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This is a homework problem. If $X_n$ converges in probability to 1, show $X_n^{-1}$ converges in probability to 1.
My attempt:
$$\begin{align*} P(|X_n^{-1}-1| > \epsilon) &= P(|X_n^{-1}-X_n + X_n-1| > \epsilon)\\ &\leq P(|X_n^{-1} - X_n| > \epsilon/2) + P(|X_n - 1| > \epsilon/2)\\ &= \end{align*} $$
I know I can bound the 2nd term, but I am not sure how to bound the first term. Perhaps another approach is necessary. Any suggestions would be appreciated.
Fix $\varepsilon\gt 0$. We have for each positive $A$, $$\{|X_n^{-1}-1|\gt \varepsilon\}=\{|X_n-1|\gt |X_n|\varepsilon\}\subset\{|X_n-1|\gt A\varepsilon\}\cup\{|X_n|\leqslant A\},$$ this because we wrote $S=(S\cap \{|X_n|\gt A\})\cup (S\cap\{|X_n|\leqslant A\})$.
Take $A:= 1/2$; then $\{|X_n|\leqslant 1/2\}\subset\{|X_n-1|\geqslant 1/2\}$ (because $(-1/2,1/2)\subset (-\infty,1/2)\cup(3/2,\infty)$). We thus obtained
$$\{|X_n^{-1}-1|\gt \varepsilon\}\subset\{|X_n-1|\gt \varepsilon/2\}\cup\{|X_n-1|\gt 1/2\}.$$ The probability of the last two events goes to $0$ as $n\to\infty$.
Consider the case $0<\epsilon<1$. Defining $\delta = \epsilon / (1+\epsilon)$, from the figure we have $$ |x-1|<\delta \Rightarrow \left| \frac{1}{x}-1\right| < \epsilon \, . $$ Hence, $$ P\left(\left| \frac{1}{X_n}-1\right| < \epsilon\right) \geq P\left(\left| X_n-1\right| < \delta\right) \to 1 \, , $$ when $n\to\infty$. The case $\epsilon\geq 1$ is easy, because $$ P\left(\left| \frac{1}{X_n}-1\right| < \epsilon\right) \geq P\left(\left| \frac{1}{X_n}-1\right| < \frac{1}{2}\right) \, , $$ and we can use the previous case.
