How can i evaluate $\lim_{n\to\infty}\left[ \left(\frac{1}{n}\right)^\frac{3}{2} \sum\limits_{i=2}^{n+1} i^{\frac{1}{2}}\right] $ I've not looked at finding the limits of sums before, and i'm interested to know how this particular example would be evaluated. The answer I believe should give $\frac{2}{3}$. 
 A: The usual way: $\sqrt{x}$ is a continuous function on $[0,1]$, hence it is Riemann-integrable and
$$\int_{0}^{1}\sqrt{x}\,dx = \lim_{n\to+\infty}\frac{1}{n}\sum_{k=1}^{n}\sqrt{\frac{k}{n}}\tag{1}$$
The RHS of $(1)$ is your limit and by the fundamental theorem of Calculus the LHS of $(1)$ is $\frac{1}{1+\frac{1}{2}}=\color{red}{\large\frac{2}{3}}$ as wanted.

The unusual way: it is enough to understand the behaviour of $\sum_{k=1}^{n}\sqrt{k}$ with decent accuracy.
For such a task, we may employ creative telescoping, since
$$ \left(k+\frac{1}{2}\right)\sqrt{k+\frac{1}{2}}-\left(k-\frac{1}{2}\right)\sqrt{k-\frac{1}{2}} = \frac{3k^2+\frac{1}{4}}{\left(k+\frac{1}{2}\right)\sqrt{k+\frac{1}{2}}+\left(k-\frac{1}{2}\right)\sqrt{k-\frac{1}{2}}}$$
differs from $\frac{3}{2}\sqrt{k}$ by less than $\frac{1}{60 k^{3/2}}$ for any $k\geq 1$. In particular,
$$ \left|\sum_{k=1}^{n}\sqrt{k}-\frac{2}{3}\left(n+\frac{1}{2}\right)^{3/2}\right|\leq \frac{1}{3\sqrt{2}}+\frac{1}{90}\sum_{k\geq 1}\frac{1}{k^{3/2}} = C $$
and the wanted limit is $\color{red}{\large\frac{2}{3}}$.
A: METHODOLOGY $1$:
Note that since $\sqrt x$ is monotonically increasing we can bound the sum of interest $S_n=\sum_{i=2}^{n+1}i^{1/2}$ by 
$$\int_1^{n+1}\sqrt{x}\,dx\le S_n=\sum_{i=2}^{n+1}i^{1/2}\le \int_2^{n+2}\sqrt x\,dx$$
whereupon evaluating the integrals gives
$$\frac23((n+1)^{3/2}-1)\le \sum_{i=2}^{n+1}i^{1/2}\le \frac23((n+2)^{3/2}-2^{3/2})\tag 1$$
Dividing $(1)$ by $n^{3/2}$ and applying the squeeze theorem yields the coveted result

$$\lim_{n\to \infty}\frac{S_n}{n^{3/2}}=\frac23$$


METHODOLOGY $2$:
This is overkill for the purpose herein, but I thought it might be instructive to present an approach that provides an expansion of $S_n=\sum_{i=2}^{n+1}i^{1/2}$.  To that end we proceed.
Using the Euler-Maclaurin Summation Formula, we can write
$$\begin{align}
S_n&=\sum_{i=2}^{n+1}i^{1/2}\\\\
&=\int_1^{n+1}\sqrt{x}\,dx+\frac12(\sqrt{n+1}-\sqrt{1})+\frac{B_2}{2!}\frac12\left((n+1)^{-1/2}-^{-1/2}\right)\\\\
&+\frac{B_4}{4!}\frac{3}{8}\left((n+1)^{-5/2}-(1)^{-5/2}\right)+C+O\left((n+1)^{-9/2}\right)\\\\
&=\frac23(n+1)^{3/2}+\frac{\sqrt{n+1}}{2}+C'+\frac{1}{24}(n+1)^{-1/2}-\frac{1}{1920}(n+1)^{-5/2}+O\left((n+1)^{-9/2}\right)
\end{align}$$
The constant $C'$ is given by the limit
$$\begin{align}
C'&=\lim_{n\to \infty}\left(\sum_{i=2}^{n+1}i^{1/2}-\frac23(n+1)^{3/2}-\frac{\sqrt{n+1}}{2}\right)\\\\
&=\zeta\left(-\frac12\right)\\\\
&\approx 0.207886225
\end{align}$$
Hence, we have
$$\begin{align}
S_n=\frac23(n+1)^{3/2}+\frac{\sqrt{n+1}}{2}+\zeta\left(-\frac12\right)+\frac{(n+1)^{-1/2}}{24}-\frac{(n+1)^{-5/2}}{1920}+O\left((n+1)^{-9/2}\right) \tag 2
\end{align}$$
Now, we see from $(2)$ that $\lim_{n\to \infty}\frac{S_n}{n^{3/2}}=\frac23$.
A: Hint:
$$\ \frac1{n^{\frac32}}\sum_{i=\color{red}1}^{\color{red}n}=\frac1n \sum_{i=1}^{n}\Bigl(\frac in\Bigr)^{\!\frac12}$$
is an upper Riemann sum for the integral of a function on the interval $[0,1]$.
