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The idea is a more convenient form for $N = 0.01001000100001000001...$ in base $r$, hopefully to show whether it is transcendental.

Sorry for brevity.

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    $\begingroup$ We can factor out an $r^c$ term from the sum. For special values of $a$ and $b$, the sum is expressible in terms of Jacobi theta functions, but I know of no closed form for the general form you have. $\endgroup$ Jul 31, 2011 at 1:29
  • $\begingroup$ let's say $r=2^{-1/2}$, $a=1$, $b=5$, $c=4$, which I think makes the series equal to $N$ in binary. Also, thanks for fixing my TeX. $\endgroup$
    – jade
    Jul 31, 2011 at 1:47
  • $\begingroup$ The $b=5$ portion would be troublesome in figuring a closed form for what you say you have... $\endgroup$ Jul 31, 2011 at 2:47
  • $\begingroup$ @J.M. I tried to complete the squares hoping to kill the linear term, but it didn't work because $5$ "happened" to be an odd number. Did you anticipate this in your comment? Also, if I gave you just $an^2+c$, then do closed-form solutions exist? $\endgroup$
    – Srivatsan
    Jul 31, 2011 at 3:01
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    $\begingroup$ Yeah, something like that @Srivatsan. ;) For your much simpler case, you have a representation in terms of the third Jacobi theta function: $\frac{r^c}{2}(1+\vartheta_3(0,r^a))$ $\endgroup$ Jul 31, 2011 at 3:04

3 Answers 3

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Of course $N$ is transcendental. [I said it is, I didn't say I can prove it.] It is conjectured that all irrational algebraic numbers are normal in all bases. If this were not transcendental, it would be a spectacular counterexample to that conjecture.

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This may come under the heading of "Siegel E-functions" or "Siegel G-functions", for which transcendence results are known.

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Here's a technique for this sort of sum (though this doesn't address the transcendentality question) I found a while ago which involves incomplete theta functions:

\begin{equation} \sum_{n=0}^{\infty} q^{an^{2}+2akn+ak^{2}+p} = q^{p}\left[\frac{\theta_{3}(0,q^{a})+1}{2} - \sum_{m=0}^{k-1} q^{am^{2}}\right] \end{equation}

for example:

\begin{equation} \sum_{n=0}^{\infty} \frac{1}{e^{7n^{2}+70n+173}} = e^{2}\left[\frac{\theta_{3}(0,1/e^{7})+1}{2} - \sum_{m=0}^{4} (1/e^{7})^{m^2}\right] \end{equation}

I'm not sure it works for arbitrary $a$, $b$, and $c$, though.

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