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Two questions about a divergent series involving Bernoulli numbers

  1. Jun 16, 2008 #1
    Recently I learned of the possibility to give a finite value to a divergent sum:
    very surprising!
    (I'm a newbie, of course;
    see renormalisation, regularisation and number theory;
    many thanks to MR Watkins and to Dan Piponi)

    Some examples
    1 + 1 + 1 + 1 + ··· = -1/2

    1 - 2 + 3 - 4 + ··· = 1/4

    1 - 2^n + 3^n - ... = ((2^(n+1)-1)/(n+1))*B_(n+1), where B_n are the Bernoulli numbers
    (http://en.wikipedia.org/wiki/1_−_2_+_3_−_4_+_·_·_·)

    1 + r^2 + r^3 + ... = 1/(1-r) (http://en.wikipedia.org/wiki/Divergent_geometric_series)

    1 - 1 + 1 - 1 + ··· = 1/2 (http://en.wikipedia.org/wiki/Grandi's_series)

    My idea is to try to evaluate a double infinite sum:

    1^0 - 2^0 + 3^0 - 4^0 + ...
    + 1^1 - 2^1 + 3^1 - 4^1 + ...
    + 1^2 - 2^2 + 3^2 - 4^2 + ...
    + 1^3 - 2^3 + 3^3 - 4^3 + ...

    Summing by columns gives:
    -1/2 +1 -1/2 +1/3 - ... = -1/2 + ln2

    Summing by rows gives:
    1/2 + 1/4 + Sum_n>=2 ((2^(n+1)-1)/(n+1))*B_(n+1)

    Equating the results one can evaluate the sum.

    Another similar sum could involve 1 -1/3 + 1/5 - ... as a result of summing by columns.

    And now my two questions:

    1) Is this a meaningful result?
    I doubt, having used different summation methods.
    In http://arxiv.org/abs/math.NT/0608675, p.3, one can read:
    "differences often happen when differing methods of summing divergent series are used"
    (other 3 links: one two three)

    2) Is there a closed form formula for Sum_n>=2 ((2^(n+1)-1)/(n+1))*B_(n+1)?
     
  2. jcsd
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