Given this series,
$ p + p(1-p)^3 + p(1-p)^6 + p(1-p)^9 + …$
This is an infinite geometric series with ratio less than 1 since it's probability.
$$ \sum_{n=0}^{\infty}p(1-p)^{3n}$$
Can you use geometric series sum formula? Is it $ p / (1-(1-p)^3)$?
How do you deal with 3 that's in front of n?
Best Answer
Your sum is correct. All you need to do is recognize that
$$\sum_{n=0}^{\infty}p(1-p)^{3n}=\sum_{n=0}^{\infty}p\left[(1-p)^3\right]^n$$
The latter series is a geometric series whose common ratio and leading term are $(1-p)^3$ and $p$, respectively, so it will converge to $\frac{p}{1-(1-p)^3}$ if $\left|(1-p)^3\right|<1$ and diverge if $\left|(1-p)^3\right|\geq 1$.