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?

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*}
