Evaluating a limit involving a sum I want to evaluate
$$ \lim_{n \to \infty} \frac{1}{n} \left( \sqrt{\frac{1}{n}} + \sqrt{\frac{2}{n}} + \cdots +\sqrt{\frac{n}{n}} \right) $$
Try:
If we write this as a summation formula we have that our limit is
$$ \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n} \sqrt{\frac{k}{n}} $$
But, isnt it just the integral $\int_0^1 \sqrt{x} \, dx $?? which is $\frac{2}{3}$??
Is this correct?

Beside the integral approach, you could also use generalized harmonic numbers since $$\sum_{i=1}^n\sqrt{\frac{i}{n}}=\frac 1{\sqrt n} H_n^{\left(-\frac{1}{2}\right)}$$ which makes $$A_n=\frac 1{n^{3/2}}\sum_{i=1}^n\ H_n^{\left(-\frac{1}{2}\right)}$$ and using the asymptotics of the harmonic numbers $$A_n=\frac{2}{3}+\frac{1}{2 n}+O\left(\frac{1}{n^{3/2}}\right)$$ which shows the limit and also how it is approched.
For example, using $n=1000$, the exact value is $\approx 0.66716013$ while the above approximation leads to $\approx 0.66716667$.
