proof that the sequence satisfies weak law of large numbers strong textHow to check if this sequence satisfies the weak law of large numbers?
$$P(X_n=1)= {{1}\over{3^n}} ,P(X_n=-1)= {{1}\over{3^n}}, P(X_n=0)= {1 -{{2}\over{3^n}}}  $$
I calculated that:
$$E(X_n)=0 $$
so according to the definition of this law: $${\sum_{n=1}^{N}{(X_n-0)}}\over{N} $$and from this one we can deduce that this expression converges to 0 so the sequence satisfies the weak law of large numbers. Is that correct?
$\sum P(X_n \neq 0)=\sum \frac 2 {3^{n}}&lt;\infty$. By Borel Cantelli Lemma $P(\lim \sup (X_n \neq 0))=0$. Hence, with probaility $1$, $X_n=0$ for all $n$ suffciently large. But this implies that the averages $\frac 1n \sum\limits_{k=1}^{n} X_k$ tends to $0$ with probability $1$. Thus, we actually have strong law for this sequence.
Simply using the definition of WLLN
$$\lim\limits_{n\to +\infty}\mathbb{P}[|X_n|&gt;\epsilon]=\frac{2}{3^n}=0$$
this means that
$$X_n\xrightarrow{\mathcal{P}}0$$
This if $\epsilon$ is chosen $|\epsilon|\leq1$. If it is chosen &gt;1 the probability is exactly zero. Anyway the sequence converges.