<s>[INST] 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?
 [/INST] 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.
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