How do I compute $\sum_{k=1}^{\infty} k \cdot p^k$

Question

I have no idea how to compute this infinite sum. It seems to pass the convergence test. It even seems to be equal to $\frac{p}{(1-p)^2}$, but I cannot prove it. Any insightful piece of advice will be appreciated.

Answer 1

Assume that $|p|<1$.

METHOD $1$:

$$p\left(\frac{d}{dp}\sum_{k=1}^\infty p^k\right)=\sum_{k=1}^{\infty}kp^k$$

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METHOD $2$:

$$\begin{align} \sum_{k=1}^\infty kp^k&=\sum_{k=1}^{\infty}p^k\sum_{j=1}^k (1)\\\\ &=\sum_{j=1}^{\infty}\sum_{k=j}^{\infty}p^k\\\\ &=\sum_{j=1}^{\infty}\frac{p^j}{1-p}\\\\ &=\frac{p}{(1-p)^2} \end{align}$$

Answer 2

Hint

$$\sum_{k=1}^\infty kp^k=p\sum_{k=1}^\infty kp^{k-1}.$$

Answer 3

$$\begin{eqnarray*}(1-p)^2\sum_{k\geq 1}kp^k &=& \sum_{k\geq 1}kp^k -2\sum_{k\geq 1}kp^{k+1} +\sum_{k\geq 1} kp^{k+2}\\&=&(p+2p^2)-(2p^2)+\sum_{k\geq 3}\left(k-2(k-1)+(k-2)\right)p^{k}\\&=&\color{red}{p}.\end{eqnarray*}$$