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?