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Define $a_n$ for $n\geq 2$ to be $1$ if the second most significant digit of $n$ in binary is $1$ and $-1$ otherwise. For example for $n=23$, in binary $n=10111_2$ and so $a_n=-1$, becuase the second digit of $n$ was $0$.

The limit in question is $\lim_{x\to 1} f(x)=\lim_{x\to 1}\sum_{n=2}^\infty a_nx^n$.

Intuition: The coefficients of this power series are only $-1$ and $+1$. There is (starting from $2$) $a_n=-1,1,-1,-1,1,1,-1,-1,-1,-1,1,1,1,1,\ldots$ followed by 8 times $-1$ and so on with the number of $\pm 1$ being increasing powers of $2$. The negative powers of $x$ therefore come earlier and should be more significant, resulting the limit in being $-\infty$.

Approaches: The limit and the sum sadly cannot be interchanged.

I tried summing the finite geometric series, which occur in the sum. This results in $$\sum_{n=1}^\infty -x^{2^n} \frac {1-x^{2^{n-1}+1}}{1-x} +x^{2^n+2^{n-1}}\frac {1-x^{2^{n-1}+1}}{1-x} $$ and does not look promising for me. Maybe this still works somehow?! (Might have mistakes, I wrote this from memory)

I found the following functional equation for $f$: $$(1+x)f(x^2)-x^2+x^3=f(x)$$ and plugging in $x=1$ yields $2f(1)=f(1)$, which leaves only $f(1)=\pm \infty,0$ . $+\infty$ is easily excluded, because $f$ is negative on $(0,1)$ and this also lacks an argument for the convergence after all.

Another approach with simply the definition of limits failed miserably. But again this might be my fault.

Any hints or maybe even solutions are appreciated.

M. Karl
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  • Maybe you can argue similarly to a lacunary series? Something like this: The radius of convergence is $1$, so there is a singularity somewhere on the unit circle; then the functional equation gives us singularities everywhere else on the unit circle? (I'm too rusty to know how to make that rigorous.) – Chris Culter Apr 02 '17 at 19:23
  • Two references to reveal that the limit and the sum can be interchanged : Commutativity of iterated limits , Iterated Limits Schizophrenia . – Han de Bruijn Apr 06 '17 at 19:47
  • @ChrisCulter I suspect, that this doesn't rule out the possibility of an essential singularity at 1, and then 0 might still be a possible limit. – M. Karl Apr 08 '17 at 00:21
  • @HandeBruijn No, if you just plug in $x=1$, the partial sums are between 0 and $2^n$, which is clearly not converging. This is indeed an example, where they can not be interchanged. Your references don't apply here. – M. Karl Apr 08 '17 at 00:24
  • On the contrary, it is standard mathematics that is full of "subtleties" that do not apply. Weird, because my arguments in the references are standard mathematics. But your conclusion is right: the partial sums are not converging. Hence there is no possible limit. – Han de Bruijn Apr 08 '17 at 15:58
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    Erm... If you, indeed, have your functional equation, what's the problem? Since $x^2\ge x^3$ on $(0,1)$, you get the inequality $f(x)\le(1+x)f(x^2)$, which you can iterate until $x^{2^k}\in(1/4,1/2]$, say. If $x$ is close to $1$, by that moment you'll accumulate quite a lot of factors close to $2$ and the last factor $f(x^{2^k})$ is a negative number separated from $0$. – fedja Apr 09 '17 at 03:51

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