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I've been thinking about this sum for a little while. I came up with it based on the observation that the product: $$\prod_{n=0}^{\infty}x^{(2^{-n})}$$ converges to $x^2$, and if we take the natural log of this product then we get: $$\sum_{n=0}^{\infty}\ln(x^{2^{-n}})$$ which converges to $2\ln(x)$. The sum of interest comes by using the linear approximation $ln (x) \approx x-1$. Intuition says that since the linear approximation becomes very good as $x \rightarrow 1$, then the sum: $$f(x) = \sum_{n=0}^{\infty}(x^{(2^{-n})}-1)$$ should also converge. This can be confirmed rigorously using the ratio test. This observation lead me to ask whether there was any way to give an elementary evaluation of this sum for any value other than one, where all there terms are clearly zero. Using Taylor expansion at one, I showed that over the open interval from 0 to 2, this is equal to the sum: $$\sum_{n=0}^{\infty}(x-1)^n \sum_{j=0}^{\infty} \prod_{i=0}^{n-1}(2^{-j}-i)$$ which though more complicated, has the advantage that the internal product and sum can be evaluated for a given value of $n$ by expanding the product, then breaking the sum into several geometric series and using the summation formula for geometric series to evaluate the individual geometric series. This altogether means that it is possible with only finitely much work to find rational estimates for $f(x)$ when $x$ is rational, which might lead to useful conjectures.

Robo300
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    Do you mean $\sum_n (x^{2^{-n}}-1)$ ? – daw Nov 09 '19 at 17:50
  • nice and easy: $$f(x)=\prod_{n\ge0}x^{1/2^n}.$$ Note that $$f(x)=x^1\prod_{n\ge1}x^{1/2^n}=x\left(\prod_{n\ge1}x^{1/2^{n-1}}\right)^{1/2}=x\sqrt{f(x)}.$$ – clathratus Nov 09 '19 at 18:19
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    It seems like $f(x)=\sum_{k=1}^\infty \frac{(\log x)^k}{k!}\frac{1}{1-2^{-k}}.$ – Kenta S Nov 09 '19 at 18:21
  • @clathratus you're correct about that function, but that function was not the one I was curious about, it was the inspiration. If you notice part way through the question I comment that the convergence of a sum derived from that product suggests the convergence of another sum which is related to it by using a linear approximation for ln(x). The function I'm looking to evaluate is not equal to that product. – Robo300 Nov 09 '19 at 18:54

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