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Could you help me prove that the following sequence of independent random variables satisfy Weak Law of Large Numbers?

\begin{equation} P(x_k=\pm 2^k)=\frac{1}{2^{2k+1}} \end{equation}

\begin{equation} P(x_k=0)=1-\frac{1}{2^{2k}} \end{equation}

I made this,

\begin{equation} E(x_k)=(2^k)(\frac{1}{2^{2k+1}})+(-2^k)(\frac{1}{2^{2k+1}})+(0)\left( 1-\frac{1}{2^{2k}}\right)=0< \infty \end{equation} also,

\begin{equation} P(|\frac{S_n}{n}-0|> \epsilon)=P(|\frac{S_n}{n}|> \epsilon)\leq \frac{V(S_n/n)}{\epsilon^2} \end{equation} also,

\begin{equation} V(\frac{S_n}{n})=\frac{E(x_k^2)}{n} \end{equation}

and then I don't know what to do.

DB1
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    Hello, Dayita, and welcome to CV! Since you’re new here, you may want to take our tour, which has information for new users. Since this looks like homework (apologies if it's not), please add the [self-study] tag and read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. If this is self-study rather than homework, let us know, and... it's still a good idea to show us what you've tried. – jbowman Aug 07 '20 at 17:20
  • The next step is to compute $E(x_k^2).$ That's super easy because there are only two possible values of $x_k^2.$ For additional guidance, emulate the answers at https://stats.stackexchange.com/questions/187285/chebychev-s-weak-law-of-large-numbers, of which this problem is a special case. – whuber Aug 07 '20 at 18:22
  • Thanks for the help @whuber – DB1 Aug 07 '20 at 19:22

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