# $\sum_{k=1}^{∞} k(1-p)^{k-1}+\sum_{k=1}^{∞} kp^{k-1}$ for $0 < p < 1$

$$\sum_{k=1}^{∞} k(1-p)^{k-1}+\sum_{k=1}^{∞} kp^{k-1}$$ where $$0

I am trying to some problems and was stuck at this step. Could someone please give me some hint on how to solve this equation?

• I think you should mention $0 < p < 1$. – Math Lover Sep 24 '18 at 17:20
• @MathLover Thank you for your suggestion... I have added this condition to my question – Chloe Zhou Sep 24 '18 at 17:22

Hint: $$\sum_{k=0}^{\infty} x^k = \frac{1}{1-x} \implies \frac{d}{dx}\left(\sum_{k=0}^{\infty} x^k\right) = \sum_{k=1}^{\infty}kx^{k-1} = \frac{d}{dx}\left(\frac{1}{1-x}\right)=\frac{1}{(1-x)^2}$$ for $$|x|<1$$.
\begin{align} \sum_{k=1}^{\infty} k(1-p)^{k-1}+\sum_{k=1}^{\infty} kp^{k-1} &= \sum_{k=1}^{\infty} \left(-(1-p)^{k}\right)'+\sum_{k=1}^{\infty} \left(p^{k}\right)' \\ &= \left(\sum_{k=1}^{\infty} -(1-p)^k+\sum_{k=1}^{\infty} p^{k}\right)' \\ &= \left(-\dfrac{1-p}{p}+\dfrac{p}{1-p}\right)' \\ &= \dfrac{-1}{(1-p)^2}-\dfrac{1}{p^2} \end{align}