How would I find the common ratio to determine the sum of the given geometric serie: How would I find the common ratio to compute the sum of the given geometric serie: $$\sum_{n=1}^\infty= \frac{(8^n+2^n)}{9^n}$$

First note the evaluation of the relevant finite summation:
$$\sum _{n=1}^{N}{c}^{n}={\frac {{c}^{N+1}}{c-1}}-{\frac {c}{c-1}} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(0)$$
And then observe that your sum is the sum of two such summations,the key difference being, these are infinite:
$$\sum _{n=1}^{\infty }{\frac {{8}^{n}+{2}^{n}}{{9}^{n}}}=\sum _{n=1}^{
\infty } \left( {\frac {8}{9}} \right) ^{n}+\sum _{n=1}^{\infty }
 \left(\frac {2}{9} \right) ^{n}
$$
We then simply set $c_1=\frac {8}{9}$ and $c_2=\frac {2}{9}$, and substitute the infinite upper bound for our sum with a variable $N$, like we see in the evaluation of the finite sum $(0)$:
$$\sum _{n=1}^{
N }  {c_1}  ^{n}+\sum _{n=1}^{N }
 c_2 ^{\,n}={\frac {{c_1}^{N+1}}{c_1-1}}-{\frac {c_1}{c_1-1}}+{\frac {{c_2}^{N+1}}{c_2-1}}-{\frac {c_2}{c_2-1}}$$
Our final step is the part for which one of the fundamental principles of calculus is required, we evaluate the limit for our variable $N$ as it becomes infinite:
$$\sum _{n=1}^{
\infty } \left( {\frac {8}{9}} \right) ^{n}+\sum _{n=1}^{\infty }
 \left(\frac {2}{9} \right) ^{n}
=\lim _{N\rightarrow \infty }\Biggl(\sum _{n=1}^{
N } \left( {\frac {8}{9}} \right) ^{n}+\sum _{n=1}^{N }
 \left(\frac {2}{9} \right) ^{n}
\Biggr)=\lim _{N\rightarrow \infty }\Biggl({\frac {{(\frac {8}{9})}^{N+1}}{\frac {8}{9}-1}}-{\frac {(\frac {8}{9})}{(\frac {8}{9})-1}}+{\frac {{(\frac {2}{9})}^{N+1}}{\frac {2}{9}-1}}-{\frac {\frac {2}{9}}{\frac {2}{9}-1}}\Biggr)=\frac{58}{7}$$
