# Show that $\lim_{n\to\infty}X_n=0$ almost surely $\implies \lim_{n\to\infty}\mathbb{E}\left[\frac{|X_n|}{1+|X_n|}\right]=0$

Let $$X$$ be a random variable and $$(X_n)_{n\geq 0}$$ be a sequence of random variables.

Show that $$\lim_{n\to\infty}X_n=0$$ almost surely $$\implies \lim_{n\to\infty}\mathbb{E}\left[\frac{|X_n|}{1+|X_n|}\right]=0$$

My thoughts: $$\lim_{n\to\infty}X_n=0 \implies \lim_{n\to\infty}1+|X_n|=1 \implies \lim_{n\to\infty}\frac{1}{1+|X_n|}=1$$ (I think?) but then I'm not sure where to go from here.

Use the dominated convergence theorem. The argument inside the expectation is bounded by $$1$$, which is integrable. So you may take the limit inside of the expectation, so it would suffice to show that the limit of the argument is $$0$$.
You are correct in stating that $$|X_n| \to 0$$ a.s. also. If your probability space is $$(\Omega, \mathscr{F}, \mathbb{P})$$, then take any $$\omega \in \Omega$$ for which $$X_n(\omega) \to0$$ as $$n \to \infty$$ to convince yourself. However, by the algebra of limits: $$$$\lim_{n \to \infty} \dfrac{|X_n(\omega)|}{1+|X_n(\omega)|} = \lim_{n \to \infty} |X_n(\omega)| \times \lim_{n \to \infty} \dfrac{1}{1+|X_n(\omega)|} = 0.$$$$
Observe that $$\frac{|X_n|}{1+|X_n|} \leq 1$$, then $$E \left[\frac{|X_n|}{1+|X_n|}\right] \leq E[1]=1$$ for any $$n \in \mathbb{N}$$. By dominated convergence you got $$E\left [\frac{|X_n|}{1+|X_n|}\right] \xrightarrow{n} 0$$