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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 $r<1$?2011-11-22
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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.2011-11-22
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    I think it is not computable. Am I wrong?2011-11-22
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    I simply don't know how to approach the problem.2011-11-22
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    It might be useful to look at [Lacunary Functions](http://en.wikipedia.org/wiki/Lacunary_function)2011-11-22
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    Nice, but in my question the term $(-1)^n$ plays an important role.2011-11-22
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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).2011-11-22
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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?2011-11-22
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    Although 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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    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.2011-11-22
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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 $\lim_{r\uparrow 1} F(r)$.2011-11-22
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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).2011-11-23
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    It is possible to avoid numerical approach if we use a Tauberian Theorem.2018-08-02

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