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I have read the following (http://www.math.harvard.edu/~elkies/Misc/sol8.html) but I dont understand the last part of the solution:

For positive $x<1$, consider the alternating sum $$S(x) = x − x^2 + x^4 − x^8 + x^{16} − x^{32} + \cdots$$ Does $S(x)$ approach a limit as $x$ approaches $1$ from below, and if so what is this limit?

Proof Since $S$ satisfies the functional equation

$S(x) = x − S(x^2)$, it is clear that if $S(x)$ has a limit as $x$ approaches $1$ then that limit must be $1/2$. But then $S(0.995) = 0.50088\cdots > 1/2$. Iterating the functional equation, we find $S(x) = x − x^2 + S(x^4) > S(x^4)$. Therefore the fourth root, 16th root, 64th root, … of 0.995 are all values of x for which $S(x) > S(0.995) > 1/2$. Since these roots approach $1$, we conclude that in fact $S(x)$ cannot tend to $1/2$ as $x$ approaches $1$, and thus has no limit at all!

My question: why the last sentence is true? (Since these roots approach $1$, we conclude that in fact $S(x)$ cannot tend to $1/2$ as $x$ approaches $1$) and is there any way to show that $S(0.995)>1/2$ without the use of computer? Thank you so much in advance.

Mark Eduardo
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2 Answers2

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We can represent $S(x)$ as $$S(x)=x-x^2+S(x^4)$$ A value for which $S(x)$ is greater than $1/2$ can be found, which is less than 1. For e.g. a simple value of $0.8$$$S(0.8)=0.8-0.8^2+0.8^4\cdots>0.5$$ If we have found a value for which $S(x)$ is greater than $0.5$, we can find others too by taking $2^n$th root of this. Taking roots makes the number as closer to $1$ as we want, thus the limit to $S(x)$ via this path has a value greater than $1/2$ and via another representation gives less than $1/2$. So there's no definite value of limit and the function oscillates infinitely near $1/2$.

RE60K
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The roots as chosen are all less than $1,$ but to tend to $1.$ Hence for any chosen $\epsilon >0,$ for sufficiently large $n,$ we have $1- r_{n} < \epsilon.$ But if $S(x) \to \frac{1}{2}$ as $x \to 1$ (from below) we should be able to to make $|S(x)- \frac{1}{2}|$ as small as we wish by choosing $x$ close enough to $1.$ Hence for large enough $n,$ we should have $|S(r_{n}) - \frac{1}{2}| < 0.00088,$ which we know we can't do.

Geoff Robinson
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