Csdrhrt

Evaluate
$$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}}.$$

I am trying to use the Sandwich principle here..

$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}}=lim_{n to infty}dfrac{sqrt{dfrac{1}{n}}+sqrt{dfrac{2}{n}}+sqrt{dfrac{3}{n}}+...+sqrt{dfrac{n-1}{n}}}{n}ge lim_{n to infty}bigg(sqrt{dfrac{1}{n}}.sqrt{dfrac{2}{n}}.sqrt{dfrac{3}{n}}....sqrt{dfrac{n-1}{n}}bigg )^dfrac{1}{n-1}=lim_{n to infty}bigg(dfrac{(n-1)!}{n^n}bigg )^dfrac{1}{2(n-1)}$

But after this I am a little in doubt.
This link may provide some light Evaluation of the limit $limlimits_{n to infty } frac1{sqrt n}left(1 + frac1{sqrt 2 }+frac1{sqrt 3 }+cdots+frac1{sqrt n } right)$ but I do not understand how it would help my problem..

In continuation of @Rebello's answer here, I would like to provide an answer for the problem given in the link

$limlimits_{n to infty } frac1{sqrt n}left(1 + frac1{sqrt 2 }+frac1{sqrt 3 }+cdots+frac1{sqrt n } right)=limlimits_{n to infty }sum_{k=1}^{n}{dfrac{1}{sqrt{kn}}}=int_{0}^{1}dfrac{1}{sqrt{x}}dx+limlimits_{n to infty }dfrac{1}{n}=2$

Hint :

$$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}} = lim_{n to infty} frac{1}{n} sum_{k=0}^{n-1} sqrt{frac{k}{n}} = int_0^1sqrt{x}mathrm{d}x$$

Just for your curiosity since you already received good answers.

We can approximate the value of
$$a_n=frac{sum_{i=1}^{n-1} sqrt i }{n sqrt n}$$
$$sum_{i=1}^{n-1} sqrt i =sum_{i=1}^{n} sqrt i -sqrt n=H_n^{left(-frac{1}{2}right)}-sqrt n$$ where appear generalized harmonic numbers.

Now, using the asymptotics
$$H_n^{left(-frac{1}{2}right)}=frac{2 nsqrt n}{3}+frac{sqrt{n}}{2}+zeta
left(-frac{1}{2}right)+frac{1}{24sqrt n}+Oleft(frac{1}{n^{5/2}}
right)$$ making, for large $n$
$$a_n=frac{frac{2 nsqrt n}{3}-frac{sqrt{n}}{2}+zeta
left(-frac{1}{2}right)+frac{1}{24sqrt n}+Oleft(frac{1}{n^{5/2}}
right)} {n sqrt n}=frac{2}{3}-frac{1}{2 n}+frac{zeta
left(-frac{1}{2}right) } {n sqrt n }+Oleft(frac{1}{n^{2}}
right)$$ which, for sure, shows the limit and also how it is approached.

But it also gives a good approximation even for small values of $n$. Using $zeta
left(-frac{1}{2}right)approx -0.207886$ and $n=10$, this would give $a_{10}approx 0.610093$ while the "exact" value is $a_{10}approx 0.610509$.

One more approach. By Stolz-Cesaro Theorem
$$begin{align}lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}}{nsqrt{n}} &= lim_{n to infty} frac{sqrt{n}}{(n+1)sqrt{n+1}-nsqrt{n}}
\&=lim_{n to infty} frac{1}{(n+1)sqrt{1+frac{1}{n}}-n}\
&=frac{1}{1+frac{1}{2}}=frac{2}{3}end{align}$$
where we used the fact that $sqrt{1+frac{1}{n}}=1+frac{1}{2n}+o(1/n)$.

It is sufficient to obtain integral of $sqrt{x}$, when $0leq xleq 1$.