<s>[INST] How to evaluate $ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$? I was given the series:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$
Making some observations I realized that the $ a_{n} $ term would be the following:
$$ a_{n} = \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right)$$
What I wanted to do is to find the result of the series, so the answer would be:
$$\sum_{n=1}^{ \infty } \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right) = \frac{3}{2} \sum_{n=1}^{ \infty } \left[ \frac{1}{3^{n} } \right]+ \sum_{n=1}^{ \infty } \left[\frac{1}{3^{n} } \frac{(-1)^{n}}{2}\right]$$
I can tell that the first term is convergent because it is a geometric series, in fact, the result is $\frac{3}{4}$. However, I have no clue in how to solve the second term series. I should say that the series given in the beginning is convert and its result is 5/8. How to arrive to it is a mystery to me.
 [/INST] In base $3$, we can say
$$N = \frac {1}{10} + \frac{2}{100} + \frac{1}{1000} + \frac{2}{10000} + \cdots
=0.121212\cdots$$
Staying in base 3, we get
\begin{array}{rcr}
   N &= &0.121212\cdots \\
100N &= &12.121212 \cdots \\
  -N &= &-0.121212\cdots \\
\hline
22N &= &12.000000\cdots \\
\end{array}
So $N = \dfrac{12}{22}$
In base $10$, this becomes $N = \dfrac 58$
Added 3/19/2021
You can also do this directly
\begin{align}
   N &= \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} 
                    + \frac{2}{3^{4}} + \frac{1}{3^{5}} + … \\
  9N &= 3 + 2 + \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} 
                    + \frac{2}{3^{4}} + \frac{1}{3^{5}} + … \\
  9N &= 5 + N \\
   N &= \frac 58
\end{align}
</s>