The series I'm working with is defined as the sum from 1 to infinity of $\sum_{n = 1}^{\infty}\dfrac{1+2^n}{3^n}$. The first terms of the sequence as defined are:
- 1
- 5/9
- 1/3
- 17/81
The trouble I'm having is that the ratio between those terms is not consistent:
- $\dfrac{5/9}{1} = r = \dfrac{5}{9}$
- $\dfrac{1/3}{5/9} = r = \dfrac{3}{5}$
- $\dfrac{17/81}{1/3} = r = \dfrac{17}{27}$
Since $r$ is inconsistent I can't find the sum of the series.
I feel like I'm probably missing something really basic here but I can't figure it out for the life of me. Any pointers?
This is not a geometric series.
Why?
Observe that the $k$th term of the sum is given by $$a_k = \dfrac{1+2^{k}}{3^{k}}$$ so that the ratio of the $(k+1)$th term to the $k$th term is $$\dfrac{a_{k+1}}{a_k}=\dfrac{(1+2^{k+1})/3^{k+1}}{(1+2^{k})/3^{k}} = \dfrac{1+2^{k+1}}{1+2^{k}} \cdot \dfrac{3^{k}}{3^{k+1}}=\dfrac{1+2^{k+1}}{1+2^{k}} \cdot \dfrac{1}{3}$$ which is not constant (because it depends on $k$). Thus, this series is not a geometric series.
However, there is a way to solve this problem.
Observe that $$a_k = \dfrac{1+2^{k}}{3^{k}}=\dfrac{1}{3^k}+\dfrac{2^k}{3^k}=\left(\dfrac{1}{3}\right)^k+\left(\dfrac{2}{3}\right)^k\text{.}$$ Thus, $$\sum_{k=1}^{\infty}a_k=\sum_{k=1}^{\infty}\left(\dfrac{1}{3}\right)^k + \sum_{k=1}^{\infty}\left(\dfrac{2}{3}\right)^k$$ This is a sum of two geometric series.