Convergence (to zero) for PDF of normal distribution.

Question

I need to prove that the PDF converges to zero when $n\to\infty$; that is,

$$\lim_{n\to \infty}f_n(x) =\lim_{n\to\infty} \frac{1}{\sqrt{2\pi n^{-3}}}\exp\left(-\frac{(x-\frac{1}{n})^2}{2n^{-3}}\right)\to 0$$

I have tried using L'Hopital and differentiate $3$ times as someone suggested on StackOverflow but I seem to be getting $\infty\cdot0$ that is $0$ for the exponential(as $n\to\infty$) and $\infty$ for the $n^6$ in the numerator both by hand and in maple.

Any other suggestions on how to show this would be much appreciated.

Answer 1

Consider the cases $x=0$ and $x \neq 0$ separately. If $x=0$ the result follows from the fact that $n^{3/2}e^{-an} \to 0$ for any $a>0$. For $x\neq 0$ use the fact $(x-\frac 1 n)^{2} \geq (\frac {|x|} {2})^{2}$ for $n$ suffciently large. Can you complete the argument now?.

Answer 2

$$\log f_n(x) \propto -3 \log n -\frac{(x-\frac{1}{n})^2}{2n^{-3}} \longrightarrow -\infty$$ So $$f_n(x) \longrightarrow e^{-\infty} = 0$$