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6 votes
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Is there any intuition of why the both, regularized logarithm of zero is $-\gamma$ and the regularized logarithm of Bernoulli umbra is $-\gamma$?

If we take the MacLaurin series for $\ln(x+1)$ and evaluate it at $x=-1$, we will get the Harmonic series with the opposite sign: $-\sum_{k=1}^\infty \frac1x$. Since the regularized sum of the ...
Anixx's user avatar
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3 votes
0 answers
171 views

The divergent sum $\sum_{n=1}^\infty (-1)^n (n^2)! x^n$

Question I'm interested in assigning a value to the divergent series $F(x)=\sum_{n=1}^\infty (-1)^n (n^2)! x^n$. I'm hoping that (1) the definition for $F(x)$ has (one-sided) derivatives of $(-1)^n (n^...
Caleb Briggs's user avatar
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3 votes
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276 views

Evaluating $\sum_{n=0}^\infty n^k n!$ in p-adics, and its connection to the summation of divergent series

Often, in the discussion of the regularization of the geometric series it is mentioned that $\sum_{n=0}^\infty p^n$ converges in the p-adics, and indeed, that it converges to $\frac{1}{1-p}$. I had ...
Caleb Briggs's user avatar
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3 votes
0 answers
358 views

New/useful method for summation of divergent series?

Questions $$ S(n,x) = x+e^x + e^{e^x} + e^{e^{e^x}} + \dots \text{$n$ times}$$ Also obeys (see background for argument): $$ \frac{1}{2 \pi i} \oint e^{S(k,x)} \frac{\partial \ln(\frac{\int_0^\...
More Anonymous's user avatar
2 votes
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Did anyone ever propose the distinction between "divergent to infinity" as opposed to "divergent but with finite average"?

There are different regularization methods that allow us to ascribe finite values to divergent integrals, series or sequences. Still, in my view there is fundamental difference between divergent ...
Anixx's user avatar
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1 vote
0 answers
106 views

What intuitive meaning "determinant" of a divergency (divergent integral, series, germ, pole or a singularity) can have?

I am working on the algebra of "divergencies", that is, infinite integrals, series, and germs. So, I decided to construct something similar to the modulus or determinant of a matrix of these ...
Anixx's user avatar
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