Expectation of a random variable that tends to infinity There is a sequence of integer-valued random variable $X_n\in\{1,2,3,\dots\}$ such that $X_n\to \infty$ in probability, that is for any given $M>0$, $P(X_n \leq M) \to 0$ as $n \to \infty$.
I am trying to figure out an accurate way to characterize $T_n = E[1/X_n]$ as $n\to \infty$. Intuitively speaking, $T_n$ is supposed to approach to $0$ as $n$ increases. My questions are

*

*How to give a rigorous way to describe this approaching limit?

*How to prove this limit?

Thanks!
 A: Here is one idea for a proof by contradiction.
First, note that $X_{n} \rightarrow \infty$ in probability implies $Y_{n}:=1/X_{n} \rightarrow 0$ in probability. This is not too hard to show.
Now, suppose for contradiction that $\mathbb{E}[Y_{n}] \not\rightarrow 0$. This means there exists an $\epsilon > 0$ such that for an infinite subsequence $n_{1},n_{2},n_{3},\dots$, we have $\mathbb{E}[Y_{n_{i}}] \geq \epsilon$ for all $i.$ The idea is now to show that, with the expectations bounded below in this way, the subsequence $Y_{n_{i}}$ cannot converge in probability to $0$ as it should. That is, there exists a fixed value of $\delta \in (0,1)$ such that $\mathbb{P}(Y_{n_{i}} \geq \delta) \not\rightarrow 0.$
For this task, you can use the Markov Inequality: for a nonnegative random variable $Z$ and $a \geq 0$, we have $\mathbb{P}(Z > a) \leq  \frac{\mathbb{E}[Z]}{a}$. Applying this to $1-Y_{n_{i}}$ (which is nonnegative) gives that for any $\delta>0$, $\mathbb{P}(1-Y_{n_{i}} > 1-\delta) \leq \frac{\mathbb{E}[1-Y_{n_{i}}]}{1-\delta} \leq \frac{1-\epsilon}{1-\delta}.$
That is, $\mathbb{P}(Y_{n_{i}} < \delta) \leq \frac{1-\epsilon}{1-\delta}$ and hence $\mathbb{P}(Y_{n_{i}} \geq \delta) \geq 1-\frac{1-\epsilon}{1-\delta}.$ As long as $0<\delta<\epsilon$, this means that $\mathbb{P}(Y_{n_{i}} \geq \delta)$ is bounded away from $0$, a contradiction.
