Find $\lim_{n\to\infty}\int_{[0,\infty[}\frac{x}{\sqrt[n]{1+x^{3n}}}$

Question

find $\lim_{n\to\infty}\int_{[0,\infty[}\frac{x}{\sqrt[n]{1+x^{3n}}}d\lambda$

does it converge or diverge?

I know I have the $3$ following options but have not been fruitful thus far:

monotone convergence I cannot prove that $f_{n}(x)$ is monotone for all $x \in [0,\infty[$.

dominating convergence here I am looking at the following: $\frac{x}{\sqrt[n]{1+x^{3n}}}\leq\frac{1}{x^{2}}$ which does not help me at all as $\frac{1}{x^2} \notin \mathcal{L}^{1}(\lambda)$. Neither does for $x>1$, $\frac{1}{x}\geq\frac{x}{\sqrt[n]{1+x^{2n}}}\geq\frac{x}{\sqrt[n]{1+x^{3n}}}$

My last hope is

Fatou's Lemma but $\liminf_{n\to\infty}\int_{[0,\infty[}\frac{x}{\sqrt[n]{1+x^{3n}}}d\lambda \geq\int_{[0,\infty[}\liminf_{n\to\infty}\frac{x}{\sqrt[n]{1+x^{3n}}}d\lambda$ and nothing meaningful comes out of this either.

Answer 1

The dominated convergence theorem is indeed useful here, but with a better dominating function. Note that $x^{-2}$ is indeed integrable on $[1, \infty)$. On the other hand,

$$\frac{x}{\sqrt[n]{1 + x^{3n}}} \le x$$

for all $x \ge 0$, so you have the (rather efficient) bound

$$0 \le \frac{x}{\sqrt[n]{1 + x^{3n}}} \le x \chi_{[0, 1]}(x) + x^{-2} \chi_{[1, \infty)}(x)$$

which is good enough.