Convergence of a sequence of reciprocals Let $X_n \geq 0$ be sequence of nonnegative random variables converging a.s. and in $L^2$ to a positive constant $c > 0$:
$0 \leq X_n \xrightarrow{a.s.,L^2} c >0$
What can we say about the reciprocal sequence $Y_n:= \frac{1}{\frac{1}{n} +X_n}$?
I think that $Y_n \xrightarrow{a.s.} \frac{1}{c}$.
Questions:


*

*(when) is the mean of $Y_n$ bounded, i.e. $\lim_{n \to \infty} \mathbb{E} Y_n < \infty$?

*(when) does $Y_n$ converge in mean or mean-square?

 A: I can provide a partial answer to my own question. 
First, consider $Z_n := \frac{1}{\frac{1}{n^2}+X_n}$, which is very similar to $Y_n$. I show that $\lim_n EZ_n$ can be infinite. Consider:
$\mathbb{P}(X_n = 0) = \frac{1}{n^{3/2}}$
$\mathbb{P}(X_n = c) = 1- \frac{1}{n^{3/2}}$
$X_n \xrightarrow{a.s.} c$ by Borel-Cantelli and in $L^p$ from $\mathbb{E} X_n^p = c^p \left(1-\frac{1}{n^{3/2}}\right) \to c^p$
However, $\mathbb{E} Z_n = \left(\frac{1}{\frac{1}{n^2}+c}\right)\left(1 - \frac{1}{n^{3/2}}\right)+\frac{n^2}{n^{3/2}}\to\infty$.
This indicates that the behavior of $\mathbb{E}Y_n$ depends on the rate of convergence of $X_n$. To clarify, for some $\epsilon > 0$:
$\mathbb{E} Y_n = \mathbb{E} Y_n \mathbf{1}_{X_n > \epsilon} + \mathbb{E} Y_n \mathbf{1}_{X_n \leq \epsilon} \leq \frac{1}{\epsilon} + n \mathbb{P}(X_n \leq \epsilon)$
Thus, the asymptotic behavior of $\mathbb{E}Y_n$ depends on the rate of convergence of $\mathbb{P}(X_n \leq \epsilon)$ to $0$. It seems that there should be counter-examples to the original question with the convergence of $P(X_n \leq \epsilon)$ being slower than $O(\frac{1}{n})$.
