Lebesgue Dominated Convergence Application

Question

I want to compute the integral

$$\lim_{n\to \infty}\int_0^\infty \frac{\sin\left(\frac{x}{n}\right)}{(1+x/n)^n}\,\mathrm{d}x$$

Since

$$\left| \frac{\sin\left(\frac{x}{n}\right)}{\left(1+\frac{x}{n}\right)^n}\right | \le \frac{1}{\left|\left(1+\frac{x}{n}\right)^n\right|}\le \frac{1}{1+x}$$

for $x\in [0,\infty)$, where I used $(1+x)^n \ge 1+xn$ to obtain the last inequality.

I have therefore found a lebesgue integrable upper bound of the series of the integrand. Since the integrand converges to zero, I would obtain the integral to be zero.

Is my reasoning correct?

EDIT: Corrected the inequality.

Answer 1

The reasoning is not correct because $$ \int_0^{+\infty} \frac{1}{1+x} = + \infty $$ And thus $\frac{1}{1+x}$ is not integrable.

Hint for a correct reasoning:

For $n \geq 2$ we have $(1+y)^n \geq 1 + n \cdot y + \frac{n \cdot (n-1)}{2} \cdot y^2 \space \space \space \forall y \geq 0$

Answer 2

A lower bound: $$ I_n = \int_{0}^{+\infty}\frac{\sin\left(\frac{x}{n}\right)}{\left(1+\frac{x}{n}\right)^n}\,dx \geq \int_{0}^{+\infty}\sin\left(\frac{x}{n}\right)e^{-x}\,dx = \frac{n}{n^2+1} $$ and a compatible upper bound: $$ I_n = n\int_{0}^{+\infty}\frac{\sin(x)}{(1+x)^n}\,dx \leq n\int_{0}^{+\infty}\frac{x}{(x+1)^n}\,dx = \frac{n}{(n-1)(n-2)} $$ prove that the wanted limit is zero and $I_n = \frac{1}{n}\left(1+o(1)\right)$.