the proof for weak law of large numbers

Question

This is the text of proof of the law

I don't understand why when $n\to \infty$, $\frac{\sigma^2}{n\epsilon^2} \to +\infty$? isn't $\frac{\sigma^2}{n\epsilon^2}=\frac{1}{k^2}$? how come $\frac{1}{k^2}\to0 $?

Answer 1

As I said, the WLLN is: under independence conditions as stated, for any fixed $\epsilon>0$ $$\lim_{n\rightarrow \infty}P\left(\left|\frac{X_1+\cdots+X_N}{n}-\mu\right|>\epsilon\right)=0.$$ Your forumla says that for any $k$: $$P\left(\left|\frac{X_1+\cdots+X_N}{n}-\mu\right|>k\frac{\sigma}{\sqrt{n}}\right)\leq\frac{1}{k^2}.$$ So let $\epsilon$ be fixed and choose $k(n)$ such that $$\epsilon=k(n)\frac{\sigma}{\sqrt{n}}$$ Then you have that $$k(n)=\sqrt{n}\frac{\epsilon}{\sigma},$$ where $\frac{\epsilon}{\sigma}$ is some positive constant. Substituting this in the above formula gives, as in the suggested solution, $$P\left(\left|\frac{X_1+\cdots+X_N}{n}-\mu\right|>\epsilon\right)\leq\frac{1}{n}{\frac{\sigma^2}{\epsilon^2},}$$ and since $\frac{\sigma}{\epsilon}$ is a bounded constant, independent of $n$, and so also in particular $\frac{\sigma^2}{\epsilon^2}$, the right hand side goes to $0$ as $n$ grows to infinity, giving the claim.

And indeed yes, $k=k(n)=\sqrt{n}\frac{\epsilon}{\sigma},$ grows to infinity for $n$ to infinity.