# Ratio test Rudin Example 3.35B

The series in question is:$$\frac{1}{2}+1+\frac{1}{8}+\frac{1}{4}+\frac{1}{32}+\frac{1}{16}+\frac{1}{128}+\frac{1}{64}...$$

where

$$\liminf\limits_{n\rightarrow \infty} \frac{a_{n+1}}{a_n}=\frac{1}{8}$$ $$\limsup\limits_{n\rightarrow \infty} \frac{a_{n+1}}{a_n}=2$$ but $$\lim \sqrt[n]a_n=\frac{1}{2}$$

I think the series is as follows: $a_{2n-1}=\frac{1}{2^n}$ and $a_{2n}=\frac{1}{2^{n-2}}$.

My idea was to say that suppose $k=2n-1$, an odd number. Then $k+1=2n$ and $$\frac{\frac{1}{2^{n-2}}}{\frac{1}{2^n}}=\frac{1}{4}$$

I'm not sure what I'm messing up in trying to fill in the details for this example.

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This is a basic shuffle of the geometric series $$\sum_{k=0}^\infty\left(\frac12\right)^k.$$ As an absolutely convergent series (since it converges and all terms are nonnegative), it is unconditionally convergent, so shuffling the values doesn't affect the sum in any way. In particular, the series sums to $2$.
Let's put $a_0=\frac12$, $a_1=1$, $a_2=\frac18$, $a_3=\frac14$, and so on. More generally speaking, $$a_n=\begin{cases}2^{-(n+1)} & n\text{ is even}\\2^{-(n-1)} & n\text{ is odd}.\end{cases}$$ Then whenever $n$ is even, we have $$\frac{a_{n+1}}{a_n}=2,$$ and whenever $n$ is odd, we have $$\frac{a_{n+1}}{a_n}=\frac18.$$