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As part of the solution of an exercise I have the following relation:

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

Where $p$ is a probability.

I don't understand where this is coming from?

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  • $\begingroup$ See this. $\endgroup$ Jun 22, 2013 at 13:11
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    $\begingroup$ Since your question is about "where this relation came from", I suggest that you write the context on which it appeared. What was the problem being solved? This information will help you get better answers. $\endgroup$
    – Gold
    Jun 22, 2013 at 13:56
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    $\begingroup$ Imo it is completely wrong to close this question "as duplicate" , since it is not the same question. Certainly in one of the answers in the other thread one can see how this question's addressed, but if the OP asked this question he'll hardly be able to recognize a different question's answer! I'm voting to reopen. $\endgroup$
    – DonAntonio
    Jun 22, 2013 at 14:45

3 Answers 3

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Hints:

For $\,x\in\Bbb R\;,\;\;|x|<1\,$ , we have the well known power series development

$$\frac1{1-x}=\sum_{n=0}^\infty x^n\stackrel{\text{differentiation}}\implies\frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}$$

But $\;p\,$ is probability, so $\;0\le p<1\;\ldots$ (if $\,p=1\,$ the claim in your question is trivial)

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Note that

$$f(p)=1+p+p^2+p^3+\ldots= 1+p(1+p+p^2+p^3+\ldots)=1+p f(p)$$

from which you get that $f(p)=\frac{1}{1-p}$.

Next notice that

$$g(p)=1+2p+3p^2+4p^3+\ldots=(1+p+p^2+p^3+\ldots)+p(1+2p+3p^2+4p^3+\ldots)=\frac{1}{1-p}+p g(p)$$

from which you get that $g(p)=\frac{1}{(1-p)^2}$.

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If $S=\sum_{1\le r\le n} r ab^{r-1}$

$b\cdot S=\sum_{1\le r\le n} r ab^r$

$$\implies S(1-b)=a(1+b+b^2+\cdots+b^{n-1})-nab^n$$

$$=a\frac{1-b^n}{1-b}- a(n b^n)$$

If $|b|<1, \lim_{n\to\infty}b^n=0$ and from this,$$\text{ if } |b|<1,\lim_{n\to\infty} nb^n=0\text{ [Proof Below] }$$

$$\implies \lim_{n\to\infty}S(1-b)=\frac a{1-b}\text{ if }|b|<1$$

Here $a=1, r=1-p$ and as $0\le p\le 1\iff 0\le 1-p\le 1$

$p=1$ will make each summand $=0$

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If $|b|<1,$ let $c=\frac1b$

$$\lim_{n\to\infty} nb^n=\lim_{n\to\infty} \frac n{c^n}=\lim_{n\to\infty} \frac1{nc^{n-1}\ln c}=0$$ as $|c|>1$

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    $\begingroup$ I don't think you applied L'Hospital's rule correctly: d/dn (1/b)^n = log(1/b) (1/b)^n $\endgroup$ Jun 22, 2013 at 13:17
  • $\begingroup$ @VincentPfenninger, please have a look into the edited answer $\endgroup$ Jun 22, 2013 at 13:30

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