Convergence in $L^1$ of a sequence of functions

Question

I have to see if the following sequence of functions is convergent in the space $L^1[(0,\infty)]$

$$f_n(x)= n\frac{\exp\left(-\frac{n}{2x^2}\right)}{x^3}$$

By definition, $f_n(x)$ is convergent in $L^1[(0,\infty)]$ if and only if:

For any $\epsilon >0, \exists n_0 \in N\,|$ any $n>n_0 ||f_n -f||_1 < \epsilon$

$$||f_n -f||_1= \int_{(0,\infty)}\left|\frac{n}{x^3}\exp\left(-\frac{n}{2x^2}\right)\right|dx= e^{-(n/2x^2)}\Big|_{(\infty,0)}$$

How can I continue?

Thank you very much

Answer 1

Note that: $$||f_n -f||_1= \int_{(0,\infty)}\left|\frac{n}{x^3}e^{\left(-\frac{n}{2x^2}\right)}\right|dx = \cdots$$

For any $n > 0$, we have that: $$||f_n -f||_1= \int_{(0,\infty)}\left|\frac{n}{x^3}e^{\left(-\frac{n}{2x^2}\right)}\right|dx = e^{0} - e^{-\infty} = 1$$

This means that $$\|f_n - f\|_1 = 1 ~~~~\forall n > 0 $$

You can conclude that $f_n$ does not converge to $0$ if one uses norm-$1$.