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: $\displaystyle\lim_{n\to\infty}\frac{1}{\sqrt{n+1}}=0$?
 A: 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
$$\displaystyle\lim_{n\to\infty} \frac{n+1}{1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}=\displaystyle\lim_{n\to\infty} \frac{n+1}{n\sqrt{n}}\frac{1}{ \frac{1}{n} \sum_{k=1}^n \sqrt{\frac{k}{n}} }=0$$
A: This is an alternative method, but you could use something like:
$$\lim_{x\to\infty}\sum_{k=0}^{n-1}\frac{n-k}{n}\cdot\Bigg(\sum_{i=1}^{n-k}\sqrt{i}\Bigg)^{-1}$$
I would definitely recommend to switch the order of the terms (although it might seem weird since you are looking $\infty\to 1$).
$\displaystyle\lim_{n\to\infty}\frac{1}{n}\Bigg(\frac{2}{1+\sqrt{2}}+\frac{3}{1+\sqrt{2}+\sqrt{3}}+\cdots+\frac{n}{1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}\Bigg)=$
$\displaystyle\lim_{n\to\infty}\frac{1}{n}\Bigg(\frac{n}{1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}+\frac{n-1}{1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n-1}}+\ldots+\frac{2}{1+\sqrt{2}}+1\Bigg)$
Now, this looks a bit 'more recursive' (:
$=\displaystyle\lim_{x\to\infty}\Big(\frac{1}{1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}}+\frac{n-1}{n}\cdot\frac{n-1}{1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n-1}}+\ldots+\frac{1}{n}\Bigg)$
$=\displaystyle\lim_{x\to\infty}\sum_{k=0}^{n-1}\frac{n-k}{n}\cdot\underbrace{\Bigg(\sum_{i=1}^{n-k}\sqrt{i}\Bigg)^{-1}}_{{\to 0}\;\forall k=0,\ldots n-1}=0$
