Derivative of power sum? In one part of a proof I'm reading I see the following:
$$ p\frac{d}{dp}\sum_{k=0}^\infty p^k = p\frac{1}{1-p} $$
However, I'm confused how the derivative of $\frac{d}{dp}\sum_{k=0}^\infty p^k = \frac{1}{1-p}$, wouldn't it equal $\frac{1}{(1-p)^2}$?
Here's a picture of the actual two lines as I've stripped them to the part I'm confused about: http://imgur.com/a/rSJS7
 A: You are right, I think, because
$$\frac{d}{dp}\sum_{k=0}^\infty p^k = (\frac{1}{1-p})'=\frac{1}{(1-p)^2}$$
A: You're right. The calculation stated in http://imgur.com/a/qJUY9 is not correct.

We obtain
  \begin{align*}
\color{blue}{E[N^2]}&=\ldots=(1-\rho)\rho\frac{d}{d\rho}\left(\rho\frac{d}{d\rho}\left(\sum_{i=0}^\infty \rho^i\right)\right)\\
&=(1-\rho)\rho\frac{d}{d\rho}\left(\rho\frac{d}{d\rho}\left(\frac{1}{1-\rho}\right)\right)\tag{1}\\
&=(1-\rho)\rho\frac{d}{d\rho}\left(\rho\frac{(1-\rho)(0)-(1)(-1)}{(1-\rho)^2}\right)\tag{2}\\
&=(1-\rho)\rho\frac{d}{d\rho}\left(\frac{\rho}{(1-\rho)^2}\right)\tag{3}\\
&=(1-\rho)\rho\cdot\frac{(1-\rho)^2(1)-(\rho)2(1-\rho)(-1)}{(1-\rho)^4}\tag{4}\\
&=(1-\rho)\rho\cdot\frac{1+\rho}{(1-\rho)^3}\\
&\color{blue}{=\frac{\rho(1+\rho)}{(1-\rho)^2}}
\end{align*}

Comment:


*

*In (1) we use   the geometric  series expansion.

*In (2) we apply  the quotient rule in the same manner as it is presented in the referred image.

*In (3) we do some simplifications and continue in (4) as we did in (2).

With $E[N]=\frac{\rho}{1-\rho}$ we finally conclude
  \begin{align*}
\color{blue}{Var(N)}&=E[N^2]-E^2[N]\\
&=\frac{\rho(1+\rho)}{(1-\rho)^2}-\frac{\rho^2}{(1-\rho)^2}\\
&=\frac{\rho+\rho^2-\rho^2}{(1-\rho)^2}\\
&\color{blue}{=\frac{\rho}{(1-\rho)^2}}
\end{align*}

