Series to infinity

Question

I am studying probability and I have come across the following:

$$p\sum\limits_{n=1}^{\infty} (1-p)^{n-1}=1$$

But I can't see why the result is equal to 1. How can the above result be proven?

Answer 1

$A= \sum_{n=1}^{\infty} a^n = a^1+\sum_{n=2}^{\infty} a^n = a^1+\sum_{n=1}^{\infty} a^{n+1}=a^1+a \times \sum_{n=1}^{\infty} a^n= a+aA$

then,

$A(1-a)=a \rightarrow A = \frac{a}{1-a}$. in your case:

$p\sum_{n=1}^{\infty} (1-p)^{n-1}=\frac{p}{1-p} \sum_{n=1}^{\infty} (1-p)^{n} = \frac{p}{1-p} \times \frac{1-p}{p}=1$

Answer 2

If $p$ is a probability, then $1-p \in [0,1]$, and so the series you've written is geometric, with value

$$ \sum_{n=1}^\infty(1-p)^{n-1} = \frac{1}{1-(1-p)} = \frac{1}{p}. $$

Answer 3

This is an infinite geometric progression

This evaluates to:

$p ( 1 + (1 - p) + (1 - p)^2 + (1 - p)^3 ..... )$

It has a common ratio of $1 - p$ and $n\rightarrow \infty$ So, it becomes

$p ( \frac{1}{1 - (1 - p)} ) = p (\frac{1}{p}) = 1$