5

So, the series's sum is of below form:

$2^{2^1}+2^{2^2}+...+2^{2^n}$

This series is an intermediate work of an bigger problem {So I am concerned with whether $2^{2^n}-\frac{\sum_{i=1}^n(2^{2^i})}{2}$ is anywhere close to $2^{2^{n-1}}$}. With limited knowledge of series, I am stuck.

Thank you.

modeller
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    Might I ask the statement of the bigger problem ? :-) – sciona Dec 18 '14 at 04:46
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    I don't think that there is any closed form for the summation. What you could notice is that the last term is extremely predominent. For $n=5$, the last term is $4294967296$ while the summation is $4295033108$; for $n=6$, the last term is $18446744073709551616$ while the summation is $18446744078004584724$. So, if $n$ is large, the summation is the last term with a very marginal error. – Claude Leibovici Dec 18 '14 at 04:49
  • @sciona So I am concerned with whether $2^{2^n}-\frac{\sum_{i=1}^n(2^{2^i})}{2}$ is anywhere close to $2^{2^{n-1}}$ – modeller Dec 18 '14 at 04:53
  • Is this related to Fermat number ? – Lucian Dec 18 '14 at 05:11
  • @ClaudeLeibovici: I think that is a good answer. It is clearly applicable to the last specific question, saying the expression is quite close to $2^{2^n}$, which is correct. The fundamental idea is that power towers only depend on the top exponent. – Ross Millikan Dec 18 '14 at 05:15
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    Pushing my calculation for $n=10$, the ratio between the summation and the last term is $1+\epsilon$ where $\epsilon \approx 7.46 \times 10^{-155}$ – Claude Leibovici Dec 18 '14 at 05:23
  • @Lucian No, it's not related to modular arithmetic – modeller Dec 18 '14 at 05:25
  • It is not close to $2^{2^{n-1}}$ but rather $2^{2^{n}-1}$ – Winther Dec 18 '14 at 05:31
  • @Winther: and even closer to $2^{2^n}$ – Ross Millikan Dec 18 '14 at 05:49
  • @RossMillikan No. My comment was related to the sum $2^{2^n} - \frac{\sum_{i=0}^n 2^{2^i}}{2}$ OP asks: "Is this close to $2^{2^{n-1}}$" – Winther Dec 18 '14 at 05:50
  • @Winther: you are right because the largest term in the sum is half $2^{2^n}$, so yours is much closer than mine. – Ross Millikan Dec 18 '14 at 05:56
  • Your sum is A060803(n)-2, related to the sequence https://oeis.org/A060803 . This doesn't help much but at least shows that you aren't missing any obvious closed form. See also https://math.stackexchange.com/q/2396259/24908 . – b_jonas Mar 22 '23 at 11:52

1 Answers1

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The simple thing is to look at the ratio of the two largest terms. $$\frac {2^{2^n}}{2^{2^{(n-1)}}}=2^{2^n-2^{(n-1)}}=2^{2^{(n-1)}}$$ which grows rapidly with $n$. The smaller terms are really negligible and the sum is very close to $2^{2^n}$. Note that you should be adding to, not subtracting from, the largest term. My argument says you don't have to bother.

Ross Millikan
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