# How to find sum of infinite geometric series with coefficient?

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?

• Use the substitution $q = (1-p)^3$. – Andy Walls Sep 30 '20 at 3:20
• Right but then n is NOT raised to q right? (1-p)^(3n) does not equal (1-p)^3^n – Jake Yoon Sep 30 '20 at 3:27
• $(1-p)^{3n}=((1-p)^3)^n$ – J. W. Tanner Sep 30 '20 at 3:44

$$\sum_{n=0}^{\infty}p(1-p)^{3n}=\sum_{n=0}^{\infty}p\left[(1-p)^3\right]^n$$
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 $$\left|(1-p)^3\right|\geq 1$$.