# Calculate $\frac13+\frac29+\frac{1}{27}+\frac{2}{81}+\frac{1}{243}+\frac{2}{729}+\ldots$

How can we calculate the formula for the following series?

$\frac13+\frac29+\frac{1}{27}+\frac{2}{81}+\frac{1}{243}+\frac{2}{729}+\ldots$

I'm not quite sure what type of series it is anyway. It doesn't look geometric. Is there any strategy/procedure to follow in order to calculate the formula for a series of numbers?

• Look at the sequence as made of two interwoven sequences, one going $1/3,1/27,1/243,...$ and the other going $2/9,2/81,2/729...$. Treat each separately using geometric series. – coffeemath Apr 4 '13 at 23:29
• @coffeemath: same idea, two ways of looking at it :-) – robjohn Apr 4 '13 at 23:38

If you are looking for the infinite sum, it is the sum of two geometric series: $$\sum_{k=1}^\infty\frac1{3^k}+\sum_{k=1}^\infty\frac1{9^k}=\frac12+\frac18=\frac58$$
If we add neighbouring pairs, we get $$\frac{5}{9}+\frac{5}{81}+\frac{5}{729}+\cdots.$$ This is an infinite geometric series with first term $\frac{5}{9}$ and common ratio $\frac{1}{9}$.
By the usual formula for the sum of an infinite geometric progression, the sum is $$\frac{5}{9}\frac{1}{1-\frac{1}{9}}.$$ This can be simplified to $\frac{5}{8}$.
$$a_n=\begin{cases}\frac{1}{3^n}&,\;\;\text{if \,n\, is odd}\\{}\\\frac{2}{3^n}&,\;\;\text{if \,n\, is even}\end{cases}\;=\ldots\;$$