$(X_n)$ is a sequence of random variables having the following distribution:
$$P(X_n=1)=1- \frac{1}{n},\; P(X_n=0)=\frac{1}{n}$$ (we don't assume that those variables are independent). $X$ is some random variable. We consider variables $Y_n=XX_n$. Does $(Y_n)$ converge almost surely? Does it converge in probability?
I think that $Y_{n} \to X$ a.s., because $X_n \to 1$ a.s.
Using Chebyshev's inequality we have for $\epsilon > 0$:
$$P(\left|X_{n}-1 \right| \ge \epsilon ) \le \frac{\text{E}(X_n-1)}{\epsilon} \le -\frac{1}{\epsilon n}$$
Seeing that $$-\sum\frac{1}{\epsilon n}=-\infty,$$ we conclude that $$\sum P(\left|X_{n}-1 \right| \ge \epsilon ) <+\infty.$$ Hence $X_n \to 1$ a.s.
Convergence in probability stems from almost sure convergence.
Is everything there okay?