TITLE: Unable to find the sum of a series
QUESTION [2 upvotes]: I am trying to find the sum of the following series:
$$\sum_{n=1}^{\infty} {\frac{1+7^n}{9^n}}$$
which I rewrote as 
$$\sum_{n=1}^{\infty} \left(\frac{1}{9^n}+
\left(\frac{7}{9}\right)^n\right)$$
I am assuming that it is a geometric series and the initial value is
$$a_1=\frac{1}{9} + \frac{7}{9}$$
I also see that 
$$a_2 = \frac{1}{9^2} + \frac{7^2}{9^2}$$
I know that in a geometric series the first term is $a$ and the second term is $ar$.
This allows me to see that 
$$\left(\frac{1}{9}+\frac{7}{9}\right)r=\frac{1}{9^2}+\frac{7^2}{9^2}$$
which when solved for $r$ gives the value $\frac{25}{36}$.
Using the formula to find the sum of a geometric series $\frac{a}{1-r}$, I find that the sum is equal to $\frac{32}{11}$.
But this value is incorrect and the sum is actually $\frac{29}{8}$. How does one find that value?

REPLY [0 votes]: \begin{align*}\sum_{n=1}^\infty\frac{1+7^n}{9^n} &= \sum_{n=1}^\infty\bigg[ \frac{1}{9^n} + \frac{7^n}{9^n}\bigg] \\ &= \sum_{n=1}^\infty\frac{1}{9}\bigg(\frac{1}{9}\bigg)^{n-1} + \sum_{n=1}^\infty \frac{7}{9}\bigg(\frac{7}{9}\bigg)^{n-1} \\ &=\frac{1}{9}\bigg(\frac{1}{1-\frac{1}{9}}\bigg)+\frac{7}{9}\bigg(\frac{1}{1-\frac{7}{9}}\bigg) \\ &= \frac{1}{8}+\frac{7}{2} \\ &= \frac{29}{8} \end{align*}
{"set_name": "stack_exchange", "score": 2, "question_id": 1006700}