Probability of sequence The question is as follow: Let $(X_{n})_{n}$ be a sequence of Random variable that is independent with the probability
$P(X_{n}=1)=1-P(X_{n}=0)=\frac{1}{n}$
Show that $P\left( \underset { n\rightarrow \infty  }\liminf{ X }_{ n }=0 \right) =1$
I tried to use the Borel-Cantelli Theorem to prove this, however I obtain that 
$P\left( \underset { n\rightarrow \infty  }\liminf{ X }_{ n }=0 \right) =0$
${ F }_{ n }:=\left\{ { X }_{ n }=0 \right\} $
$\sum _{ n=1 }^{ \infty  }{ P\left( { F }_{ n }^{ C } \right)  } =\sum _{ n=1 }^{ \infty  }{ 1-P\left( X_{ n }=0 \right)  } =\sum _{ n=1 }^{ \infty  }{ \frac { 1 }{ n }  } =\infty$
By Borel-Cantelli, we know that $P\left( { \underset { n\rightarrow \infty  }\limsup  }{ F }_{ n }^{ C } \right) =1$
Since ${ \left( { \underset { n\rightarrow \infty  }{ \liminf }  }{ F }_{ n } \right)  }^{ C }=\left( { \underset { n\rightarrow \infty  }{\limsup}  }\quad { F }_{ n }^{ C } \right) $,
(ie) $P\left( { \underset { n\rightarrow \infty  }{ \liminf }  }\quad { F }_{ n } \right) =1-P\left( { \underset { n\rightarrow \infty  }{ \limsup }  }\quad { F }_{ n }^{ C } \right)$,
I obtain that $P\left( { \underset { n\rightarrow \infty  }{ \liminf }  }\quad { X }_{ n }=0 \right) =0$  
Since this is the opposite of what I want to prove and the statement is said to be true, I can't see where I went wrong in my argument. 
 A: You have a mistake in the last line of your argument
$$ \liminf_{n \to \infty} F_n = \liminf_{n \to \infty} \lbrace X_n = 0 \rbrace \neq \lbrace \liminf_{n \to \infty} X_n = 0 \rbrace$$
The event $\liminf F_n$ is the event that eventually $F_n$ will be true. That is from some moment $k$ for every $n > k$, $F_n$ will happen. In this case this is equivalent to say that the sequence will be constant and equal to zero.
Independent Borel-Cantelli will assure you that both $\limsup F_n$ and $\limsup F_n^c$ will be true, since both probability series diverge. Remember that the interpretation of limsups of events is that the event will happen infinitely often (i.o). That means that both $\lbrace X_n = 0 \rbrace$ and $\lbrace X_n = 1 \rbrace$ will happen i.o. which implies that almost surely $\liminf X_n = 0$ and $\limsup X_n = 1$
A: In a nutshell, for $\{0,1\}$-valued random variables $(X_n)$,

$$[\liminf X_n=0]=\limsup\ [X_n=0]$$
  $$[\limsup X_n=0]=\liminf\ [X_n=0]\quad(=[\lim X_n=0])$$

The event $A=[\liminf X_n=0]$ is $A=\limsup A_n$ with $A_n=[X_n=0]$. Now, $\limsup A_n=\bigcap\limits_nB_n$ with $B_n=\bigcup\limits_{k\geqslant n}A_k$. For every $n$ and $k\geqslant n$, $A_k\subseteq B_n$ and $P[A_k]=1-1/k$ hence $P[B_n]\geqslant1-1/k$ hence $P[B_n]=1$. Thus, $P[A]=1$.
This does not use the independence property, only the hypothesis that $P[X_n\ne0]\to0$.

Regarding your proof, note that you are asked to compute the probability of $[X=0]$ with $X=\liminf X_n$, and that you compute the probability of $\liminf [X_n=0]$ instead. In other words, your problem is that you assumed (wrongly) that the events
$[\liminf X_n=0]$ and $\liminf [X_n=0]$ coincide.
A: Note $1-P(X_n=0)=1/n$ implies $P(X_n=0)=1-1/n$ or $P(X_n=0)=(n-1)/n$ and apply Borel-Cantelli.  Hope this helps.
A: Are you certain about the statement of the problem? I can't see what is wrong with your proof,  and using Borel 0-1 law (in French, but this is basically a statement similar to, while slightly easier to use than, the Borel-Canetelli lemma; see also Pro. 2.2 from these lecture notes with $F_n$ (independent events, with $\mathbb{P} F_n=\frac{1}{n}$), one also gets that
$$
 \mathbb{P}\{ \limsup_{n\to+\infty} F_n \} = 1
$$
since $\sum_{n=1}^\infty \mathbb{P} F_n = +\infty$. In other terms, infinitely many of the $F_n$'s occur, i.e. $X_n=0$ for infinitely many values of $n$. But then, one cannot have $X_n=0$ ultimately (that is, $\forall n \geq N$ for some $N$) w.p. $1$, which is the result you are apparently asked to show.
Edit: what I wrote is wrong (from the beginning), in the sense that it is not the right quantity that is considered. See Did's answer.
A: I think using Fatou's lemma instead of Borel-Cantelli can prove to be useful. Fatou's lemma is applicable to $\{X_n\}$ since they are non-negative. By Fatou's lemma, \begin{equation}
\begin{split}
\ &\mathbb{E}(\liminf_{n\rightarrow \infty} X_n)\leq \liminf_{n\rightarrow \infty} \mathbb{E}(X_n)=\liminf_{n\rightarrow \infty}\frac{1}{n}=0\\
\ & \Rightarrow P(\liminf_{n\rightarrow \infty} X_n=1)\leq 0\\
\ & \Rightarrow P(\liminf_{n\rightarrow \infty} X_n=0)=1\hspace{0.6cm} \Box
\end{split}
\end{equation}
Otherwise, if you really want to use BC lemma then the answer given by @Did is useful.
