Ashlyn Krause

2022-07-23

How to find sum of infinite geometric series with coefficient?
Given this series, $p+p\left(1-p{\right)}^{3}+p\left(1-p{\right)}^{6}+p\left(1-p{\right)}^{9}+...$
This is an infinite geometric series with ratio less than 1 since it's probability.
$\sum _{n=0}^{\mathrm{\infty }}p\left(1-p{\right)}^{3n}$
Can you use geometric series sum formula? Is it $p/\left(1-\left(1-p{\right)}^{3}\right)$?
How do you deal with 3 that's in front of n?

Hassan Watkins

Expert

Step 1
Your sum is correct. All you need to do is recognize that $\sum _{n=0}^{\mathrm{\infty }}p\left(1-p{\right)}^{3n}=\sum _{n=0}^{\mathrm{\infty }}p{\left[\left(1-p{\right)}^{3}\right]}^{n}$
Step 2
The latter series is a geometric series whose common ratio and leading term are $\left(1-p{\right)}^{3}$ and p, respectively, so it will converge to $\frac{p}{1-\left(1-p{\right)}^{3}}$ if $|\left(1-p{\right)}^{3}|<1$ and diverge if $|\left(1-p{\right)}^{3}|\ge 1$

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