Considering now a random variable X with a geometric distribution with parameter p > 0, we have $$E(X) =\sum_{k=1}^\infty kp(1-p)^{k-1} $$ We already shown that $$\sum_{k=1}^\infty p(1-p)^{k-1} =1 $$ Which implies $$\sum_{k=1}^\infty (1-p)^{k-1}=1/p $$ If we take the derivative on p we get $$\sum_{k=1}^\infty(k-1)(-1)(1-p)^{k-2}=-1/p^2$$ which is equivalent to $$\sum_{k=1}^\infty k(1-p)^{k-1}=1/p^2 $$
I know this is a silly question, but I couldnt get the equivalent part how is$$\sum_{k=1}^\infty k(1-p)^{k-1}$$ is equivalent to$$\sum_{k=1}^\infty(k-1)(-1)(1-p)^{k-2}$$ it will be much appreciated if someone explain on how they are equivalent?
\begin{align} \sum_{k=1}^\infty k(1-p)^{k-1} = {} & 1(1-p)^0 + 2(1-p)^1 \\ & {} + 3(1-p)^2 + 4(1-p)^3+\cdots \\[12pt] \sum_{k=1}^\infty (k-1)(1-p)^{k-2} = {} & 0(1-p)^{-1} \\ & {} + 1(1-p)^0 + 2(1-p)^1 \\[8pt] & {} + 3(1-p)^2 + 4(1-p)^3 + \cdots \end{align}