<s>[INST] the limit of $\frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt2}+\frac{3}{1+\sqrt2+\sqrt3}+\dots+\frac{n}{1+\sqrt2+\sqrt3+\dots+\sqrt n})$ as $n\to\infty$ I need to find:
$$\lim_{n \to +\infty} \frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt{2}} + \frac{3}{1+\sqrt{2}+\sqrt{3}} + \ldots + \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}), n \in \mathbb{N}$$
Looking at denominators, I see that [(...) represents any element between] : $$ 1 \le (\ldots) \le 1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}$$
Then I take reverses and get:
$$ 1 \ge (\ldots) \ge \frac{1}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$
Then I put the other sequence on top of the former one (I see that the rightmost element is still the smallest one)
$$ 1 \ge (\ldots) \ge \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$
Then I take the sum of n elements on every end of inequality (to sum up n times the biggest element and n times the smallest element) and get:
$$ n \ge (\ldots) \ge \frac{n^2}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$
Ultimately I take into consideration $\frac{1}{\sqrt{2}}$ and get:
$$ \frac{n}{\sqrt{n}} \ge (\ldots) \ge \frac{n^2}{\sqrt{n}(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})}$$
Now I can use the squeeze theorem and get:

*

*$\lim_{n \to +\infty} \frac{n}{\sqrt{n}} = \infty$

*$\lim_{n \to +\infty} \frac{n^2}{\sqrt{n}(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})} = \lim_{n \to +\infty} \frac{\frac{n^2}{\sqrt{n}}}{(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})} \implies Stolz = \lim_{n \to +\infty} \frac{\frac{(n+1)^2}{\sqrt{n+1}}-\frac{(n)^2 }{\sqrt{n}}}{\sqrt{n+1}}$
And that is pretty disappointing - I think that the solution is wrong. Does anybody see an error in my way of thinking?
Unfortunately, I can not use integrals while doing that exercise.
 [/INST] The Stolz–Cesàro theorem indeed suffices, with a derivative-like limit (the last one below).
Let $a_n=\sum\limits_{k=1}^n\sqrt{k}$ and $b_n=\sum\limits_{k=1}^n k/a_k$; we're computing
\begin{align*}
\lim_{n\to\infty}\frac{b_n}{\sqrt{n}}
&=\lim_{n\to\infty}\frac{b_n-b_{n-1}}{\sqrt{n}-\sqrt{n-1}}
\\&=\lim_{n\to\infty}\frac{n}{a_n}(\sqrt{n}+\sqrt{n-1})
\\&=2\lim_{n\to\infty}\frac{n^{3/2}}{a_n}
\\&=2\lim_{n\to\infty}\frac{n^{3/2}-(n-1)^{3/2}}{a_n-a_{n-1}}
\\&=2\lim_{n\to\infty}\frac{n^{3/2}-(n-1)^{3/2}}{\sqrt{n}}
\\&=\color{gray}{2\lim_{n\to\infty}\frac{1-(1-1/n)^{3/2}}{1/n}}=\mathbf{3}.
\end{align*}
</s>