limit of $c_n$ when $\sqrt{n}*(c_n - 1)$ converges to a real number $c$

Question

I need to show that $c_n*\bar{X}$ converges in probability to $\theta$, where $X$ is from an EXP~($\theta$).

It is given that $\sqrt{n}*(c_n - 1)$ converges to a constant $c$ . I know that $\bar{X}$ converges in probability to $\theta$ by WLLN, but where does $c_n$ converge to? I need to know this limit to check wheter I can use Slutsky's theorem to prove the statement.

Answer 1

$c_n$ converges to 1. You can show this using a simple $\epsilon$ argument. Pick $N$ large enough that

$$|c - \sqrt{N}*(c_N - 1)| < \epsilon.$$

Then,

$$\left|\frac{c}{\sqrt{N}} - (c_N - 1)\right|< \frac{\epsilon}{\sqrt{N}}.$$

By applying the reverse triangle inequality ($|a-b| \geq |a| - |b|$), $$|c_N - 1| < \frac{|c| + \epsilon}{\sqrt{N}}.$$

This proves that $c_n \rightarrow 1$.