Evaluation of the Lebesgue Integral I'm having some trouble with the second point of this question and I'm not completely sure that I made the first right so:
For $x\in\mathbb{R} - \{0\}$, let $f_n(x) = \frac{1}{(1 + n^2 + x^2)(\arctan (x^2))^{1/n}}$, $n \in\mathbb{N} - \{0\}$
(i) determine for which $n ∈ \mathbb{N}, f_n$ ∈ $L^1(\mathbb{R})$
(ii) evaluate  $\lim_{n\to \infty}$ $\int_{\mathbb{R}} f_n \,dm$;
For the fist part I notice that the $f_n$ are a.e. continuous so that implies that they are measurable, and that the functions are even, so I can narrow my analysis to $[0, \infty )$.
Then when $x\rightarrow\infty, \, f_n \sim \frac1{x^2\bigl({\pi\over2} \bigr)^{1 \over n}} \le {1 \over {x^2}}$ so I can say that the integral converges for every $n\in\mathbb{N} - \{0\}$
Instead when $x \rightarrow 0$,  $f_n \sim \frac1{(1+x^2)x^{2 \over n}} \le {1 \over {x^{2\over n}}}$ that converges for every $n \gt 2
$.
Then we came to the second point, here I could not find a an integral majorant to use Lebesgue's dominated convergence theorem and I have not even been able to determine if there is a chain of the type $f_1 \le f_2 \le f_3 \le \cdots \le f_n \le \cdots$ in order to use the monotone convergence theorem in order to pass the limit under sign of integral and evaluate it.
As an addition I think that the integral in the end is zero because $\lim_{n\to \infty} f_n = 0$.
Thank you very much.
 A: Just note that for any fixed $\epsilon \in(0,1)$ then $0\leqslant f_n(x)\leqslant \frac1{(1+x^2)\arctan (\epsilon ^2) }$ for all $x\in \mathbb{R}\setminus (-\epsilon ,\epsilon )$ and all $n\in \mathbb{N}$ so the dominated convergence theorem show us that
$$
\lim_{n\to\infty}\int_{\mathbb{R}\setminus (-\epsilon ,\epsilon )}f_n\mathop{}\!d \lambda =0
$$
Now note that $|x|/2\leqslant |\arctan x|$ for all $x\in (-\epsilon ,\epsilon )$. Therefore
$$
0\leqslant \int_{(-\epsilon ,\epsilon )}f_n\mathop{}\!d \lambda \leqslant \int_{(-\epsilon ,\epsilon )}\frac2{n^2|x|^{2/n}}\mathop{}\!d x =\int_{(0,\sqrt[n]{\epsilon })}\frac4n y^{n-3}\mathop{}\!d y\leqslant \int_{(0,1)}\frac4n\epsilon ^{1-3/n}\mathop{}\!d y
$$
Then the dominated convergence theorem show us that
$$
\lim_{n\to\infty}\int_{(0,\sqrt[n]{\epsilon })}\frac4n y^{n-3}\mathop{}\!d y=0\implies \lim_{n\to\infty}\int_{(-\epsilon ,\epsilon )}f_n\mathop{}\!d \lambda =0
$$
so we conclude that $\lim_{n\to\infty}\int_{\mathbb{R}}f_n\mathop{}\!d \lambda =0$.
