Compute the mean and variance given a probability mass function I'm given the formula:
$(1-p)^{x-1}p, x = 1, 2, ..., \infty$
and I'm asked to find the mean and variance.
I know the mean is represented by $\sum_{i=1}^n p_ix_i$ and the variance by $\sum_{i=1}^n p_i(x_i-\mu)^2$, but I'm not really sure how to get from those formulas to a generalized answer for the mean and variance. I've looked around online, but nearly everything seems to involve the use of a finite set of numbers. Those examples make sense, but I'm not sure how to translate that understanding into the more general solution asked for above. An explanation or a link to reading on the topic would help greatly!
 A: Calculating the expected value
$$E(X)=p\sum_{x=1}^{\infty} x\cdot (1-p)^{x-1} $$


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*Indexshift: $x=k+1 \Rightarrow k=x-1$


$$E(X)=p\sum_{k=0}^{\infty} (k+1)\cdot (1-p)^{k}=p\sum_{k=0}^{\infty} k\cdot (1-p)^{k}+p\sum_{k=0}^{\infty} (1-p)^{k}$$


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*The first summand of $p\sum_{k=0}^{\infty} k\cdot (1-p)^{k}$ is 0.


Thus $$p\sum_{k=0}^{\infty} k\cdot (1-p)^{k}=p\sum_{k=1}^{\infty} k\cdot (1-p)^{k}$$
$$E(X)=p\sum_{k=1}^{\infty} k\cdot (1-p)^{k}+p\sum_{k=0}^{\infty}  (1-p)^{k}$$


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*Multiplying out (1-p) at the first sigma sign.


$$E(X)=(1-p) \underbrace{p\sum_{k=1}^{\infty} k\cdot (1-p)^{k-1}}_{E(x)}+\underbrace{p\sum_{k=0}^{\infty}  (1-p)^{k}}_{=1}$$
Thus $E(x)=(1-p)\cdot E(x)+1 \Rightarrow E(x)=\frac{1}{p}$
A: 
I'm given the formula:
  $$(1−p)^{x−1} p \quad,x=1,2,...,∞ $$

That's not a formula.  A formula would be: $$P_X(x)=(1-p)^{x-1}p\quad:x\in\{1,2\ldots,\infty\}$$

and I'm asked to find the mean and variance.
I know the mean is represented by $\sum_{i=1}^n p_i x_i$   and the variance by $\sum^n_{i=1} p_i (x_i −μ)^2$.

In this case $$\begin{align}
\mathsf E(X) & = \sum_{x=1}^\infty x\mathsf P_X(x)
\\[2ex]
\mathsf {Var}(X) & =\sum_{x=1}^\infty (x-\mathsf E(X))^2\mathsf P_X(x)
 \\ & = \sum_{x=1}^\infty x^2\mathsf P_X(x) - (\mathsf E(x))^2
\end{align}$$
Now $$\begin{align}
\mathsf E(x) & = \sum_{x=1}^\infty x\mathsf P_X(x)
\\ & = \sum_{x=1}^\infty xp(1-p)^{x-1}
\\ & = p(1+\frac 2{(1-p)}+\frac 3{(1-p)^2} + \frac 4{(1-p)^3} + \cdots)
\end{align}$$
Which is based on an order -1 polylogarithm (an extension of geometric series). $\sum_{k=1}^\infty k r^k = r/(1-r)^2, |r|<1$
Can you complete?
