# The series of independent random variables and LLW.

Determine whether for the series of independent random variables any (either weak or strong) Law of large numbers is true. The variables are defined as below: $$P(X_k=2^k)=P(X_k=-2^k)=\frac{1}{2}.$$ I thought of using the following theorem: $$\text{For the series of random variables }\{X_n\}_{n=1}^{\infty}\text{, where }\mathbf EX_n<\infty\text{, the strong / weak law of large numbers is true}\iff \frac{\sum_{i=1}^{n}X_i-\sum_{i=1}^n\mathbf EX_i}{n}\text{ converges almost surely / converges in probability towards 0}$$

Firstly, I found the expected value of each variable and it's $$0$$, therefore $$\sum_{i=1}^n\mathbf EX_i=0$$ for every natural n, and $$\mathbf Var(X_k)=2^{2k}.$$

For the weak law I tried using Markov's inequality, to simply determine that $$\forall \epsilon >0\quad \lim \limits_{n \to \infty }P(\lvert\frac{1}{n}\sum_{i=1}^{n}X_i\rvert>\epsilon)=0,$$ but all I got was that's just smaller than infinity, which gave me nothing. For the strong law I tried finding the limit of $$\frac{1}{n}\sum_{i=1}^{n}X_i,$$ but can't seem to prove or disprove that it's 0.

It would be great to receive some help from you.

Let $$S_n=X_1+X_2+...+X_n$$. If $$\frac {S_n} n$$ converges in probability then $$\frac {X_n} n$$ converges to $$0$$ in probability. But $$P\{|\frac {X_n} n|>1\} \geq P\{X_n =2^{n}\} =1/2$$ if $$n$$ is so large that $$2^{n} >n$$. Hence weak law does not hold. [For the first step use: $$\frac {X_n} n=\frac {S_n-S_{n-1}} n= \frac {S_n} n- \frac {n-1} n \frac {S_{n-1}} {n-1}$$].