Find $\lim_\limits{x\to\infty} \frac{1+\sqrt2+\sqrt3+...+\sqrt{n}}{n\sqrt{n}} $ Evaluate the limit:

$$\lim_{x\to \infty} \frac{1+\sqrt2+\sqrt3+...+\sqrt{n}}{n\sqrt{n}} $$

Am I right in saying that we can apply Stolz-Cesaro to find limit of $\lim_{x\to\infty} \frac{\sqrt{n+1}}{(n+1)\sqrt{n+1} - n\sqrt{n}}$ which is $\lim_{x\to\infty} \frac{1}{(n+1) + n\sqrt\frac{n}{n+1}}$ = $0$ ?
My bad. $\lim_{x\to\infty} \frac{\sqrt{n+1}}{(n+1)\sqrt{n+1} - n\sqrt{n}}$ =
$\lim_{n \to \infty} \dfrac{\sqrt{n+1}[(n+1)\sqrt{n+1}+n\sqrt n]}{(n+1)^3-n^3}$
= $\lim_{n \to \infty} \dfrac{(n+1)^2+n\sqrt{n(n+1)}}{3n^2+3n+1}$
=$\lim_{n \to \infty} \dfrac{n^2+2n+1 + n^2\sqrt{1+\frac{1}{n}}}{3n^2+3n+1}$
=$\frac{2}{3}$
Yes that's fine, for the last step it is better to add this one also
$$ \dfrac{n^2+2n+1 + n^2\sqrt{1+\frac{1}{n}}}{3n^2+3n+1}= \dfrac{1+\frac2n+\frac1{n^2} + \sqrt{1+\frac{1}{n}}}{3+\frac3n+\frac1{n^2}} \to \frac{1+1}{3}=\frac23$$
Yes that's fine, for the last step it is better to add this one also
$$ \dfrac{n^2+2n+1 + n^2\sqrt{1+\frac{1}{n}}}{3n^2+3n+1}= \dfrac{1+\frac2n+\frac1{n^2} + \sqrt{1+\frac{1}{n}}}{3+\frac3n+\frac1{n^2}} \to \frac{1+1}{3}=\frac23$$