# What is the sum of the following infinite series?

$$\frac{1}{3} + \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \frac{1}{243} + \frac{2}{729} + \cdots$$ So basically I separated it into two series

where:

one of them is $$\left(\frac{1}{3}\right)^n$$

And I use geometric series formula to find that this series equals $$\frac{1}{2}$$.

But I can't figure out the series of the other one.

Apparently the answer for the series combined is: $$\frac{5}{8}$$

What is the other series?

It seems you have

\begin{aligned} & \frac{1}{3} + \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \frac{1}{243} + \frac{2}{729} + \ldots \\ & = \left(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + \frac{1}{243} + \frac{1}{729} + \ldots\right) + \left(\frac{1}{9} + \frac{1}{81} + \frac{1}{729} + \ldots\right) \\ & = \sum_{i=1}^{\infty}\left(\frac{1}{3}\right)^i + \sum_{i=1}^{\infty}\left(\frac{1}{9}\right)^i \\ & = \frac{\frac{1}{3}}{1 - \frac{1}{3}} + \frac{\frac{1}{9}}{1 - \frac{1}{9}} \\ & = \frac{\frac{1}{3}}{\frac{2}{3}} + \frac{\frac{1}{9}}{\frac{8}{9}} \\ & = \frac{1}{2} + \frac{1}{8} \\ & = \frac{5}{8} \end{aligned}\tag{1}\label{eq1A}

Note I was able to split the sum into $$2$$ parts in the second line due to the series being absolutely convergent, with details about this in the Rearrangements and unconditional convergence section. Also note I used, such as described in Geometric series, that for $$|r| \lt 1$$, you have

$$\sum_{i=0}^{\infty}ar^i = \frac{a}{1 - r} \tag{2}\label{eq2A}$$

$$9\left(\frac{1}{3} + \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \frac{1}{243} + \frac{2}{729}+\cdots\right)=3+2+\frac{1}{3} + \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \cdots$$

so that

$$9S=5+S.$$

Denote by $$S$$ the value of the infinite sum:

$$S=\frac13+\frac29+\frac1{27}+\frac2{81}+\frac1{243}+\frac2{729}+\cdots$$

Some rearranging of terms lets us write

$$S=\frac23-\frac19+\frac2{27}-\frac1{81}+\frac2{243}-\frac1{729}+\cdots$$

That is, $$\frac29=\frac39-\frac19=\frac13-\frac19$$, and so on.

\begin{align*} 2S&=1+\frac29+\frac2{81}+\frac2{729}+\cdots\\[1ex] S&=\frac12+\sum_{n\ge1}\frac1{9^n}\\[1ex] &=\frac58 \end{align*}

• what a beasttt and different way of doing it, any benefits in the long ruun? or is splitting better Mar 17 '20 at 19:49
• My first instinct would have been the same as yours/John's. I just happened to notice that parts of this sum could be condensed using a Gauss-like argument Mar 17 '20 at 19:52
• Also $= \frac{5}{9} + \frac{5}{81} + \frac{5}{729} ...$ Mar 17 '20 at 19:54

Here is a trick that I like to use:

Recall that if you wanted to find the value of the infinite repeating decimal $$0.\overline{12}$$, you would get $$\frac{12}{99}$$. The $$99$$ in the denominator comes from the fact that we use a base $$10$$ number system, and $$10^2-1=99$$.

Create the decimal in base $$3$$, and you get $$\frac{12_3}{3^2-1}=\frac{5}{8}$$.

• wow thats amazing never THOT of t liek that Mar 20 '20 at 0:51

Note that when you group adjacent terms, $$\frac{1}{3} + \frac{2}{9} = \frac{5}{9}$$. Therefore, the sum of the series becomes:

$$\frac{5}{9} + \frac{1}{9} \cdot \frac{5}{9} + \left(\frac{1}{9} \right)^2 \cdot \frac{5}{9} + \cdots$$ $$= \frac{\frac{5}{9}}{1 - \frac{1}{9}}$$ $$= \frac{5/9}{8/9} = \frac{5}{8}$$