Sum of n geometric series
\[\sum_{0}^{n}\frac{1}{b^{n}}= \frac{a(1-r^{n})}{1-r} \]
For example
\[\sum_{0}^{n}\frac{1}{3^{n}}= \frac{1(1-(1/3))^{n})}{1-(1/3))} \]
The above equation can also be written as
\[\sum_{0}^{n}\frac{1}{3^{n}}= \frac{1}{3^{0}}+\frac{1}{3^{1}}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+.......\frac{1}{3^{n}} \]
where first term is 1, 2nd term is 1/3 and so on
Common ratio- Each term after the first term is found by multiplying its preceding term by a non zero constant which is called common ratio.
In this case common ratio is \[\frac{\frac{1}{3^{2}}}{\frac{1}{3}} \] is \[\frac{1}{3} \]
