Let $X$ be a distribution over $N$ (the set of non-negative numbers), with mass $P(X=i) = \frac{a}{2^i}$ . (More details in discription)

Question

This question consists of $4 $ parts in which the first part of calculating $a$ has been done by me and my answer is $1/2$.
The second part asks to find the $E[X]$ i.e.-

$$\sum_{i=0}^n i*P(i)$$

According to my calculation the answer is $0$, but unfortunately, it's not correct.
Can anybody help me with this?
The two more parts which I am unable to solve are-
$P[Y=1]$ , where $Y=Xmod3$ $$E[Y]$$

Answer 1

A small hint for one of the parts. You're looking for something like this: $$\sum_{n = 0}^\infty \frac{n}{2^{n+1}} = \frac{1}{4} \sum_{n = 0}^\infty \frac{n}{2^{n-1}}$$ To calculate it, first, find out what is $f(x)$ here: $$f(x) = \sum_{n=0}^\infty n x^{n-1}$$ Notice that the initial sum is equal to $\frac{1}{4} f\left(\frac{1}{2}\right)$ and that $f(x) = g'(x)$ where $g(x) = \sum_{n = 0}^\infty x^n$.