Q: Consider the following random variable $Y$. It takes only values of the form $\frac 1{2^k}$ for positive integers $k$ and $P\left(Y= \frac 1{2^k}\right) = \frac 1{2^k}$ for each $k$. Find the expected value of this random variable.
I did find out that $a = \frac12$ in the geometric series but I am unable to find the proper result
General expression for expected value of a discrete random variable $X$. Let $a_n$ be a value and let $p_n=P(X=a_n)$. Then $E(X)=\sum a_np_n$ In your case $E(Y)=\sum_{k=1}^\infty \frac{1}{2^{2k}}=\frac{1}{3}$.