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.
 A: 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*}
A: If you are familiar with generalized harmonic numbers.
The expression you are considering is
$$a_n=\frac{1}{\sqrt{n}}\sum _{p=1}^n \frac{p}{H_p^{\left(-\frac{1}{2}\right)}}$$ and, for large $p$, we have
$${H_p^{\left(-\frac{1}{2}\right)}}=\frac{2 p^{3/2}}{3}+\frac{\sqrt{p}}{2}+\zeta
   \left(-\frac{1}{2}\right)+\frac{1}{24 \sqrt{p}}+O\left(\frac{1}{
   p^{5/2}}\right)$$
$$\frac{p}{H_p^{\left(-\frac{1}{2}\right)}}=\frac {3}{2\sqrt{p}}-\frac {9}{8p\sqrt{p}}+\cdots$$
$$\sum _{p=1}^n \frac{p}{H_p^{\left(-\frac{1}{2}\right)}}=3 \sqrt{n}+\left(\frac{3 \zeta \left(\frac{1}{2}\right)}{2}-\frac{9 \zeta
   \left(\frac{3}{2}\right)}{8}\right)+O\left(\frac{1}{n^{1/2}}\right)$$
$$a_n=3+\left(\frac{3 \zeta \left(\frac{1}{2}\right)}{2}-\frac{9 \zeta
   \left(\frac{3}{2}\right)}{8}\right)\frac 1 {\sqrt n}+O\left(\frac{1}{n^{3/2}}\right)$$ which shows the limit and how it is approached.
A: $\sum_{k=1}^n k^{1/2}
\approx \int_0^n x^{1/2} dx
=\dfrac{n^{3/2}}{3/2}
=\dfrac23 n^{3/2}
$
and
$\sum_{k=1}^n k^{-1/2}
\approx \int_0^n x^{-1/2} dx
=\dfrac{n^{1/2}}{1/2}
=2n^{1/2}
$
so
$\begin{array}\\
s(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}})\\
&=\sum_{k=1}^n \dfrac{k}{\sum_{j=1}^k j^{1/2}}\\
&\approx\sum_{k=1}^n \dfrac{k}{\dfrac23 k^{3/2}}\\
&=\dfrac32\sum_{k=1}^n k^{-1/2}\\
&\approx\dfrac32(2n^{1/2})\\
&=3n^{1/2}\\
\end{array}
$
so the limit is $3$.
