Consider the sequence of random non-negative values $X_1...X_n$ where $E[X_n], Var[X_n] > 0$ for all $n \in \mathbb{N}$ and: $$\lim_{n\rightarrow \infty} \frac{Var[X_n]}{E[X_n]^2} = 0$$ Prove: $$\lim_{n\rightarrow \infty} P(X_n >0) = 1$$
How would I prove this using Chebyshev's inequality?
Chebyshev's inequality is in the right direction, but before trying to apply it without a plan, take a look at how assumption looks like: it says that $$ \lim_{n\rightarrow \infty} \frac{Var[X_n]}{E[X_n]^2} = 0. $$ If one looks at the assumption carefully, one may notice that $\frac{1}{E[X_n]}$ plays a role of normalizing constant. So, by letting $$X_n' :=\frac{X_n}{E[X_n]},$$ one gets $$ E[X_n'] = 1,\quad \lim_{n\rightarrow \infty} Var[X'_n] = 0. $$ The problem gets simpler now. Chebyshev's inequality implies that $$ P(X_n>0) = P(X'_n>0)\geq P(|X'_n -1|\leq \frac{1}{2}) \geq 1- 4Var[X'_n]\to 1, $$ (in fact, $X'_n \to_p 1$ in probability) and hence that $\lim_{n\to\infty}P(X_n>0)=1$ as desired.