I am trying to prove that if $X_n\rightarrow 0$ in probability, then $1/(1+X_n)$ is bounded in probability. My attempt is:
$$P(\frac{1}{1+X_n}<\frac{1}{1-\epsilon})=P(|1+X_n|>1-\epsilon)\\ \geq P(|1-|X_n||>1-\epsilon)\\ \geq P(1-|X_n|\geq 1- \epsilon)\\=P(|X_n|\leq \epsilon)>1-\epsilon$$ However, $\frac{1}{1-\epsilon}$ can go to infinity when $\epsilon$ goes to 1. The definition of bounded in probability requires that $M_{\epsilon}$ is finite, so the attempt seems to be incorrect. I tried calculating $P(\frac{1}{1+X_n}< \frac{1}{1+\epsilon})$ but didn't get the desired result either. Anyone can help me? Thanks in advance.
I have never heard of bounded in probability before this so I googled and am going off of this pdf. Here is a screenshot of the relevant definition(I assume the bent equals sign is meant to be $Θ$?):
It felt unlikely that $\frac{1}{1+X_n}$ will be converging to 0 in probability so I decided to try to prove the other case.$\newcommand{\prob}{\mathbb{P}}$
Lets try (say) 2 for the upper bound. Suppose it didn't work; that would mean that for any $1>c>0$, and any $N$ we can find an $n>N$ such that \begin{equation} \prob\left[ \left|\frac{1}{1+X_n}\right| \geq 2 \right] > c > 0 \end{equation}
But note that \begin{equation} \left|\frac{1}{1+X_n}\right| \geq 2 \iff |1+X_n|<\frac{1}{2} ⇔ -3/2 < X_n < -1/2 ⟹ |X_n| > 1/2 \end{equation}
So this would mean that $\prob\left[|X_n| > 1/2 \right]$ was not null, a contradiction. The lower bound is easy to compute as $|X_n| > ε \iff \frac{1}{1+|X_n|} < \frac{1}{1+ε} $, so that \begin{equation} \prob\left[\frac{1}{1+|X_n|} \geq \frac{1}{1+ε}\right] > 1-ε \end{equation} for large $n$, and $\frac{1}{1+|X_n|} < \frac{1}{|1+X_n|}$ so \begin{equation} \prob\left[\left|\frac{1}{1+X_n}\right| \geq \frac{1}{1+ε}\right] > 1-ε \end{equation}
To conclude, we show the complementary event is null; [$m<A_n$ and $A_n<M$] has complement [$m≥A_n$ or $M≤A_n$] whose probability is bounded by the sum of the two. Thus, \begin{equation} \prob\left(\left[\frac{1}{1+ε} \leq \left|\frac{1}{1+X_n}\right| \leq 2 \right]^c\right) \leq \prob\left[ \left|\frac{1}{1+X_n}\right| < \frac{1}{1+ε} \right] + \prob\left[ \left|\frac{1}{1+X_n}\right| > 2\right] \end{equation}
Both of which we have shown to be null.