Consider the following random variable $Y$. It takes only values of the form $\frac{1}{2^a}$ for positive integer $a$ and $P(Y=\frac{1}{2^a})=\frac{1}{2^a}$ for each $a$. Find expected value of this random variable. Enter answer as a proper fraction.
My compute is:
a. Expectation is $$E(x)=\sum _{k=1}^{\infty} P_{k}X_{k}\ \Rightarrow\ E(x)=\sum _{k=1}^{\infty} \frac{1}{2^a}\frac{1}{2^a}= \sum_{k=1}^{\infty} \frac{1}{2^{2a}} = \sum_{k=1}^{\infty} \frac{1}{2^{2}}\frac{1}{2^{a}}$$
$$a=\frac{1}{4} ;\quad r = \frac{1}{2};\quad \frac{a}{1-r}=2.$$
Apparently I made an error but I am not able to see it.
Thanks in advance
edit: attached the image of the statement
add sugestion platform.
You have to use $k$ as for the index.
$$\mathbb E(X)=\sum_{k=1}^{\infty} \frac1{2^k}\cdot \frac1{2^k}=\sum_{k=1}^{\infty} \frac1{ 2^k\cdot 2^k}=\sum_{k=1}^{\infty} \frac1{ 4^k}=\sum_{k=1}^{\infty} \left(\frac1{ 4}\right)^k$$
Therefore here $r=\frac1{ 4}$. Using the formula of the geometric series we have
$$\mathbb E(X)=\frac1{ 4}\cdot \frac{1}{1-\frac1{ 4}}=...$$