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I am revising for an exam and I am facing the following statement: $\frac{1}{1-p}=\sum_{k=0}^\infty {p^k}$

Unfortunately, I was not able to figure out why this is the case. May somebody of you quickly help me? I appreciate it a lot for a detailed explanation!

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    Presumably $|p|<1$? If so then this is just the formula for the sum of a geometric series. – 2017-01-06
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    https://en.wikipedia.org/wiki/Geometric_series – 2017-01-06
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    perfect, that's about it. Thanks to you both a lot! – 2017-01-06

3 Answers 3

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From the equaility $(1-p)(1+p+p^2+ \dots + p^k)=1-p^{k+1}$ we deduce $\frac{1-p^{k+1}}{1-p}=1+p+\dots+p^k$

So if $p<1$ we have $\frac{1}{1-p}=\lim\limits_{k\to \infty} \frac{1-p^{k+1}}{1-p}=\lim\limits_{k\to \infty}1+p+p^2+\dots+ p^k$

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Since$$\frac{1-p^n}{1-p}=1+p+p^2+...+p^{n-1}=\sum_{i=1}^{n-1}p^k$$

If $|p|<1$, $n\to \infty$, then $$\lim_{n\to \infty}\frac{1-p^n}{1-p}=\frac{1}{1-p}=\sum_{i=1}^{\infty}p^k$$

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HINT: consider the finite sum $$\sum_{k=0}^np^k={\frac {{p}^{n+1}}{p-1}}- \left( p-1 \right) ^{-1}$$ and compute the limit for $n$ tends to infinity.

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    Dr. Sonnhard, thank you for your fast response! – 2017-01-06