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Does, there exist some closed-form solution of the following finite-series ?

$ S_n = 2^n p^2 + 2^{n-1} p^4 + 2^{n-2} p^8\cdots + 2p^{2^n}, $ where $n$ is a Positive Integer and $0<p<1$ .

Note that number of terms in a series is $n$. So length of series varies according to value of $n$.

TZakrevskiy
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kaka
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    Technically, it is not a series, it is a partial sum of a series. – Samrat Mukhopadhyay Jul 29 '13 at 14:36
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    Above series is not part of a sequence, so it is not a partial sum of series. As, I mentioned its length is exactly equal to n. – kaka Jul 29 '13 at 14:43
  • As this is a finite sum then the value of $p$ would not be a significant issue. – Maesumi Jul 29 '13 at 14:43
  • the starting point should be the recurrence $S_{n+1}=2S_n+2p^{2^{n+1}}$ – Andrea Mori Jul 29 '13 at 14:46
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    @kaka: "Above series is not part of a sequence, so it is not a partial sum of series"... But each $S_n/2^{n-1}$ is a partial sum of the series $s(x,p)=\sum\limits_{k\geqslant1}x^kp^{2^k}$ for $x=1/2$... for which no closed formula I would be aware of is available (and in general getting closed forms for partial sums is even more complicated). – Did Jul 29 '13 at 14:49
  • The recurrence is $a_0=p^2, a_{n+1}=a_n^2$, and $S_0=0, S_{n+1}=2(S_n+a_n)$. I also think no closed form is known. – Maesumi Jul 29 '13 at 14:49

2 Answers2

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The infinite series $\sum_{i=0}^\infty 1/k^{2^i}$ is transcendental for integers $k > 1$. Thus, if there is a simple (defined by simple operations) general closed-form formula for your partial sums, then it almost certainly must involve transcendental numbers like $\pi$ or $e$, because when you divide your sums by $2^n$ and take the limit of the "simple" formula you must be able to get transcendental answers even when $p$ is rational. But then how would such a sum formula involving transcendental numbers always give a rational or integer answer for partial sums for example when $p = 1/2$? Thus I take this as strong evidence there is no simple closed-form partial sum formula, either for your case or the simpler specialized case of $\sum_{i=0}^n 1/k^{2^i}$ for integer $k > 1$.

user2566092
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  • I didnt constrain for rational or integer answer particularly. Otherwise, I am agree with you. Would you please let me know how $\sum_{i=0}^\infty 1/k^{2^i}$ is transcendental? – kaka Aug 04 '13 at 01:58
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    This was an answer to a question posted here http://math.stackexchange.com/questions/459124/for-integer-k-1-is-sum-i-0-infty-1-k2i-transcendental-or-algebr – user2566092 Aug 08 '13 at 19:45
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Your question reminded me of an even simpler question I encountered about closed form formula for partial sums, posted at Is there a closed form solution for partial sums of $1/(2^{2^0}) + 1/(2^{2^1}) + 1/(2^{2^2}) + \ldots$ . It seems pretty clear to me that if no one knows a formula for that simpler case, then there is almost certainly no known formula for your case either because of the similarity with your special case when $p = 1/2$. On the other hand, if someone can point to a formula and proof for that simpler partial summation problem, then maybe the formula and proof idea can be adapted to give the partial sum formula you requested. I'll update if such a formula and proof for the simpler problem is pointed out, hopefully with the required adaptation to give a generalized formula for your problem. At the very least there would be a hope for a solution when $p = 1/2$.

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user2566092
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