So, I am currently trying to study the nature of divergent summations, and I often came upon their negative rational values in their regularizations. Such as this classic:
$$\sum_{n=1}^\infty n = -\frac{1}{12}$$
I was wondering, would it be possible to apply Borel Regularization to all summations. If Not, why?
Yours Sincerely,
Aster17
On
I don't know, whether Konrad Knopp's book on series (including divergent series) has been translated to english (from german, from the 1920'ies if I recall this correctly). I like this book very much, he is much explanative, has many examples & exercises. (But G.H.Hardy's book should be available everywhere in english spoken countries, however it is much more involved and likely not so easy to digest as the Knopp's book.)
One thing what I've got from this, that the various known (and accepted!) methods of summation are of different "strength". So Cesaro-summation cannot sum geometric series (even when alternating), but Abel- and Euler-summation can (when signs of series are alternating; it has been designed for this task). But Euler-summation cannot sum factorial series ("hypergeometric") even when signs are alternating, but Borel can. And so on. There is a hierarchy of methods.
Even there are series, which can be summed by Cesaro, but not by Euler-summation, so the "strength" which I alluded to above is not a onedimensional aspect with a simple order.
It would be too much to explain, not well suited for Q&A on MSE, so I'd recommend you go further to collect introductory material before stepping more forward.
Partial answer regarding the Abel method.
The power series $$A(z) = \sum_{j=1}^{\infty} j z^j$$ is the Taylor series of $\frac{1}{(1-z)^2} - \frac{1}{1-z} = \frac{z}{(1-z)^2}$ at $z_0 = 0$. Its radius of convergence equals 1.
The Abel summation method gives: $$ Abel(\sum_{j=1}^{\infty}j) = \lim_{z \to 1^{-}} A(z) = \lim_{z \to 1^{-}} \frac{z}{(1-z)^2} = + \infty. $$