Deciding whether series containing $a_n$ are convergent knowing that $\lim_{n\rightarrow\infty}\frac{\frac{(-1)^n}{\sqrt{n}}}{a_n}=1$ We know the following thing about sequence ${a_n}$:
$$\lim_{n\rightarrow\infty}\frac{\frac{(-1)^n}{\sqrt{n}}}{a_n}=1$$
And now the problem asks us whether it's true for every such $a_n$ that:


*

*$\sum_{n=1}^{+\infty} a_n$ is convergent

*$\sum_{n=1}^{+\infty} \frac{a_n}{n}$ is convergent

*$\sum_{n=1}^{+\infty} \frac{a_n}{n}$ is absolutely convergent


So even though the upper statement looks like we should do something with limit comparison test, I think we cannot do that as we know  that $a_n > 0$ is not true for every such $a_n$. Intuitevely all of these points should be true but what should I use in order to prove that?
 A: $\boldsymbol{1.}$ is false:
Let
$$
a_n=\frac{(-1)^n}{\sqrt{n}}+\frac1n\tag{1}
$$
Because $\dfrac1{\sqrt{n}}$ monotonically decreases to $0$, the Alternating Series Test says
$$
\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{n}}\text{ converges}\tag{2}
$$
Because $\dfrac1n$ monotonically decreases to $0$ and $\displaystyle\int_1^\infty\frac1x\,\mathrm{d}x$ diverges, the Integral Test says
$$
\sum_{n=1}^\infty\frac1n\text{ diverges}\tag{3}
$$
Therefore, since the sum of a convergent series and a divergent series diverges,
$$
\sum_{n=1}^\infty a_n\text{ diverges}\tag{4}
$$
However,
$$
\begin{align}
\lim_{n\to\infty}\frac{\frac{(-1)^n}{\sqrt{n}}}{a_n}
&=\lim_{n\to\infty}\frac{\frac{(-1)^n}{\sqrt{n}}}{\frac{(-1)^n}{\sqrt{n}}+\frac1n}\\
&=\lim_{n\to\infty}\frac1{1+\frac{(-1)^n}{\sqrt{n}}}\\[4pt]
&=1\tag{5}
\end{align}
$$

There are some claims that $\sum\limits_{n=1}^\infty a_n$ should converge by the Alternating Series Test.
Although
$$
|a_n|=\frac1{\sqrt{n}}+\frac{(-1)^n}n\tag{6}
$$
tends to $0$, it does not do so monotonically. To see this, note that
$$
\begin{align}
\frac1{\sqrt{n}}-\frac1{\sqrt{n+1}}
&=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}\\
&=\frac1{\sqrt{n}\sqrt{n+1}(\sqrt{n+1}+\sqrt{n})}\\
&\le\frac1{2n^{3/2}}\tag{7}
\end{align}
$$
Thus, for positive odd $n$
$$
\begin{align}
&\left[\frac1{\sqrt{n}}+\frac{(-1)^n}n\right]-\left[\frac1{\sqrt{n+1}}+\frac{(-1)^{n+1}}{n+1}\right]\\
&=\left[\frac1{\sqrt{n}}-\frac1{\sqrt{n+1}}\right]-\left[\frac1n+\frac1{n+1}\right]\\
&\le\frac1{2n^{3/2}}-\frac1n\\[12pt]
&\lt0\tag{8}
\end{align}
$$
Thus, the terms increase from odd $n$ to even $n+1$.
Here is a plot of $\displaystyle\frac1{\sqrt{n}}+\frac{(-1)^n}n$ showing the non-monotonicity:
$\hspace{3cm}$

$\boldsymbol{2.}$ and $\boldsymbol{3.}$ are true:
Since
$$
\left|\frac{a_n}n\right|\sim\frac1{n^{3/2}}\tag{9}
$$
$\displaystyle\sum_{n=1}^\infty\frac{a_n}n$ converges absolutely by comparison to $\displaystyle\sum_{n=1}^\infty\frac1{n^{3/2}}$ .
