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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 8:19
    
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 8:19
    
I think it is not computable. Am I wrong? –  bx1 Nov 22 '11 at 8:20
    
I simply don't know how to approach the problem. –  bx1 Nov 22 '11 at 8:21
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Actually, 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 $\lim_{r\uparrow 1} F(r)$. –  sos440 Nov 22 '11 at 13:18
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2 Answers 2

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

enter image description here

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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What does it mean for a function to tend toward being periodic? –  Did Nov 24 '11 at 15:48
    
@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
    
In this sense, every function with a finite limit at infinity tends toward being periodic... –  Did Nov 24 '11 at 21:24
    
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 9:46
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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.

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