For $r<1$ define $F(r)=\sum_{n\in\mathbb N}(-1)^nr^{2^n}$. Does $F$ have a limit as $r\nearrow 1$?
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Did you try to compute the series for fixed $r<1$? – Rasmus Nov 22 '11 at 08:19
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In order for the community to better assist you, it is helpful if you provide what you have tried so far and indicate precisely where you are having difficulties. – Austin Mohr Nov 22 '11 at 08:19
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I think it is not computable. Am I wrong? – bx1 Nov 22 '11 at 08:20
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I simply don't know how to approach the problem. – bx1 Nov 22 '11 at 08:21
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It might be useful to look at Lacunary Functions – robjohn Nov 22 '11 at 09:11
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Nice, but in my question the term $(-1)^n$ plays an important role. – bx1 Nov 22 '11 at 09:18
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There is a general theorem, called High-Indice Theorem, that gives an answer to this problem, but you may solve it without aid of this theorem (which is of course too strong to refer to). – Sangchul Lee Nov 22 '11 at 09:24
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Google did not give me a universal link, however some pages show that this may be what I need. Do I understand correctly that the answer is affirmative? – bx1 Nov 22 '11 at 09:29
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Although some of the examples on that page have coefficients of $1$, that is not necessary. Note the examples under An elementary result, Lacunary trigonometric series, and on the referenced page about the Ostrowski-Hadamard gap theorem have more general coefficients. So your series fits right in. – robjohn Nov 22 '11 at 12:16
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However, you are only looking at real $r$ and not complex $z$, so there may be a limit for this particular point on the unit circle. – robjohn Nov 22 '11 at 12:36
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2Actually, the answer is 'no'. The High-Indice Theorem tells us that whenever $0 \leq n_0 < n_1 < n_2 < \cdots$ satisfies $n_{j+1}/n_{j} \geq \rho > 1$ for all $j$ for some constant $\rho$, then $\lim_{r\uparrow 1} \sum_{j=0}^{\infty} a_j r^{n_j} = A$ if and only if $\sum_{j=0}^{\infty} a_j = A$. In particular, since $\sum_{n=0}^{\infty} (-1)^n$ does not converge, neither is $\lim_{r\uparrow 1} F(r)$. – Sangchul Lee Nov 22 '11 at 13:18
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While investigating this problem, I got a deeper understanding of the issues than I did when I had complex analysis 30 years ago (+1). – robjohn Nov 23 '11 at 15:31
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It is possible to avoid numerical approach if we use a Tauberian Theorem. – Sungjin Kim Aug 02 '18 at 04:27
2 Answers
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Note that $$ F(r)=r-F(r^2)\tag{1} $$ Thus, if $a=\lim\limits_{r\to1^-}F(r)$ exists, then $$ a=\lim_{r\to1^-}F(r)=\lim_{r\to1^-}r-\lim_{r\to1^-}F(r^2)=1-a\tag{2} $$ Therefore, if the limit exists then it is $a=\frac{1}{2}$.
Applying equation $(1)$ twice, we get $$ F(r)=r-r^2+F(r^4)\tag{3} $$ As $r\to1$, $(3)$ indicates $F$ tends toward being periodic in $-\log(-\log(r))$ with period $\log(4)$. Note that as $r\to1^-$, $-\log(-\log(r))\to\infty$. $F(r)$ is the sum of the lengths of the intervals in the following animation
The value of the sum oscillates between $0.49728$ and $0.50272$ over each period. Therefore, $\lim\limits_{r\to1^-}F(r)$ does not exist.
robjohn
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@Didier: Let $t=-\log(\log(r))$ and define $G(t) = F(\exp(-\exp(-t)))$; i.e. $G(t)=F(r)$. Then, equation $(3)$ becomes $$G(t)=\left(e^{-e^{-t}}-e^{-e^{-t+\log(2)}}\right)+G(t-\log(4))$$ and as $t\to\infty$, $e^{-e^{-t}}-e^{-e^{-t+\log(2)}}\to0$. In other words, $$\lim_{t\to\infty}G(t)-G(t-\log(4))=0$$ That is the sort of almost periodicity I was meaning. – robjohn Nov 24 '11 at 17:19
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In this sense, every function with a finite limit at infinity tends toward being periodic... – Did Nov 24 '11 at 21:24
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Yes, every function with a finite limit at infinity would tend toward being a constant function (and constant functions are indeed periodic). However, my statement is attempting to describe how the limit fails to exist. $F$ oscillates between $0.49728$ and $0.50272$ infinitely often as $r\to1^-$. – robjohn Nov 25 '11 at 09:46
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(+1), This is a really interesting behavior. Even if I understand the calculation in the comment under your post, I don't understand how you can directly see from (3) that $F$ tends toward being periodic in $-\log(-\log(r))$. I mean when I am looking at the functional equation : $F(r) = r-r^2+F(r^4)$ it doesn't strike me that $F$ tends toward being periodic in $-\log(-\log(r))$. So is there a way to directly see this (as your post suggest) ? Thank you ! – Thinking Feb 02 '19 at 17:40
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1Suppose that $F(r)=F!\left(r^4\right)$. $$ r_2=r_1^4\iff-\log(r_2)=-4\log(r_1)\iff\log(-\log(r_2))=\log(4)+\log(-\log(r_1)) $$ so $F$ has period $\log(4)$ in $\log(-\log(r))$. That is, every time $\log(-\log(r))$ increases/decreases by $\log(4)$, $F(r)$ repeats. – robjohn Feb 03 '19 at 02:17
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My question was connected with this one. Namely, consider the sequence $(1, -1, -1, 1, 1, 1, 1, -1, \dots)$, where $(-1)^{k}$ stands for indices from $2^{k-1}$ to $2^k-1$. The Cesaro means can easily be calculated and they don't have a limit. The function $F$ here corresponds to the Abel means, and the equivalence of these summation methods for the bounded sequence implies that the Abel means diverge, too.
