How to check the Lebesgue integrability?

Question

I have an elementary problem about Lebesgue integrability. How to check the Lebesgue integrability?

I tried to check it by definition, but this definition, you know, is very constructable. So I cannot use it.

Here is my example:

For fixed $n\ge 2$, is $f_n(x)=(1+\frac{x}{n})^{-n}$ Lebesgue-integrable over $(1,\infty)$?

I cannot start to prove this problem. I'm waiting your hint or solution. Thanks in advance.

Answer 1

For $n\geq 2, x> 1$, $f_n(x) \leq (\frac{x}{n})^{-n} = \frac{n^n}{x^n}$ which is integrable on $(1,\infty)$ for $n\geq 2$.

Answer 2

For all $n\geqslant2$ and $x\in(1,\infty)$ we have $0\leqslant1+\frac xn\leqslant e^{\frac xn}$ and hence $$0\leqslant\left(1+\frac xn\right)^{-n} \leqslant \left(e^{\frac xn}\right)^{-n}=e^{-x}. $$ It follows that $$\int_1^\infty \left(1+\frac xn\right)^{-n}\,\mathsf dx \leqslant\int_1^\infty e^{-x}\,\mathsf dx = e^{-1}<\infty, $$ so the function $x\mapsto \left(1+\frac xn\right)^{-n}$ is integrable on $(1,\infty)$.

Note that this also allows us to use the dominated convergence theorem to compute $$\lim_{n\to\infty}\int_1^\infty\left(1+\frac xn\right)^{-n}\,\mathsf dx = \int_1^\infty \lim_{n\to\infty}\left(1+\frac xn\right)^{-n}\,\mathsf dx=\int_1^\infty e^{-x}\,\mathsf dx = e^{-1}. $$