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#1 | |
Dec 2002
Frederick County, MD
2×5×37 Posts |
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Alrighty,
At http://www.mersenne.org/math.htm, we are told: Quote:
Tim |
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#2 |
Dec 2002
Frederick County, MD
17216 Posts |
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In a posibly related note, is the following infinite series equal to 1?
[code:1]inf Σ 1/(n(n+1)) = 1 ? n=1[/code:1] |
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#3 |
Jan 2003
far from M40
53 Posts |
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Yes, it is.
1/(n*(n+1)) = (n+1-n)/(n*(n+1)) = (n + 1)/(n*(n+1)) -n/(n*(n+1)) = 1/n - 1/(n+1) So, the kth partial - sum has the value 1 - 1/(k+1) which's limit for k to infinity is 1 q.e.d. Sums like these, where a summand is (partly) nihilized by its successor are called telescope sums. Benjamin |
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#4 | |||
"Richard B. Woods"
Aug 2002
Wisconsin USA
22×3×641 Posts |
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Quote:
That is, it refers to the probability of the existence of a factor (known or unknown) of 2^p-1 within the specified range, not necessarily the probability of discovering a previously-unknown factor there after some factors of 2^p-1 are already known. |
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#5 |
"Richard B. Woods"
Aug 2002
Wisconsin USA
22×3×641 Posts |
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For a thorough earlier discussion, see the thread "Does the LL test:s factorization save or waste CPU time?" in The Software forum at http://www.mersenneforum.org/viewtopic.php?t=78, especially svempasnake's table comparing the 1/n prediction to the actual number of factors found, on page 3 at Mon Sep 16, 2002 9:32 pm (http://www.mersenneforum.org/viewtop...highlight=#895).
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