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.
 A: Your sum can be put as
$$(1+\frac 23)\Bigl(\frac 13+\frac{1}{3^3}+\frac{1}{3^5}...\Bigr)=$$
$$\frac 53\frac 13\Bigl(1+\frac 19+\frac{1}{9^2}+...\Bigr)=$$
$$\frac 53\frac 13 \frac{1}{1-\frac 19}=\frac 58$$
A: 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 $\frac{1}{2} +\frac{1}{8}$.
A: The series can also be written as
$$
\begin{align}
\overbrace{\ \sum_{k=1}^\infty\frac1{3^k}\ }^{\substack{\text{each power}\\\text{of $3$}}}+\overbrace{\ \sum_{k=1}^\infty\frac1{9^k}\ }^{\substack{\text{even powers}\\\text{of $3$}}}
&=\frac{\frac13}{1-\frac13}+\frac{\frac19}{1-\frac19}\\
&=\frac12+\frac18\\[6pt]
&=\frac58
\end{align}
$$
A: One more option that apparently hasn't been mentioned yet: in your original question you mentioned being stuck on the term
$$\sum_{n=1}^{ \infty } \left[\frac{1}{3^{n} } \frac{(-1)^{n}}{2}\right]$$
This can be rewritten as
$$\frac{1}{2}\sum_{n=1}^{ \infty } \left(-\frac{1}{3}\right)^n$$
which can also be evaluated as a geometric series.
A: 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}
A: 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:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...=\sum_{n=1}^{\infty} \frac{1}{3^{2n-1}}+\sum_{n=1}^{ \infty }\frac{2}{3^{2n} }=3\sum_{n=1}^{\infty} \frac{1}{3^{2n}}+2\sum_{n=1}^{ \infty }\frac{1}{3^{2n} }=5\sum_{n=1}^{ \infty }\frac{1}{9^{n} }=\frac{5}{8}$$
A: HINT:
$\frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$
Separate odd and even terms.
$\frac{1}{3} +\frac{1}{3^{3}}  + \frac{1}{3^{5}} + ...$
$ \frac{2}{3^{2} } + \frac{2}{3^{4}} +\frac{2}{3^{6}}  ...$
