Let $f_n (x)= \dfrac{1}{x^{\frac{1}{n}}(1+\frac{x}{n})^n }$. Show that $f_n$ is integrable for $n\geq 2$ For $n\in \mathbb{N}$ such that $n\geq 2$, consider the functions
$f_n : (0,+\infty) \to \mathbb{R}$ given by
$$f_n (x)= \dfrac{1}{x^{\frac{1}{n}}(1+\frac{x}{n})^n }$$


*

*Show that $f_n$ is integrable (Lebesgue) for $n\geq 2$.

*Show that for $n\geq 2$, $x\geq 1$, $f_n (x)\leq \dfrac{4}{x^2}$.

*Calculate $\lim_{n\to \infty} \int_{(0,+\infty)} f_n (x) dx$.
I think for part 3 I should use the Dominated Convergence Theorem by using part 2 for the interval $[1, \infty)$, but then I should prove that $\dfrac{1}{x^2}$ is integrable in that interval right?
For part 1 and 2 I don't have any idea, so any help would be appreciated, thank you so much!
 A: Assume that $n \geq 2$ and notice that
\begin{align*}
0 < x < 1
&\quad \Rightarrow \quad x^{1/n} \geq \sqrt{x} \quad\text{and}\quad \left(1+\frac{x}{n}\right)^n \geq 1 \\
&\quad \Rightarrow \quad f_n(x) \leq \frac{1}{\sqrt{x}}
\end{align*}
and that
\begin{align*}
x \geq 1
&\quad \Rightarrow \quad x^{1/n} \geq 1 \quad\text{and}\quad \left(1+\frac{x}{n}\right)^n \geq 1 + \binom{n}{1}\frac{x}{n}+\binom{n}{2}\frac{x^2}{n^2} \geq \frac{x^2}{4} \\
&\quad \Rightarrow \quad f_n(x) \leq \frac{4}{x^2}.
\end{align*}
This already proves part (2). Moreover, we obtain the dominating function
$$ g(x) = \begin{cases} x^{-1/2}, & 0 < x < 1 \\ 4x^{-2}, & x \geq 1 \end{cases} $$
which is integrable:
$$ \int_{(0,\infty)} g(x) \, dx
= \int_{0}^{1} \frac{dx}{\sqrt{x}} + \int_{1}^{\infty} \frac{4}{x^2} \, dx
= 2 + 4 = 6 < \infty. $$
Since $0 \leq f_n \leq g$ and $g$ is integrable, the same is true for $f_n$ when $n \geq 2$. This resolves part (1). Finally, applying the dominated convergence theorem
$$ \lim_{n\to\infty} \int_{(0,\infty)} f_n(x) \, dx
= \int_{(0,\infty)} \lim_{n\to\infty} f_n(x) \, dx
= \int_{(0,\infty)} e^{-x} \, dx
= 1. $$
