Computing $ \sum_{k\ge2}k(1-p)^{k-2}$ Computing $\displaystyle \sum_{k\ge2}k(1-p)^{k-2}$, $p\in ]0,\space1[$
WolframAlpha says it is $\cfrac {p+1}{p^2}$ but I couldn't get that value but anyway here is what I did:
$$\displaystyle \sum_{k\ge2}k(1-p)^{k-2} = (1-p)^{-1}\sum_{k\ge2}k(1-p)^{k-1} = (1-p)^{-1} \left (-\sum_{k\ge1}(1-p)^{k} \right)'= (1-p)^{-1} \left (-\cfrac {1-p}{p}\right)' = (1-p)^{-1} \left (1-p^{-1}\right)'=(1-p)^{-1} p^{2} = \cfrac 1{p^2(1-p)}$$ 
Please tell me what I'm doing wrong or is WolframAlpha wrong on this one?
 A: $$f(x):=\sum_{n=0}^\infty x^n=\frac{1}{1-x}\,\,\,,\,\,|x|<1\Longrightarrow$$
$$\Longrightarrow f'(x)=\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2}\Longrightarrow $$
$$\Longrightarrow \frac{1}{x(1-x)^2}-\frac{1}{x}=\sum_{n=2}^\infty nx^{n-2}$$
so substituting $\,x=1-p\,$ (why can we?), we get
$$\frac{1}{p^2(1-p)}-\frac{1}{1-p}=\sum_{n=2}^\infty n(1-p)^{n-2} $$
And since
$$\frac{1}{p^2(1-p)}-\frac{1}{1-p}=\frac{1+p}{p^2}$$
we're then done.
A: You shifted the index when you should not have done so:
$$\begin{align*}
\sum_{k\ge 2}k(1-p)^{k-1}&=-\left(\sum_{k\ge 2}(1-p)^k\right)'\\
&=-\left(\frac{(1-p)^2}p\right)'\\
&=\frac{(1-p)^2+2p(1-p)}{p^2}\\
&=\frac{1-p^2}{p^2}\\
&=\frac{(1-p)(1+p)}{p^2}\;.
\end{align*}$$
You had $-\left(\sum_{k\ge 1}(1-p)^k\right)'$, but if you write out the first couple of terms, you’ll see that this gives you a term that isn’t present in the original summation: the $k=1$ term, after differentiation, is $-1\cdot(1-p)^0(-1)=1$, while $\sum_{k\ge 2}k(1-p)^{k-1}$ has no constant term.
If you keep the right range of indices, you get
$$(1-p)^{-1}\cdot\frac{(1-p)(1+p)}{p^2}=\frac{1+p}{p^2}\;,$$ just as Wolfram|Alpha does.
By the way, you could reduce the chance of making computational errors by letting $q=1-p$ and working with $q$; dealing with $\sum_{k\ge 2}kq^{k-1}$ is easier than dealing with $\sum_{k\ge 2}k(1-p)^{k-1}$, especially when you get to the differentiations.
