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A175409
Successive numbers of consecutive positive terms to add when rearranging the alternating harmonic series to sum to log[7/3].
0
1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2
OFFSET
1,2
COMMENTS
Let s = log(7/3). Add a(n) positive terms 1 + 1/3 + 1/5 + ... + 1/(2a(n)-1)until their sum is greater than s, then subtract negative terms 1/2, 1/4, ... until the sum drops below s. Continue alternating in this way, adding a(2) consecutive positive terms, subtracting negative terms, and so on. The numbers a(n) are the terms of the sequence.
Since x = (1/4)exp(2s) = 49/36 is rational, a result in the reference shows that this sequence is eventually periodic.
REFERENCES
Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185.
LINKS
Francisco J. Freniche, On Riemann's Rearrangement Theorem for the Alternating Harmonic Series, Amer. Math. Monthly 117(2010), 442-448.
EXAMPLE
s = log(7/3) = 0.847298. The first term, 1, of the alternating harmonic series already exceeds s, so a(1)=1. Subtracting negative terms, we get 1-1/2 = 1/2, which is less than s. Then, adding 1/3 gives 0.833333, which is less than s, so we also add a second term, 1/5, to get 1.03333 which exceeds s. Thus a(2)=2.
CROSSREFS
Sequence in context: A161113 A161048 A152830 * A161073 A161112 A161047
KEYWORD
nonn
AUTHOR
John W. Layman, May 04 2010
STATUS
approved