Find $\lim\limits_{n \rightarrow \infty}(\frac{1}{\sqrt{n}}\times\sum_{k=1}^n\left|\frac{a_k}{\sqrt{k}}\right|)$, $a_n$ converges to $a$ $a_n$ converges to $a \in \mathbb{R}$, find $\lim\limits_{n \rightarrow  \infty}\frac{1}{\sqrt{n}}(\frac{a_1}{\sqrt{1}} + \frac{a_2}{\sqrt{2}} +\frac{a_3}{\sqrt{3}} + \cdots + \frac{a_{n-1}}{\sqrt{n-1}} + \frac{a_n}{\sqrt{n}})$
I've tried to use Squeeze theorem. Nothing good. Maybe I've tried it in a wrong way. According to my assumptions, this sequence diverges (I mean "goes" to infinity).
Can we solve this using Stolz–Cesàro theorem? Is there easier/better/more correct approach? If there is, please, share your thoughts
 A: Define $$A_n=\sum_{k=1}^{n}\frac{a_k}{\sqrt{k}}$$ and $B_n=\sqrt n.$
The we want to find the limit of $\frac{A_n}{B_n}.$
Then Stolz-Cesàro requires us to find the limit:
$$\frac{A_{n+1}-A_n}{B_{n+1}-B_n}=\frac{a_{n+1}}{\sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)}$$
But $$\frac{1}{\sqrt{n+1}-\sqrt n}=\sqrt{n+1}+\sqrt n.$$
So you get $$\frac{\sqrt{n+1}+\sqrt n}{\sqrt {n+1}}\cdot a_{n+1}\to 2a.$$
Now, Stolz requires $A_n\to \infty$ and $B_n\to\infty,$ or both converge to $0.$ So to apply Stolz, you'll probably need $a$ to be non-zero.
So you might have to prove the case $a=0$ separately.
A: Given $\varepsilon>0$, there is $N$ such that $\sup_{n\geq N}|a_n-a|<\varepsilon$
$$\frac{1}{\sqrt{n}}\sum^n_{k=1}\frac{1}{\sqrt{k}}=\frac{1}{\sqrt{n}}\sum^n_{k=1}\frac{a_k-a}{\sqrt{k}} +\frac{a}{\sqrt{n}}\sum^n_{k=1}\frac{1}{\sqrt{k}}$$
The second term is the Riemann sum of the convergent integral $a\int^1_0\frac{1}{\sqrt{x}}$ and thus,  as $x\mapsto\frac{1}{\sqrt{x}}$ is monotone nondecreasing,
$$\frac{a}{\sqrt{n}}\sum^n_{k=1}\frac{1}{\sqrt{k}}\xrightarrow{n\rightarrow\infty}2a$$
The first term can be estimated similarly.
$$\Big|\frac{1}{\sqrt{n}}\sum^n_{k=1}\frac{a_k-a}{\sqrt{k}}\Big|\leq\frac{1}{\sqrt{n}}\sum^N_{k=1}\frac{|a_k-a|}{\sqrt{k}}+\varepsilon\frac{1}{\sqrt{n}}\sum^n_{k=N+1}\frac{1}{\sqrt{k}}$$
The  term $\frac{1}{\sqrt{n}}\sum^N_{k=1}\frac{|a_k-a|}{\sqrt{k}} \xrightarrow{n\rightarrow\infty}0$ (sińce $N$ is fixed), the second one is bounded by $\frac{\varepsilon}{n}\sum^n_{k=1}\frac{\sqrt{n}}{\sqrt{k}}\xrightarrow{n\rightarrow\infty}\varepsilon\int^1_0\frac{1}{\sqrt{x}}=2\varepsilon$.
Putting things together,
$$\limsup_n\Big|\frac{1}{\sqrt{n}}\sum^n_{k=1}\frac{a_k}{\sqrt{k}}  -2a\Big|\leq2\varepsilon$$
Therefore  limit of the expression in the OP exists and is $2a$.
A: *

*If $a \neq 0$ then $a_n \sim a$ so $\dfrac{a_n}{\sqrt{n}} \sim \dfrac{a}{\sqrt{n}}$.
The series $\sum \dfrac{a}{\sqrt{n}}$ diverges then :
$$\sum_{k = 1}^n \dfrac{a_k}{\sqrt{k}} \sim \sum_{k = 1}^n \dfrac{a}{\sqrt{k}}$$
We deduce that :
$$\dfrac{1}{\sqrt{n}} \sum_{k = 1}^n \dfrac{a_k}{\sqrt{k}} \sim \dfrac{1}{\sqrt{n}} \sum_{k = 1}^n \dfrac{a}{\sqrt{k}} = \dfrac{a}{n} \sum_{k = 1}^n \dfrac{1}{\sqrt{\dfrac{k}{n}}} \to a \int_0^1 \sqrt{x} \mathrm{d}x = 2 a$$
Finally :
$$\dfrac{1}{\sqrt{n}} \sum_{k = 1}^n \dfrac{a_k}{\sqrt{k}} \to 2a$$

*If $a = 0$ then $a_n = o(1)$ so $\dfrac{a_n}{\sqrt{n}} = o \left(\dfrac{1}{\sqrt{n}}\right)$.
The series $\sum \dfrac{1}{\sqrt{n}}$ diverges then :
$$\sum_{k = 1}^n \dfrac{a_k}{\sqrt{k}} = o\left(\sum_{k = 1}^n \dfrac{1}{\sqrt{k}}\right)$$
We deduce that :
$$\dfrac{1}{\sqrt{n}} \sum_{k = 1}^n \dfrac{a_k}{\sqrt{k}} = o \left(\dfrac{1}{\sqrt{n}} \sum_{k = 1}^n \dfrac{1}{\sqrt{k}}\right) = o\left(\dfrac{1}{n} \sum_{k = 1}^n \dfrac{1}{\sqrt{\dfrac{k}{n}}}\right) = o(1)$$
because :
$$\dfrac{1}{n} \sum_{k = 1}^n \dfrac{1}{\sqrt{\dfrac{k}{n}}} \to \int_0^1 \sqrt{x} \mathrm{d}x = 2$$
Finally :
$$\dfrac{1}{\sqrt{n}} \sum_{k = 1}^n \dfrac{a_k}{\sqrt{k}} \to 0$$
