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?
$\boldsymbol{1.}$ 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:
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$\boldsymbol{2.}$ 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 $\displaystyle\sum_{n=1}^\infty\frac1{n^{3/2}}$ .