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Topic: ***Spoiler*** inequality (Read 880 times) 

anonymous
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\[ \begin{split} \sum_{n=1}^{\infty} (a_na_{n+1})a_{n+2} \\ = \sum_{n=1}^{\infty} (a_1^{2^{n1}}a_1^{2^n})a_1^{2^{n+1}} \\ \leq \sum_{i=1}^{\infty} (a_1^ia_1^{i+1})a_1^{2i+2} \\ = (1a_1)a_1^2\sum_{i=1}^{\infty} a_1^{3i} \\ = (1a_1)a_1^5/(1a_1^3) \\ = a_1^5/(1+a_1+a_1^2) \\ < 1/3 \end{split} \] The second to third line uses telescopic sum, and there is a geometric series in the fourth line.


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wowbagger
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Re: ***Spoiler*** inequality
« Reply #1 on: Jun 24^{th}, 2003, 5:17am » 
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For those who aren't used to reading LaTeX:

« Last Edit: Jun 24^{th}, 2003, 5:17am by wowbagger » 
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James Fingas
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Re: ***Spoiler*** inequality
« Reply #2 on: Jun 24^{th}, 2003, 9:26am » 
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How do we justify the third line? It's not true that every term in the new sum is larger than the corresponding term in the old sum. For instance, with i=n=2 and a_{1}=0.99, the terms are 0.017997 for the old and 0.009415 for the new. I am assuming that you have a good justification ... maybe I don't understand telescopic sums like I thought I did?


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anonymous
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Re: ***Spoiler*** inequality
« Reply #3 on: Jun 24^{th}, 2003, 10:27am » 
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I'll take n=3 as an example when n=3, \[ \begin{split} (a^{2^{n1}}a^{2^n})a^{2^{n+1}} \\ =(a^4a^8)a^{16} \\ =(a^4a^5)a^{16}+(a^5a^6)a^{16}+(a^6a^7)a^{16}+(a^7a^8)a^{16} \\ <(a^4a^5)a^{10}+(a^5a^6)a^{12}+(a^6a^7)a^{14}+(a^7a^8)a^{16} \\ =\sum_{i=2^{n1}}^{2^n1}(a^ia^{i+1})a^{2i+2} \end{split} \] When n is summed from n=1 to infinity, then i is also summed from i=1 to infinity.

« Last Edit: Sep 7^{th}, 2003, 2:12pm by Icarus » 
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James Fingas
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Re: ***Spoiler*** inequality
« Reply #4 on: Jun 24^{th}, 2003, 11:24am » 
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Okay, I see where you're coming from now. Too nonobvious for me though...


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towr
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Re: ***Spoiler*** inequality
« Reply #5 on: Jun 24^{th}, 2003, 11:29am » 
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for those who'd rather not read latex, even though/if they can/could

« Last Edit: Jun 24^{th}, 2003, 11:30am by towr » 
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