any hints would be greatly appreciated.
Thanks,
 A: If you partition the interval $[0,N]$ to $N$ equal length subintervals, then those subintervals all have length $1$. Not $1/N$ as you seem to think. 
I think that you are expected to observe that the sum
$$
\frac1{\sqrt{N}}\,\frac{\sqrt1+\sqrt2+\cdots+\sqrt{N}}N
$$
is a Riemann sum related to the definite integral
$$
\int_0^1\sqrt x\,dx.
$$
In other words, scale everything by a factor of $\sqrt{N}$.
Standard error estimate available for all increasing functions is good enough unless I made a mistake.
A: I believe the proper Riemann Sum is
$$
\begin{align}
\frac1n\sum_{k=1}^nk^{1/2}
&=n^{1/2}\sum_{k=1}^n\color{#C00000}{\left(\frac kn\right)^{1/2}}\,\color{#00A000}{\frac1n}\\
&=n^{1/2}\int_0^1\color{#C00000}{x^{1/2}}\,\color{#00A000}{\mathrm{d}x}+O\left(n^{-1/2}\right)\\[3pt]
&=\frac23n^{1/2}+O\left(n^{-1/2}\right)
\end{align}
$$
since the error estimate for the Riemann Sum is
$$
\begin{align}
\frac1n\int_0^1\left|\,f'(x)\right|\,\mathrm{d}x
&=\frac1n\int_0^12x^{-1/2}\,\mathrm{d}x\\
&=\frac1n
\end{align}
$$
A: We can also do this with Stolz-Cesaro. Let $S_n = \sum_{k=1}^{n}k^{1/2}.$ We are looking at
$$\frac{S_n -(2/3)n^{3/2}}{n}.$$
S-C says to look at
$$(n+1)^{1/2}+ (2/3)[n^{3/2} - (n+1)^{3/2}].$$
By the MVT, the second term equals $-c_n^{1/2},$ where $c_n \in (n,n+1).$ So we have
$$(n+1)^{1/2}-c_n^{1/2} = \frac{n+1-c_n}{(n+1)^{1/2}+c_n^{1/2}},$$
which is positive and less than $1/ (n+1)^{1/2} \to 0.$
