How to show $\lim_{n \to \infty} \frac{1}{\sqrt{n}} \frac{1}{n} \Sigma_{k=1}^n \sqrt{k-1} = \frac{2}{3}$? How to show $$\lim_{n \to \infty} \frac{1}{\sqrt{n}} \frac{1}{n} \sum_{k=1}^n \sqrt{k-1} = \frac{2}{3}$$
In fact, this is the lower sum of the integral of $\sqrt{x}$ from 0 to 1. So the value of the above must be $\frac{2}{3}$. But how to show?
 A: $f(x)=\sqrt{x}$ is an increasing and continuous function over $[0,1]$: this gives
$$ \frac{1}{n}\left[f(0)+f\left(\tfrac{1}{n}\right)+\ldots+f\left(\tfrac{k-1}{n}\right)\right]<\int_{0}^{1}\sqrt{x}\,dx < \frac{1}{n}\left[f\left(\tfrac{1}{n}\right)+f\left(\tfrac{2}{n}\right)+\ldots+f(1)\right]. $$
The middle term equals $\frac{2}{3}$ and the difference between the RHS and the LHS is $\frac{1}{n}$.
This proves the wanted identity through elementary manipulations.

Alternative approach, without integrals. Let us consider $g(x)=x\sqrt{x}$. We may notice that
$$\begin{eqnarray*} g(n+1/2)-g(n-1/2) &=&\frac{(n+1/2)^3-(n-1/2)^3}{(n+1/2)^{3/2}+(n-1/2)^{3/2}}\\&=&\frac{3n^2+\frac{1}{4}}{2n^{3/2}+O\left(\frac{1}{\sqrt{n}}\right)}\\&=&\frac{3}{2}\sqrt{n}+O\left(\frac{1}{n\sqrt{n}}\right).\end{eqnarray*}$$
Summing both the RHS and the LHS for $n=1,\ldots,N$ leads to
$$ (N+1/2)^{3/2}-O(1) = \frac{3}{2}\sum_{n=1}^{N}\sqrt{n}+O(1) $$
or
$$ \sum_{n=1}^{N}\sqrt{n} = \frac{2}{3}(N+1/2)^{3/2}+O(1). $$
If you divide both sides by $N^{3/2}$ and consider the limit as $N\to +\infty$ you have the claim.
This technique is known as creative telescoping.
A straightforward generalization gives that for any $\alpha>0$ the sum $\sum_{n=1}^{N}n^\alpha$ and the integral $\int_{0}^{N}x^\alpha\,dx $ share the leading term of the asymptotic expansion, i.e. $\frac{N^{\alpha+1}}{\alpha+1}$.
A: Using right Riemann sums, \begin{align*}
\frac{1}{{\sqrt n }}\frac{1}{n}\sum\limits_{k = 1}^n {\sqrt {k - 1} } & = \frac{1}{{\sqrt n }}\frac{1}{n}\sum\limits_{k = 1}^{n - 1} {\sqrt k }  = \frac{1}{n}\sum\limits_{k = 1}^{n - 1} {\sqrt {\frac{k}{n}} } \\ & = \left( {\frac{1}{n}\sum\limits_{k = 1}^n {\sqrt {\frac{k}{n}} } } \right) - \frac{1}{n} \to \int_0^1 {\sqrt x dx}  - 0 = \frac{2}{3}.
\end{align*}
A: I'll provide a proof that do not use integrals (and will probabily make you apreciate FTC). We can add an extra $\sqrt{n}$ without changing the limit. Will show now that the following limit exists.
\begin{gather*}
L=\lim_{n \to +\infty}\frac{\sum_{i=1}^n \sqrt{i}}{n\sqrt{n}}
\end{gather*}
To show the limit exist, notice that, notice that
\begin{align*}
\frac{i}{\sqrt{i+1} + \sqrt{i}} &\leq \frac{i}{2\sqrt{i}} \leq \frac{i}{\sqrt{i} + \sqrt{i-1}} \\
2i(\sqrt{i+1} - \sqrt{i}) &\leq \sqrt{i} \leq 2i(\sqrt{i} - \sqrt{i-1})
\end{align*}
Now
\begin{gather*}
\sum_{i=1}^n 2i(\sqrt{i+1} - \sqrt{i}) = 2\sum_{i=1}^n (i+1)\sqrt{i+1} - i\sqrt{i} - \sqrt{i+1} = \\
= 2 \left( (n+1)\sqrt{n+1} - 1 - \sum_{j=2}^{n+1}\sqrt{j} \right) = 2\left( n\sqrt{n+1} - \sum_{j=1}^{n}\sqrt{j} \right)
\end{gather*}
so
\begin{align*}
2n\sqrt{n+1} - 2\sum_{i=1}^{n}\sqrt{i} &\leq \sum_{i=1}^n \sqrt{i} \\
\frac{2}{3}n\sqrt{n+1} &\leq \sum_{i=1}^n \sqrt{i}
\end{align*}
Proceding similarly with the other inequality yields
\begin{gather*}
\sum_{i=1}^n 2i(\sqrt{i} - \sqrt{i-1}) = 2\sum_{i=1}^n i\sqrt{i} - (i-1)\sqrt{i-1} - \sqrt{i-1} = \\
= 2 \left( n\sqrt{n} - \sum_{j=0}^{n-1}\sqrt{j} \right) = 2\left( (n+1)\sqrt{n} - \sum_{j=1}^{n}\sqrt{j} \right)
\end{gather*}
so
\begin{align*}
\sum_{i=1}^n \sqrt{i} &\leq 2(n+1)\sqrt{n} - 2\sum_{i=1}^{n}\sqrt{i} \\
\sum_{i=1}^n \sqrt{i} &\leq \frac{2}{3}(n+1)\sqrt{n} 
\end{align*}
Finally, we can apply squeeze theorem as
\begin{gather*}
\frac{2}{3} \sqrt{1 + \frac{1}{n}} \leq \frac{\sum_{i=1}^n \sqrt{i}}{n\sqrt{n}} \leq \frac{2}{3} (1 + \frac{1}{n})
\end{gather*}
A: A solution based on Stolz-Cesàro theorem. Let $f(x)=x^{\frac{3}{2}}$
\begin{align*}
\lim_{n \to \infty} \frac{\sum_{i=1}^{n-1}\sqrt{i}}{n\sqrt{n}} = \lim_{n \to \infty} \frac{\sqrt{n}}{(n+1)\sqrt{n+1} - n\sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{n}}{(1 + \frac{1}{n})^{\frac{3}{2}} - 1} = \frac{1}{f'(1)} = \frac{2}{3}
\end{align*}
