Can you help me solve this series: $\sum_{k=0}^\infty p^kk$?

Question

I have the infinite series: $$\sum_{k=0}^\infty p^kk$$ where $0<p<1$. I can see from a computer calculation that the series converges to $\frac{p}{(1-p)^2}$, but I can't see why. Thanks in advance for any help.

Answer 1

So notice that $$ \sum_{k=0}^\infty p^kk = \sum_{k = 0}^\infty p^k + \sum_{k = 1}^\infty (k-1)p^k - 1 $$ which gives $$ \sum_{k=0}^\infty p^kk = \sum_{k = 0}^\infty p^k + p\sum_{k=0}^\infty p^kk - 1 $$

Edit: Namely, writing $S$ for your sum, $$ S = \sum_{k = 0}^\infty p^k + pS - 1 $$ or $$ (1-p) S = \sum_{k = 0}^\infty p^k - 1 $$ the sum in the right hand side of which has the well-known value $\frac{1}{1-p}$ for $|p| < 1$. Therefore $$ S = \frac{p}{(1-p)^2} $$

Edit 2: Whoops, I should have noticed that multiplying by the index makes the zeroth term vanish. Fixed to reflect that (just a minor issue).

Answer 2

Here's another way if you remember the properties of power series:

$$\sum_{k=0}^{\infty}p^kk = p\sum_{k=0}^{\infty}p^{k-1}k = p\sum_{k=0}^{\infty}\dfrac{d}{dp}p^k$$

$$ = p\cdot\dfrac{d}{dp}\sum_{k=0}^{\infty}p^k = p\cdot \dfrac{d}{dp}\dfrac{1}{1-p} = p\cdot \dfrac{1}{(1-p)^2}$$

Answer 3

Sometimes when the answer is given it doesn't hurt to start from the other side:

$\displaystyle \begin{aligned} \frac{1}{(1-p)}\cdot \frac{1}{(1-p)} & = \sum_{j=0}^{\infty}p^j \cdot \sum_{j=0}^{\infty}p^j = \sum_{k=0}^{\infty} \sum_{j=0}^{k}p^{j}p^{k-j} = \sum_{k=0}^{\infty} \sum_{j=0}^{k}p^{k} =\sum_{k=0}^{\infty}(k+1)p^k \end{aligned} $

So $\displaystyle \frac{p}{(1-p)^2} = \sum_{k=0}^{\infty}(k+1)p^{k+1}$ and adjusting the index this is equal to $\displaystyle \sum_{k=0}^{\infty}kp^{k}$.