Let $X_1,X_2,...X_n,...$ are random variables satisfying $0\leq\dfrac{X_n}{n}\leq C,$ and $Var(X_n)\leq C'n$ for some constants $C, C'>0.$ Furthermore, $X_n(\omega)$ is non-decreasing i.e. $X_n(\omega)\leq X_{n+1}(\omega).$ Then, how do I show that $X_n\longrightarrow\infty$ in probability, provided that $EX_n\longrightarrow\infty?$
For given $a>0$,I have tried bounding: $$P(X_n\leq a) = P(X_n-EX_n\leq a- EX_n) = P((X_n-EX_n)^2\geq (a-EX_n)^2)\leq\dfrac{VarX_n}{(a-EX_n)^2},$$
for large enough $n$. But this is not enough to conclude that $P(X_n\leq a)\longrightarrow 0$, as there is no information about the how fast $EX_n$ converges to infinity.
Are there smarter ways to use Chebyshev's inequality?
I think the statement is false as written.
Counterexample: Let $X_n$ be $0$ or $\sqrt n$, each with probability $1/2$. Then $0 \leq X_n / n \leq 1$ holds; also, $Var(X_n) = \frac 1 4 \cdot n$. Further, $\mathbb{E}[X_n] = \sqrt n / 2$, so this expectation goes to infinity, and yet $\mathbb{P}(X_n = 0) = 1/2$ for all $n$, so $X_n \not \rightarrow \infty$.