Clarifications on $\sum_{k=1}^{+\infty} a_n$ with $a_n=\int_0^{\frac{1}{n^a}}\sin{\sqrt[3]{x}}\, dx$ with $a>0$ I do this question as a sort of continuation of this previuous one (Study the series of $\int_0^{\frac{1}{n^a}}\sin{(\sqrt[3]{x})}\,dx$ with respect to $a>0$), I mean I have other extra doubts related to this question that I have aked in comments but maybe it is the case to ask a question apart.
If I have $\sum_{k=1}^{+\infty} a_n$ with $a_n=\int_0^{\frac{1}{n^a}}\sin{\sqrt[3]{x}}\, dx$ with $a>0$. I have $2$ questions:
$1)$ This is a positive terms series? I think yes since $\sin{(\sqrt[3]{x})}>0$ in $(0,n^{-a})\subseteq (0,1)$ and in $(0,1)$ I have $\sin{\sqrt[3]{x}}>0$
$2)$I have not understood why I use the Mclaurin polynomial of $\sin{\sqrt[3]{x}}$. I mean I have to consider that $x\to 0^+$ since when $n\to \infty$ I am in a neighbourhood of $0$?
$3)$When I have two asymptotic functions $f\sim g$ as $x\to 0$ then it is not necessarily true that $\int_0^{n^{-a}}f\sim \int_0^{n^{-a}}g$ as $n\to \infty$, right? Can I consider $f(x)=\frac{1}{\sqrt{x}}$ and $g(x)=\frac{1}{\sqrt{x}}+\frac{1}{\sqrt[3]{x}}$ as a counterexample?
 A: *

*Yes, indeed.


*Yes: if $\alpha\downarrow 0$, then since $0 \leq x\leq  \alpha$, $x$ goes to $0$ as well. (This is a little handwavy phrased like this, but that's the gist of it).


*Let $\alpha \to 0^+$, and $f,g\geq 0$ such that $f\sim_0 g$. By definition, this means that there exists $\varepsilon\colon \mathbb{R}\to\mathbb{R}$ such that $f(x) = g(x)+\varepsilon(x)g(x)$ for all $x$ and $\lim_{x\to 0} \varepsilon(x) = 0$. Now,
$$
\int_0^{\alpha} f(x)dx
= \int_0^{\alpha} g(x)dx + \int_0^{\alpha} \varepsilon(x) g(x)dx
$$
Now, since
$$
\left|\int_0^{\alpha} \varepsilon(x) g(x)dx\right| \leq \sup_{x\in(0,\alpha)} |\varepsilon(x)| \cdot \int_0^{\alpha} g(x)dx
$$
and $\sup_{x\in(0,\alpha)} |\varepsilon(x)| \xrightarrow[\alpha\to 0]{} 0$, we do have
$$
\int_0^{\alpha} f
= \int_0^{\alpha} g + o\!\left(\int_0^{\alpha} g \right)
$$
as $\alpha\to 0$, which means
$$
\int_0^{\alpha} f \operatorname*{\sim}_{\alpha \to 0}\int_0^{\alpha} g
$$
Note that in the example you gave,
$$
\int_0^\alpha f = 2\alpha^{1/2}, \qquad \int_0^\alpha g = 2\alpha^{1/2} + \frac{3}{2}\alpha^{2/3} = 2\alpha^{1/2} + o(\alpha^{1/2})
$$
so there is no contradiction.
