Is this a valid proof for the divergence of $E\left( \frac{1}{\bar{X}} \right)$ for a Normal distribution?

Question

I have come across this proof and I was wondering if it is valid. What troubles me is the inequality $e^{-\frac{n}{2} \theta^2} > \frac{1}{2}$ as it's not obvious to me how one gets there. I know that $e^{-x} \geq 1-x$ but other than that I am stuck.

All help would be greatly appreciated, thank you.

Answer 1

So, since $e^{-x}\ge 1-x$ (as you pointed), for fixed $n$ and $0<θ<ε$ you know that $$e^{-\frac{n}{2}θ^2}>e^{-\frac{n}{2}ε^2}\ge 1-\frac{n}{2}ε^2$$ where the first inequality is due to the monotonicity of the exponential function. So, if you could show that $1-\frac{n}{2}ε^2\ge \frac12$ the proof would be valid. You have that $$1-\frac{n}{2}ε^2\overset{?}\ge \frac12 \iff nε^2\overset{?}\le 1$$ but since $n$ is fixed and $0<ε$ is arbitrarily small (or "sufficiently small" as the author of the proof indicates) then this holds for any $ε\le \frac{1}{\sqrt{n}}$.

Answer 2

Hint: What is $$\lim\limits_{\theta \to 0} e^{-\frac{n}{2}\theta^2}?$$

What does this tell us about when $\theta$ is sufficiently small?