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Not sure if this is research level? While analyzing a particular algorithm, I came across the following series: $\sum_{k=1}^{m-1}\frac{3^{k}}{\prod_{i=1}^{k-1}2^{2^{i}}}$

Is there a closed form expression that illuminates asymptotic values in terms of $m$ accurately? In particular, I am interested in how slow this function grows compared to $\frac{3^{m}-1}{2}$ (which is the summation if one sets the denominator in each of the sum term to $1$).

Possibly is there a function that approximates $f(m)$ where $f(m)$ is such that $S_{m} =(\frac{3^{m}-1}{2})^{f(m)}$?

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For large enough k, the denominator can be rewritten more simply. The result converges quickly, but otherwise gives me no clue as to a closed form. Gerhard "Ask Me About System Design" Paseman, 2012.06.14 – Gerhard Paseman Jun 14 '12 at 22:34

Presumably the sum $S_m$ can be given in a more compact way as $S_m = 4\sum_{k=1}^{m-1} \frac{3^k} {2^{2^k}}.$ The limit should be a transcendental number, since the sum is extremely lacunary, so it seems unlikely that a closed for exists (since most closed form transcendental numbers are "named" constants").

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$\sum_{n=0}^{\infty}(1/n!)$ is a lacunary sum with a transcendental value, yet it has what would generally be accepted as a closed form; I'm not objecting to your conclusion, just to the word, "so". – Gerry Myerson Jun 14 '12 at 22:56
Hi Igor, I am only interested in something that grows as fast as $S_{m}$ and not S_{m}$ accurately. – Turbo Jun 14 '12 at 23:51
The sequence $S_m$ converges to some number between $2.83$ and $2.84$, so it doesn't grow very much... – François G. Dorais Jun 15 '12 at 0:23
$S_{m} −S_{m−1} = \frac{3^{m}}{2^{2^{m}−1}}$ – Turbo Jun 15 '12 at 1:16
Basically Gerry's and Igor's answer says it converges to a constant. ok. – Turbo Jun 15 '12 at 1:50

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