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.

It is acceptable to split one convergent series in two as long as both are convergent.
$$ 0 < \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ... < \frac{2}{3} + \frac{2}{3^{2} } + \frac{2}{3^{3}} + \frac{2}{3^{4}} + \frac{2}{3^{5}} + ...$$
and since
$$\frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$ is monotonically increasing as you are adding terms, and it is bounded by another convergent series (above), it is convergent itself.
Now:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...=\sum_{n=1}^{\infty} \frac{1}{3^{2n-1}}+\sum_{n=1}^{ \infty }\frac{2}{3^{2n} }=3\sum_{n=1}^{\infty} \frac{1}{3^{2n}}+2\sum_{n=1}^{ \infty }\frac{1}{3^{2n} }=5\sum_{n=1}^{ \infty }\frac{1}{9^{n} }=\frac{5}{8}$$
