Expected value of PMF I am stuck on this question, would appreciate any hints for how to start:
Let X have PMF
$$P(X = k) = \frac{cp^k}{k} for\:k = 1,2,...,$$
where p is a parameter with 0 < p < 1 and c is a normalizing constant.
*We have c = −1/ log(1 − p), as seen from the Taylor series:
$$−log(1−p)=p + \frac{p^2}{2} + \frac{p^3}{3} +....$$
This distribution is called the Logarithmic distribution. Find the mean and variance of X.
Now, this might be silly - but I'm a little confused by the statement 
"as seen from the Taylor series". However, I proceeded as such:
$$E(X) = cp + \frac{cp^2}{2} + \frac{cp^3}{3} + \frac{cp^4}{4} + ... $$
If we factor out c and p, we get:
$$E(X) = cp \sum_{k=1}^{\infty} (1+\frac{p^{k-1}}{k}) $$
Does anyone have tips for proceeding?
Update
I computed E(X) as such:
$$E(X) = \frac{-1}{log(1-p)}\frac{p}{1-p}$$
Now I'm having trouble with finding Var(X), namely the first component, E(X^2). Here are the steps I've done:
$$E(X^2) = \sum_{k=1}^{\infty}k^2P(X=k) =  \sum_{k=1}^{\infty}k^2\; \cdot c \; \cdot\frac{p^k}{k} = \sum_{k=1}^{\infty}kcp^k $$
$$cp\sum_{k=1}^{\infty}kp^{k-1} = \frac{1}{1-p}$$
$$cp\sum_{k=1}^{\infty}k^2p^{k-2} = \frac{1}{(1-p)^2}$$
$$E(X^2) = c\sum_{k=1}^{\infty}k^2p^{k} = \frac{p}{(1-p)^2}$$
Therefore our entire expression is:
$$E(X^2) = \frac{-1}{log(1-p)}\frac{p}{(1-p)^2}$$
Is this the correct expression for E(X^2)? Thanks so much.
 A: Recall that for a pmf we need
$$\sum_{k=1}^\infty P(X=k) =1$$
Plugging the pmf we get
$$cp+\frac{cp^2}2+\frac{cp^3}3+\cdots=1$$
$$\implies c\left[p+\frac{p^2}2+\frac{p^3}3+\cdots\right]=1$$
Now we need the logarithmic that you mentioned to conclude that
$$c[-\log (1-p)]=1\implies c=-\frac1{\log (1-p)}$$
Note that we are using properties of pmf not expectation.
A: Already you have some confusion on the definition of expected value. You write
$$"E(X)=cp+cp^2/2+\dotsc+cp^k/k"$$
First, was it a typo to leave off the rest of the infinite sum? Second, you aren't even abiding by the definition of expectation. We know $P(X=k)=cp^k/k$ for $k=1,2,\dotsc$. So, by definition, 
$$E(X)=\sum_{k=1}^\infty kP(X=k)=c\sum_{k=1}^\infty k \cdot \frac{p^k}{k}=c\sum_{k=1}^\infty p^k$$
$$=c(p+p^2+p^3+\dotsc)=cp(1+p+p^2+\dotsc),$$
and if you recognize the final series and are aware of its closed form expression then you can easily conclude. If not, then you need to review geometric series. 
The other answer shows you how to find $c$ so I have omitted that part. Finding the variance amounts to computing 
$$E(X^2)=c \sum_{k=1}^\infty kp^k$$ and using the fact that the variance is equal to $E(X^2)-E(X)^2$. To compute the new series, pull out a $p$ and recognize it as the derivative of the series $p+p^2+\cdots$. Is this doable?
Comment if you need clarification. 
