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

Does anyone know how to find the exact sum of

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

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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 $r=1-p$,

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

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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 $|1-p| < 1$.

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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 $1/p.$

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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 $\sum_{k=0}^\infty (k+1)x^k$

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