For $r<1$ define $F(r)=\sum_{n\in\mathbb N}(-1)^nr^{2^n}$. Does $F$ have a limit as $r\nearrow 1$?
Does the family of series have a limit?
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0Did you try to compute the series for fixed $r<1$? – 2011-11-22
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0In 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. – 2011-11-22
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0I think it is not computable. Am I wrong? – 2011-11-22
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0I simply don't know how to approach the problem. – 2011-11-22
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0It might be useful to look at [Lacunary Functions](http://en.wikipedia.org/wiki/Lacunary_function) – 2011-11-22
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0Nice, but in my question the term $(-1)^n$ plays an important role. – 2011-11-22
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0There 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). – 2011-11-22
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0Google 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? – 2011-11-22
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0Although some of the examples on [that page](http://en.wikipedia.org/wiki/Lacunary_function) have coefficients of $1$, that is not necessary. Note the examples under [An elementary result](http://en.wikipedia.org/wiki/Lacunary_function#An_elementary_result), [Lacunary trigonometric series](http://en.wikipedia.org/wiki/Lacunary_function#Lacunary_trigonometric_series), and on the referenced page about the [Ostrowski-Hadamard gap theorem](http://en.wikipedia.org/wiki/Ostrowski-Hadamard_gap_theorem) have more general coefficients. So your series fits right in. – 2011-11-22
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0However, 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. – 2011-11-22
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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 $\lim_{r\uparrow 1} F(r)$. – 2011-11-22
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0While investigating this problem, I got a deeper understanding of the issues than I did when I had complex analysis 30 years ago (+1). – 2011-11-23
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0It is possible to avoid numerical approach if we use a Tauberian Theorem. – 2018-08-02
2 Answers
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.
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0What does it mean for a function to *tend toward being periodic*? – 2011-11-24
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0@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](http://en.wikipedia.org/wiki/Almost_periodic_function) I was meaning. – 2011-11-24
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0In this sense, every function with a finite limit at infinity *tends toward being periodic*... – 2011-11-24
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0Yes, 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 $r\to1^-$. – 2011-11-25
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.