Sum of $p\sum_{i = 1}^{\infty} i(1 - p)^{i - 1}$

Question

Does anyone know how to find the exact sum of

$$ p\sum_{i = 1}^{\infty} i(1 - p)^{i - 1} $$

Answer 1

Note that

$$\sum_{i=1}^{\infty} i \, r^{i-1} = \frac{d}{dr} \sum_{i=0}^{\infty} r^i = \frac{d}{dr} \frac{1}{1-r}=\frac{1}{(1-r)^2}$$

So your sum is, letting $r=1-p$,

$$p \frac1{p^2} = \frac1{p}$$

Answer 2

I'm going to do this my favorite way instead of the same way as everyone else.

Let $S$ be the sum in question. Then we can multiply $S$ by $\displaystyle \frac{1}{1-p}$ to shift the sum and then subtract, obtaining $$\frac{p}{1-p}S = \frac{S}{1-p} - S = \frac{p}{1-p} + p\sum_{i=0}^\infty (1-p)^i = \frac{p}{1-p} + p(\frac{1}{p}) = \frac{1}{1-p}.$$ Solving for $S$ we obtain $\displaystyle S = \frac{1}{p}$.

Of course, the sum only converges when $|1-p| < 1$.

Answer 3

The series is related to the expectation value of a geometric distributed random variable. Let the RV be T: \begin{align} E[T] &= \sum_{i=1}^\infty i p (1-p)^{i-1} \quad \| (1-p) = x\\ &= p \sum_{i=1}^\infty i x^{i-1} \\ &= p \sum_{i=0}^\infty (i+1) x^{i} \\ &= p \big[ \frac{d}{dx}\sum_{i=0}^\infty \frac{(i+1)}{(i+1)} x^{i+1} \big] \\ &= p \big[\frac{d}{dx} \big(x \sum_{i=0}^\infty x^i \big) \big] \quad \| \text{geometric series}\\ &= p \big[ \frac{d}{dx} \big(x \frac{1}{1-x} \big)\big]\\ &= p \big[ \frac{1}{1-x} + x(1-x)^{-2} \big] \quad \| (1-p) = x \\ &= p \big[ \frac{1}{p} + \frac{1-p}{p^2} \big]\\ &= \frac{1}{p} \end{align}

Answer 4

Yes. Integrate the sum termwise with respect to $p,$ to get $$-\sum_{i=1}^\infty(1-p)^i = -(p-1) \sum_{i=0}^\infty (1-p)^i = -\frac{1}{p}.$$ Differentiate back with respect to $p,$ to get $1/p^2.$ Multiply by $p$ to finally get $1/p.$

Answer 5

Given that $|x|< 1$, $$\frac 1{(1-x)^2}=\sum_{k=0}^\infty (k+1)x^k$$

So $$p^{-1}=\frac{p}{(1-(1-p))^2}=p\sum_{k=0}^\infty (k+1)(1-p)^k$$

Reasoning: $(1-x)(1+x+x^2+\ldots)=1+x+x^2+\ldots-(x+x^2+\ldots)=\lim_{n\to\infty}1-x^n=1$ so $\frac{1}{1-x}=\sum_{k=0}^\infty x^k$. Note that $$\frac{1}{(1-x)^2}=\sum_{k=0}^\infty \frac{x^k}{1-x}=\sum_{k=0}^\infty\left(\sum_{\ell=0}^\infty x^{k+\ell}\right)$$ For any $c$, there are $c+1$ pairs $(k,\ell)$ such that $k+\ell=c$, so the sum reduces to $\sum_{k=0}^\infty (k+1)x^k$