Rudin Example 3.35B Why the $n$th root of $a_n$ is $1/2$? For Baby Rudin Example 3.35(b), I understand how the $\liminf$ and $\limsup$ of the ratio test were found, but I am not clear why $\ \lim \sqrt[n]{a_n } = \frac{1}{2} $.
Please help.

The sequence in question is 
$$\frac{1}{2} + 1 + \frac{1}{8} + \frac{1}{4}+ \frac{1}{32}+ \frac{1}{16}+\frac{1}{128}+\frac{1}{64}+\cdots$$
In case the pattern is not clear, we double the first term, the divide the next by $8$, the double, then divide by $8$, and so on.
The general formula for an odd term is $a_{2k-1}=\frac{1}{2^{2k-1}}$. The formula for an even term is $a_{2k}=\frac{1}{2^{2k-2}}$. In the first case, $\sqrt[n]{a_n}=\frac{1}{2}$. In the second, the limit is $\frac{1}{2}$. Since $n$th roots of both the even and odd terms converge to $\frac{1}{2}$, you have your desired result.
