Each integer $n \in \mathbb{N}$ has a unique representation $n = i + 2j$ for integers $j \geq 0$ and $0 \leq i < 2j$. Set $\Omega := (0, 1]$ and let P be Lebesgue measure on $\mathcal{F} = \mathcal{B}(\Omega)$. Define random variables $$\Large{X_n(\omega)} := \Large1_{{\frac{i}{2^j} < \omega \leq \frac{i+1}{2^j}}}$$.
For which real numbers $\alpha \in \mathbb{R}$ do the random varibales $Y_{n} = n^{\alpha}X_{n}$ converge to zero in probability and almost surely?
My approach: For all $\epsilon >0$
$P(Y_{n} > \epsilon) = P(X_{n}>\epsilon/n^\alpha) = P(X_{n}=1)= 1/2^j$.
Now, $2^j = n-i$. So, $1/2^j = 1/(n-i).$
As, $n \to \infty, 1/(n-i) \to 0$.
All we assumed was $\epsilon/n^\alpha >0$. This will be true for all values of $\alpha$, hence $Y_{n}$ converges to zero in probability for all values of $\alpha$. Is this approach correct?
For the almost sure convergence, $$\sum_{n=1}^\infty 1/(n-1)$$ is not summable. So, $P(Y_{n} > \epsilon$ $i.o) = 1$. Hence, $Y_{n}$ does not converge to zero $a.s.$ for any value of $\alpha$.
My confusion is how to change the index from $j$ to $n$ because we get the probabilities in terms of $j$ but the limit is in terms of $n$.
Any help will be appreciated.