<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] There is a much simpler way to solve this.
You know how to compute the geometric series
$$\frac{1}{3} + \frac{1}{3^2}+ \frac{1}{3^3}+ \frac{1}{3^4}+ \cdots = \frac{1}{2}$$
you also know how to compute the geometric series
$$\frac{1}{9} + \frac{1}{9^2}+ \frac{1}{9^3}+ \frac{1}{9^4}+ \cdots= \frac{1}{8}$$
But this can be rewritten in terms of $3^2$, so that you get
$$\frac{1}{3^2} + \frac{1}{3^4}+ \frac{1}{3^6}+ \frac{1}{3^8}+ \cdots$$
Now the answer to your question is simply $\frac{1}{2} +\frac{1}{8}$.
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