# How to express this series as a geometric series

How do I express $\frac{1}{3^n}$, or $3^{-n}$ as a typical geometric series of the form: $ar^{n-1}$ ?

This is a small part of a larger question, in which I'm trying to find the convergence of $\sum_{n=0}^\infty \frac{1+2^n}{3^n}$.

I broke the series into two separate series', $\sum_{n=0}^\infty \frac{1}{3^n} + \sum_{n=0}^\infty (\frac23)^n$.

However I am stuck trying to represent the first term as a geometric series. I know that it is a geometric series only because Wolfram Alpha told me so here.

• Not exactly sure what is confusing you here. $\frac{1}{3^n} = (\frac{1}{3})^n = 0.333...^n$ which obviously converges since $0.333... < |1|$. Jan 8, 2016 at 15:13
• Thanks for the tip here. I oversaw the $\frac{1}{3^n} = (\frac{1}{3})^n$. I forgot (stupidly) that $1^n$ is, of course, just $1$. Jan 8, 2016 at 15:18

HINT: Notice, $$\frac 1{3^{n}}=\frac{1}{3\cdot 3^{n-1}}=\frac{1}{3}\cdot \frac{1}{3^{n-1}}=\frac{1}{3}\cdot \left(\frac{1}{3}\right)^{n-1}$$ comparing with typical form of G.P. $ar^{n-1}$, one should have $a=\frac 13$ & $r=\frac 13$
In the expression $a r^{n-1}$, we think of $a$ as representing the first term of the sequence, and $r$ as the ratio between common terms.
• the first term is $\frac{1}{3^1}$
• the ratio of terms is $\frac{\frac{1}{3^{n+1}}}{\frac{1}{3^n}}=1/3$
So, $a=r=\frac{1}{3}$.