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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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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. – J. M. Jul 31 '11 at 1:29
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. – jon Jul 31 '11 at 1:47
The $b=5$ portion would be troublesome in figuring a closed form for what you say you have... – J. M. Jul 31 '11 at 2:47
@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? – Srivatsan Jul 31 '11 at 3:01
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Yeah, something like that @Srivatsan. ;) For your much simpler case, you have a representation in terms of the third Jacobi theta function: $\frac{r^c}{2}(1+\vartheta_3(0,r^a))$ – J. M. Jul 31 '11 at 3:04
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2 Answers

up vote 4 down vote

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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