Find the sum of the series with unordered powers of $3$ 
Consider the following series: $$\sum_{n=1}^{\infty} a_n =  1/3+1+1/3^3+1/3^2+1/3^5+1/3^4+1/3^7+1/3^6 +\dots$$
Determine  if it converges, and find the sum.

Here is what I got:
a) $\lim_{n\to\infty} \frac{a_{n+1}}{a_n}  =  (\frac{1}{3})^3 =1/27$
b) $\lim_{n\to\infty} \sqrt[n]{a_n}  = 1/3$
Is it right?
 A: HINT:
The sum $$=\sum_{r=0}^\infty\left(\frac13\right)^{2r+1}+\sum_{r=0}^\infty\left(\frac13\right)^{2r}$$
Check the proof of the convergence of infinite geometric progression
A: Since $a_n=\dfrac{1}{3^n}$ when n is odd and $a_n=\dfrac{1}{3^{n-2}}$ when n is even,
A)$\;\;$ $\dfrac{a_{n+1}}{a_n}=\begin{cases}\frac{1}{27},&\mbox{if n is odd}\\\; 3,&\mbox{if n is even}\end{cases}.$  Therefore $\displaystyle\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n}$ does not exist.
B) $\;\;$$\displaystyle\left(a_n\right)^{\frac{1}{n}}=\begin{cases}\frac{1}{3},&\mbox{if n is odd}\\\left(\frac{1}{3}\right)^{1-\frac{2}{n}},&\mbox{if n is even}\end{cases}.$ $\;\;\;$ Therefore $\displaystyle\lim_{n\to\infty}\left(a_n\right)^{\frac{1}{n}}=\frac{1}{3}$.

As pointed out by Ross Millikan, this is just a rearrangement of the absolutely convergent series 
$\;\;\;\displaystyle\sum_{n=0}^{\infty}\frac{1}{3^n}=\sum_{n=0}^{\infty}\left(\frac{1}{3}\right)^n$, so it converges and has the same sum.
A: The limit $\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}$ certainly does not exist, since even terms are $3$ times the preceding one any odd terms are $\frac{1}{27}$ the preceding even one. So, that ratio alternates between those two values. However, you could compute
$$\lim_{n\rightarrow \infty}\frac{a_{n+2}}{a_{n}}=\frac{1}9$$
and that this is less than $1$ suffices to show convergence - it still implies that $|a_n|$ is bounded above by a geometric series. Of course, the calculation you did for the root test is correct in itself and already suffices to show convergence.
I would proceed algebraically to find the sum, since you know it exists. In particular, let $$L=\frac{1}3+1+\frac{1}{3^3}+\frac{1}{3^2}+\frac{1}{3^5}+\frac{1}{3^4}+\ldots$$
which is the sum of the sequence. Notice that if we multiply through by $9$ we get:
$$9L=3+9+\frac{1}{3}+1+\frac{1}{3^3}+\frac{1}{3^2}+\ldots$$
but notice that, excepting the first two terms, we have the sum of $L$ on the left side - so
$$9L=12+L$$
which can be solved to give the only possible value for $L$ and, knowing that $L$ exists, what must be the limit for it.
