Unable to find the sum of a series

Question

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?

Answer 1

This is not a geometric series.

But it is the sum of two geometric series:

$$\sum_{n=1}^{\infty} {\frac{1+7^n}{9^n}} = \sum_{n=1}^{\infty} {\frac{1}{9^n}}+ \sum_{n=1}^{\infty} {\frac{7^n}{9^n}} =\frac 19 \frac 1{1-\frac 19} +\frac 79\frac 1{1-\frac 79}$$

because $$ \left|\frac 19\right|<1 \\\left|\frac 79\right|<1 $$

Answer 2

Hint: Try re-doing it as two separate series, $\sum(1/9)^n$ and $\sum(7/9)^n.$ The first has first term $1/9$ and common ratio $1/9$ while the second has first term $7/9$ with common ratio $7/9.$

Answer 3

$$ \frac{\frac19}{1-\frac19}+\frac{\frac79}{1-\frac79}=\frac18+\frac72=\frac{29}{8}. $$

Answer 4

\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*}