<s>[INST] Applying root test to sequence $\frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots$ The following is an example from Principles of Mathematics, by Rudin. I've been trying to understand the example but haven't quite grasped it because it seems I can solve it differently.
Given the following sequence: $\displaystyle \frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \cdots$
Using the Ratio Test:
$$\lim \inf_{n \to \infty} \frac{a_{n+1}}{a_{n}} = \lim_{n \to \infty} 
(\frac{2}{3})^n = 0$$
$$\lim \sup_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} 
\frac{1}{2}(\frac{3}{2})^n = +\infty$$
Using the Root Test:
$$\lim \inf_{n \to \infty} (a_n)^{\frac{1}{n}} = \lim_{n \to \infty} (\frac{1}{3^n})^{\frac{1}{2n}} = \frac{1}{\sqrt{3}}$$
$$\lim \sup_{n \to \infty} (a_{n})^{\frac{1}{n}} = \lim_{n \to \infty} (\frac{1}{2^n})^{\frac{1}{2n}} = \frac{1}{\sqrt{2}}$$
What I don't understand is how to find the $\lim \sup$ and $\lim \inf$ for the ratio test. I also don't understand why for the root test, we are looking at the $2n^\text{th}$ root. Where does this 2 come from? Furthermore, are we looking at $a_n$ as alternating between $\frac{1}{2^m}$ and $\frac{1}{3^m}$ or is $a_{n}$ actually $\frac{1}{2^m} + \frac{1}{3^m}$?
As a side note, I do know how to solve this question if asked whether or not this series converges. I simply don't understand the book went around solving it.
 [/INST] As Abstraction says, the book uses $n$ for two different things.
$a_n$ alternates between $\frac1{2^m}$ and $\frac1{3^m}$.  So the first, third, fifth... terms are powers of $1/2$, and the second,fourth,sixth... are powers of $1/3$.
For the root test, for example, $\frac1{3^6}$ is the twelfth term, so you look at $\left(\frac1{3^6}\right)^{1/12}=\frac1{\sqrt{3}}$.  For the eleventh term, you should look at $\left(\frac1{2^6}\right)^{1/11}$, but the book looks at $\left(\frac1{2^6}\right)^{1/12}=\frac1{\sqrt2}$.  The true root is $\left(\frac1{\sqrt2}\right)^{12/11}=\left(\frac1{\sqrt2}\right)^{1+\frac1{
11}}$.  This root approaches $\frac1{\sqrt2}$ as you go further in the sequence.
For the ratio test, $\frac{a_{2n}}{a_{2n-1}}$ approaches zero, and $\frac{a_{2n+1}}{a_{2n}}$ grows without limit.  
</s>