Given $P(X_n=0)=1-\frac{1}{n}$ and $P(X_n=1)=\frac{1}{n}$ with $E(X_n)=\frac{1}{n}$ and $Var(X_n)=\frac{n-1}{n^2}$
Show that $X_n$ converges to zero in probability.
I do it by using Chebyshev's inequality $$P\left(\left|X_n-E(X_n)\right| \geq \epsilon\right) \leq \frac{Var(X_n)}{\epsilon^2}$$
Then $$P\left(\left|X_n-\frac{1}{n}\right| \geq \epsilon\right) \leq \frac{n-1}{n^2 \epsilon^2}$$
Now I need to work on the left hand side of inequality where I replace $\epsilon$ with $(\epsilon+\frac{1}{n})$: $$P\left(|X_n| \geq \epsilon\right) \leq P\left(\left|X_n-\frac{1}{n}\right| \geq \epsilon+\frac{1}{n}\right)$$
So we have that $$P\left(\left|X_n-\frac{1}{n}\right|\geq \epsilon+\frac{1}{n}\right) \leq \frac{n-1}{n^2 \cdot (\epsilon+\frac{1}{n})^2} = \frac{n-1}{n^2} \cdot \frac{1}{\left(\epsilon+\frac{1}{n}\right)^2} \overset{n \rightarrow \infty}{\rightarrow} 0 \cdot \frac{1}{\epsilon^2} = 0$$
Is it really correct like that? Because that's how I would do it in the exam. But maybe it's possible to do it much more quickly / shorter?
I believe you're overcomplicating it.
For any $\varepsilon < 1$.
$$P(|X_n| \ge \varepsilon)=P(X_n=1)=\frac{1}{n} \longrightarrow 0.$$