let $X_1,X_2,....$ be independent real-valued random variables, all uniformly distributed over $(0,1)$.
Let $Y_n=\prod_{i=1}^{n}X_i$.
I want to show that $Y_n \longrightarrow 0$ almost surely.
I know theres many different ways to show that $Y_n$ converges to 0. I have worked on one way using Borel-Cantelli condition for a.s convergence. It state that: $$\forallε>0 : \sum_{n=1}^{\infty}P(|Y_n-0|\geq\epsilon)<\infty \implies p(|Y-0|\geq\epsilon\ \ i.o\ )=0$$ and hence: $Y_n\longrightarrow0$ almost surely.
$\bullet$ My attempt is the following :
we can write the event $(|Y_n-0|\geq\epsilon)=(Y_n\geq\epsilon)$ since $Y_n\in (0,1)$.
so for all $ε>0$
$$\sum_{n=1}^\infty p(Y_n\geq\epsilon)\leq\sum_{n=1}^\infty \frac{E(Y_n)}{\epsilon}=\sum_{n=1}^{\infty}\frac{E(X^n)}{\epsilon} \leq \sum_{n=1}^\infty \frac{1}{(n+1)\ \epsilon}<\infty$$ Where i used "markov inequality" in the first first inequality.
In the second equality $X_i's$ are iid unif(0,1)".
is therefore by Borel-Cantelli this implies $\Longrightarrow P(|Y_n-0|\geq\epsilon\ \ i.o)=0$. Which are equivalent to $P(|Y_n-0|<\epsilon \ \ evt.)=1$
And hence: $Y_n\longrightarrow 0$ a.s.
Is it right? im sure there is some mistake!
Edit: (update:)
$\bullet $ The second equality. is not right. Since $X_1,X_2,...$ all are uniformly distributed. then the we get that: $$\sum_{n=1}^\infty p(Y_n\geq\epsilon)\leq\sum_{n=1}^\infty \frac{E(Y_n)}{\epsilon}=\sum_{n=1}^{\infty}\frac{E(\prod_{i=1}^nX_i)}{\epsilon}=\sum_{n=1}^\infty\frac{\prod_{i=1}^n E(X_i)}{\epsilon}=\sum_{n=1}^\infty\frac{E(X_1)^n}{\epsilon} =\sum_{n=1}^\infty \frac{1}{2^n\ \epsilon}<\infty$$
and now: $\sum_{n=1}^\infty p(Y_n\geq\epsilon)<\infty \implies p(Y_n<\epsilon \ \text{evt.})=1 $ Hence, $Y_n\longrightarrow 0$ almost surely.
There is a fatal mistake in your proof. The harmonic series is divergent!
$\frac 1 n \sum\limits_{k=1}^{n} ln (X_k) \to \int_0^{1} \ln x dx=-1$ almost surely by SLLN's. Hence, $\sum\limits_{k=1}^{n} ln (X_k) \to -\infty$ a.s and taking exponential finishes the proof.