 Solution

Markov's inequality states that for $Y\ge 0$ and $t>0$, $\mathbb{P}(Y\ge t)\le \mathbb{E}Y/t$. It follows by taking expectations in $Y\ge t{1}_{\{Y\ge t\}}$.

If $X$ has finite mean $\mu$ and variance ${\sigma }^2$, apply Markov to $(X-\mu {)}^2$ to obtain Chebyshev's inequality:For the first moment method, assume ${\mu }_n=\mathbb{E}{X}_n\to 0$. Nonnegative integer values give $\{X_n>0\}=\{X_n\ge 1\}$, soFor a standard second moment method, suppose ${\mu }_n>0$ eventually, the second moments are finite, and $\mathbb{E}{X}_n^2/{\mu }_n^2\to 1$. Equivalently, $\mathrm{Var}(X_n)/{\mu }_n^2\to 0$. The zero event is contained in $\{|X_n-{\mu }_n|\ge {\mu }_n\}$, soHence $\boxed{\mathbb{P}(X_n>0)\to 1}$. The second-moment condition controls the fluctuations relative to the mean; a large mean alone is insufficient.