Using divergent sums to define other divergent sums

Question

In this post, Annix describes an extension to the real numbers that includes a logarithm of 0, which could allow the definition of several divergent integrals and sums. However, this does not include all divergent sums. This made me think about the possibility of defining any divergent sum. In order to start this process, let us create a divergent sum then convert its terms into a Taylor Series: $$\sum_{n=0}^{\infty}a_n=\sum_{n=0}^{\infty}f(n)=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\alpha_k n^k$$ Since rising factorials can be denoted as polynomials, we can then use the Stirling Numbers ($s(n,k)$) to write: $$n^{\bar{k}}=\sum_{p=0}^{k} s(k,p) n^p\rightarrow \sum_{k=0}^{\infty}\alpha_k n^k=\sum_{k=0}^{\infty}\beta_k n^{\bar{k}}=\sum_{k=0}^{\infty}\beta_k \sum_{p=0}^{k} s(k,p) n^p$$ Since this is a divergent sum, we should not be able to switch the sum operators, but let's ignore that rule: $$\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\beta_k n^{\bar{k}} =\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}\beta_k n^{\bar{k}}= \sum_{k=0}^{\infty}\beta_k \sum_{n=0}^{\infty} n^{\bar{k}}$$ Based on an earlier equation, we can substitute $\sum_{n=0}^{\infty} n^{\bar{k}}$ with $\omega^k$ to find: $$\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\beta_k n^{\bar{k}} =\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}\beta_k n^{\bar{k}}= \sum_{k=0}^{\infty}\beta_k \sum_{n=0}^{\infty} n^{\bar{k}}=\sum_{n=0}^{\infty}\beta_k \omega^k$$ Therefore, we can define any divergent sum as: $$\sum_{n=0}^{\infty}a_n=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\alpha_k n^k=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\beta_k n^{\bar{k}}=\sum_{n=1}^{\infty} \beta_k \omega^k=\beta_1\omega+\beta_2\omega^2+\beta_3\omega^3+...$$ Once we find all of our $\beta_k$'s, we can define any divergent sum using this method. I used Wolfram to find another representation of $\omega^k$ to be: $$\sum_{n=0}^{\infty} n^{\bar{k}}=\omega^k=\sum_{n=0}^{\infty} \frac{(n+k)!}{n!}=\lim_{m\rightarrow \infty} \frac{(k+m+1)!}{k!(m+1)}$$ That being said, how do I find all of the $\beta_k$ from all of the $\alpha_k$?
Update
I found another solution to this divergent sum problem, $\sum_n a_n=a_0+a_1\omega$, where $\omega$ is a general infinity. While it does not take any later terms into account, I found that this approach does work for continued fractions. I will try to find another approach that involves all $a_n$.

Answer 1

This is a comment as I have yet to verify your Stirling number procedure but a closely related procedure involving the Riemann Zeta function does not sum the geometric series correctly.

We recall that

$$\sum_{n=0}^{\infty} e^n = \frac{1}{1-e}$$

Note that $e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}$ and therefore:

$$ \sum_{n=0}^{\infty} e^n = \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{(n)^k}{k!} $$

Doing the "questionable rearrangement" we conjecture that:

$$ \sum_{n=0}^{\infty} e^n = \sum_{k=0}^{\infty} \frac{1}{k!} \sum_{n=0}^{\infty} n^k $$

And therefore:

$$ \sum_{n=0}^{\infty} e^n = 1 +\sum_{k=0}^{\infty} \frac{\zeta(-k)}{k!} $$

But this is wrong. We are off by a $-1$. The Euler Maclaurin formula correctly expresses:

$$ \frac{1}{1-e^x} = \mathbf{-\frac{1}{x}} +1 + \sum_{k=0}^{\infty} \frac{\zeta(-k)}{k!} $$

A moral of the story here is that these kind of re-arrangements often get you "symbolically close" to where you want to go but they aren't correct (of course your stirling number formulation might work I haven't shown it doesn't yet). More specifically the parts that you did re-arrange are often symbolically correct BUT there is extra stuff that you lose by naively re-arranging sums.

A better approach to understanding these things is instead to consider the function $\hat{f}(x) = \sum_{n=0}^{\lfloor x \rfloor } a_n$ and calculate the average value of the $O(1)$ terms of $\hat{f}(x)$. This post explores that and it works surprisingly well on everything from convergent sums, to the harmonic series to the geometric series and many things in between including the infamous $1+2+3... = -\frac{1}{12}$.

Unfortunately this too has limitations for example it fails to assign divergent continued fractions correctly so a more general technique needs to be invented to unify all these cases.

Answer 2

To compare a divergent series to another one, the easiest way is to map the both to the countable surreal numbers.

There is quite developed theory of mapping Hardy fields to surreals, and basically considering surreals as an H-field (H-field is a Hardy field with a unity).

In simple language, Hardy fields are the fields of germs (asymptotic behaviors) of functions at infinity. For instance, the germ of the function $f(x)=x$ is greater than any real numbers but smaller than the germs of $x+1$, $2x$ or $x^2$.

The surreal numbers also can be seen as a Hardy field, with canonical embedding of the germs of smooth functions into surreals being equalizing $\omega$ with the germ of the identity function $f(x)=x$.

This way, you can embeed germs at infinity into surreals, and with some other limitations, divergent improper integrals (at least those which correspond to surreals with no infinitesimal part).

With series the matter is a bit more complicated, since you need to convert the discrete partial sum into a smooth germ. But this formula does the trick:

$$\sum_{k=0}^\infty a_k=\int_{\omega }^{\omega +1} \left(\sum _{k=0}^{x-1} a_k\right) \, dx$$

This way you can get the expression of a divergent series as a surreal number, compare them, extract the finite part (that is regularized value), etc.

For instance, you can see that $\sum_{k=0}^\infty e^k=e^{\omega }+\frac{1}{1-e}$.

You can see that $e^\omega$ is a log-atomic number, so it has zero finite part, so the second term is the regularized value.

In a similar way, you can see that $\sum_{k=0}^\infty k=\frac{\omega ^2}{2}-\frac{1}{12}$ and so on.

For the sum of harmonic series you will get $\sum_{k=1}^\infty \frac1k=\ln \omega+\gamma$. For some reason, online Wolfram Alpha is too weak for this, but Wolfram Mathematica does just well. Using $\lambda=-\ln \omega-\gamma$ from the linked post, the harmonic series is equal to $-\lambda$.