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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.

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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: $\frac {X_n} n=\frac {S_n-S_{n-1}} n= \frac {S_n} n- \frac {n-1} n \frac {S_{n-1}} {n-1}$].

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