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I am aware of the generalized Bernoulli numbers, but these are not what I'm looking for. I was wondering if there exists such a thing as fractional, real or even complex Bernoulli numbers ( $B_z$ for $z \in \mathbb{C}$).

My motivation comes from the Ramanujan Summation, as the Bernoulli numbers are involved in it. I was hoping that, if the Bernoulli numbers could be extendend, so could perhaps the Ramanujan Summation, allowing it to assign a sum to a wider class of divergent series.

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How would writing down fractional, real, or complex Bernoulli numbers help you extend Ramanujan summation? –  Qiaochu Yuan Jun 8 '11 at 12:47
    
@Qiaochu The Bernoulli numbers are involved in the Ramanujan summation function. I suppose that I'm hoping that if every component of the Ramanujan summation formula could be extended to higher argument analogous (So this includes replacing the $(k+1)!$ term by $\Gamma(z)$), an extension of Ramanujan summation could be made possible. I'm not really sure the extension of functions to larger sets of arguments works like that, though. –  Max Muller Jun 8 '11 at 13:10
    
@Qiaochu I guess more work needs to be done to extend Ramanujan summation. For example, I think you also need to know how to differentiate some function fractionally (see the $f^{k}(n)$ term). I should not that the motivation not only stems from extending the Ramanujan summation, but I think a generalization of Bernoulli numbers is also interesting on its own right. –  Max Muller Jun 8 '11 at 13:29
    
Do you think there is a chance that Ramanujan summation could be extended by means of some ideas that are loosely based on the things I just said in the comments? –  Max Muller Jun 8 '11 at 13:31

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up vote 10 down vote accepted

Ramanujan himself gave a definition of Bernoulli numbers for a complex index - see Ramanujan's Notebooks - Part 1 by Bruce C. Berndt, Chapter 5 equation (25.1) with further results in Chapter 7.


Added by J. M.:

Ramanujan's definition of the Bernoulli numbers, as given in Berndt's book, is

$$B_s^\ast=\frac{2\Gamma(s+1)}{(2\pi)^s}\zeta(s)$$

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This is a nice precise answer, thanks! –  Max Muller Jun 8 '11 at 14:17

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