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


$$\liminf\limits_{n\rightarrow \infty} \frac{a_{n+1}}{a_n}=\frac{1}{8}$$ $$\limsup\limits_{n\rightarrow \infty} \frac{a_{n+1}}{a_n}=2$$ 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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up vote 2 down vote accepted

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 $$\frac{a_{n+1}}{a_n}=\frac18.$$

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I understand that much, but I'm not sure how Rudin got his numbers. – emka Oct 26 '12 at 1:21
My apologies. I didn't read the title of your post. (Incidentally, many on this site don't read titles, but only the posts, so it's a good idea to but the question in the body of the post, even if you've also put it in the title.) – Cameron Buie Oct 26 '12 at 1:27
No worries. This is more of a problem that I apparently understood a few years ago, as I made notes in the margin, but I now cannot solve. I understand the last number of 1/2, but now the previous numbers of 1/8 or 2. – emka Oct 26 '12 at 1:29
Hopefully, the updated answer will fix your issues. – Cameron Buie Oct 26 '12 at 1:46
Yes, thank you. I guess I tried to follow how I worked out example 3.35a (I'm not sure if you have the book handy, but someone posted a similar work out here:…) – emka Oct 26 '12 at 1:55

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