Why does the CLT suggest that $\frac{1}{\sqrt{n}}$ is the right scale

Question

Let $(X_n)_{n \geq 1}$ be i.i.d. square-integrable random variables with $E(X_1) = 0$ and $Var(X_1) = 1$. Assume that $(c_n)_{n \geq 1}$ is a sequence of constants such that $\lim_{n \to \infty} \frac{\sqrt{n}}{{c_n}} = \infty$. Why does now directly from the central limit theorem follows that $$ \bigg \vert \frac{X_1 + \ldots + X_n}{c_n} \bigg\vert \longrightarrow \infty $$ in probability? I cannot see this immediately.

Answer 1

you can write $$\lvert \frac{X_1+...+X_n}{c_n} \rvert = \lvert \frac{\sqrt{n}}{c_n} \frac{X_1+...+X_n}{\sqrt{n}} \rvert = \lvert \frac{\sqrt{n}}{c_n}\rvert \lvert \frac{X_1+...+X_n}{\sqrt{n}} \rvert.$$

Work in the extended real axis. The first factor converges to $\infty$ while the second converges to a Gaussian in distribution. By Slutsky it follows that the product converges to $\infty$ in distribution (which is a deterministic r.v. in the extended setting) and hence in probability.