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.
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what is $a_n$? try to make the problem self contained. – GA316 Nov 15 '13 at 08:54
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ah, sorry. The $ a_n $ is not given but here is another page that has the full example 3.35B typed out http://math.stackexchange.com/questions/221234/ratio-test-rudin-example-3-35b – user108986 Nov 16 '13 at 21:58
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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.
Potato
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Thank you for the explanation. I wonder if you can help me understand a bit better. Why is the limit of the second term 1/2? My calculations say it goes to 0 as k goes to infinity. Also, why does the limit of the second term being 1/2 imply that the nth root of it converges to 1/2? I realize I am asking very rudimentary questions, but your time will be really appreciated! – user108986 Nov 15 '13 at 09:05
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Thanks, yes it does help! However, I think it should be $ \frac{2^{1/k}}{2} $ instead of $\frac{1}{2} \cdot \frac{1}{2^{1/k}} $ as you have written since the second term is $ a_{2k} = \frac{1}{2^{2k} \cdot 2^{-2}} $. Let me know if that's wrong.Either way, you're right it does go to $\frac{1}{2} $ . Thanks a lot! – user108986 Nov 15 '13 at 15:25
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