Applying the Cesaro-Stolz Theorem recursively I have fo find: $$\displaystyle\lim_{n\to\infty}\frac{1}{n}\Bigg(1+\frac{2}{1+\sqrt{2}}+\frac{3}{1+\sqrt{2}+\sqrt{3}}+\cdots+\frac{n}{1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}\Bigg)$$
using the Cesaro-Stolz Theorem.
Applying what the theorem states once i get: $\displaystyle\lim_{n\to\infty} \frac{n+1}{1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}$
My question is: can i apply it again to get: $\displaystyle\lim_{n\to\infty}\frac{1}{\sqrt{n+1}}=0$?

S-C is the easiest way to calculate 
$$\displaystyle\lim_{n\to\infty} \frac{n+1}{1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}$$
If you want an alternate solution, note that
$$1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}=\sqrt{n} \sum_{k=1}^n \sqrt{\frac{k}{n}} \,,$$
and 
$$\lim_n \frac{1}{n}  \sum_{k=1}^n \sqrt{\frac{k}{n}} =\int_0^1 \sqrt{x} dx \,.$$
Then
$$\displaystyle\lim_{n\to\infty} \frac{n+1}{1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}=\displaystyle\lim_{n\to\infty} \frac{n+1}{n\sqrt{n}}\frac{1}{ \frac{1}{n} \sum_{k=1}^n \sqrt{\frac{k}{n}} }=0$$
