2

I'm hoping to find a list of divergent sums where the assigned value is generally accepted. For instance $\sum_{n=0}^\infty (-1)^n$ is generally accepted to be $\frac{1}{2}$. Moreover, its agreed upon that $\frac{1}{1-x} =\sum_{n=0}^\infty x^n$. I'm looking for series that must be summed with something more powerful than an analytical continuation. For instance, some of the series I have compiled so far are (I'm not sure if these are all generally accepted, but they are in the spirit of the type of series I'm interested in).

$$\sum_{n=0}^\infty (-1)^n \left(\alpha n\right)! x^n = \int_{0}^\infty \frac{e^{-t}}{1+xt^\alpha }$$ $$\sum_{n=0}^\infty (-1)^n \ln(n+1) = \frac{1}{2}\ln\left(\frac{2}{\pi}\right)$$

Approximations are also gladly accepted. For example, I have this table for a divergent sum $$s(x) = \sum_{n=0}^{\infty}\frac{2^{\frac{1}{2}n\left(n-1\right)}}{n!}x^{n} $$\begin{array}{|c|c|c|c|} \hline x& -\frac{1}{10} & -\frac{2}{10} & -\frac{3}{10} \\ \hline s(x)& 0.90897 & 0.8323 & 0.76639\\ \hline \end{array}

For my own purposes, I'm especially interested in series of the form sum $\sum_{n=0}^\infty f(n)x^n$. Furthermore, I'm interested primarily in sums that aren't made up of only positive terms. Nonetheless, series that don't fit in either of these categories are still welcome.

Thanks in advance!

Caleb Briggs
  • 1,662
  • 1
    You may need to explain in which sense the divergent series have a value. For example, $\sum_{n=0}^{\infty}(-1)^n$ is usually understood as Cesaro summation, and $\sum_{n=0}^{\infty} n$ is usually understood as analytical continuation of zeta function. You have said you are interested in series like $\sum_{n=0}^\infty f(n)x^n$, so you can read more about analytic continuation of complex power series. – user779130 Aug 10 '21 at 02:42
  • I'm looking for series that have a value that agrees across different methods. For instance, $\sum_{n=0}^\infty (-1)^n$ agrees using Abel summation, analytical continuation, regularization, etc. I'm looking at summing series for which analytical continuation is too weak, for instance, when the radius of convergence is 0. – Caleb Briggs Aug 10 '21 at 03:00
  • If you are interested not only in series but also integrals, these links may be useful: https://mathoverflow.net/questions/388628/a-set-of-divergent-integrals-that-i-think-equal-to-gamma ; https://mathoverflow.net/questions/389694/regularization-of-the-area-under-hyperbola ; https://mathoverflow.net/questions/394326/where-do-these-divergent-integrals-appear-in-mathematics-and-physics – Anixx Aug 10 '21 at 07:11

1 Answers1

1

Here is a list of some divergent integrals and series (some of them have the both forms) that I composed (it includes some other properties that you likely not need). "Finite part" in the table means "the regularized value":

$$\small \begin{array}{cccccc} \text{Delta form} & \text{In terms of } \tau, \omega_+,\omega_- & \text{Finite part} & \text{Integral or series form} & \text{Germ form} &\text{Determinant}\\ \pi \delta (0) & \tau & 0 & \int_0^{\infty } \, dx;\int_0^{\infty } \frac{1}{x^2} \, dx & \underset{x\to\infty}{\operatorname{germ}} x;\underset{x\to0^+}{\operatorname{germ}}\frac1x&\frac{e^{-\gamma}}4 \\ \pi \delta (0)-\frac{1}{2} & \omega _-;\tau-\frac{1}{2} & -\frac{1}{2} & \sum _{k=1}^{\infty } 1; \int_{1/2}^\infty dx & \underset{x\to\infty}{\operatorname{germ}} (x-1/2) &e^{-\gamma} \\ \pi \delta (0)+\frac{1}{2} & \omega _+;\tau+\frac{1}{2} & \frac{1}{2} &\sum _{k=0}^{\infty } 1; \int_{-1/2}^\infty dx & \underset{x\to\infty}{\operatorname{germ}} (x+1/2) & e^{-\gamma} \\ 2 \pi \delta (i) & e^{\omega_+}-e^{\omega_-}-1 & 0 & \int_{-\infty }^{\infty } e^x \, dx & \underset{x\to\infty}{\operatorname{germ}} e^x &(1)\\ & \frac{\tau ^2}{2}+\frac{1}{24};\frac{\omega_+^3-\omega_-^3}6 & 0 & \int_0^{\infty} x \, dx;\int_0^\infty \frac2{x^3}dx & \underset{x\to\infty}{\operatorname{germ}}\frac{x^2}2;\underset{x\to0^+}{\operatorname{germ}} \frac1{x^2}&(2)\\ & \frac{\tau ^2}{2}-\frac{1}{24} & -\frac1{12} & \sum _{k=0}^{\infty } k & \underset{x\to\infty}{\operatorname{germ}} \left(\frac{x^2}2-\frac1{12}\right)&(3) \\ -\pi \delta''(0) &\frac {\tau^3}3 +\frac\tau{12};\frac{\omega_+^4-\omega_-^4}{12}& 0 & \int_0^\infty x^2dx;\int_0^\infty\frac6{x^4}dx&\underset{x\to\infty}{\operatorname{germ}}\frac{x^3}3;\underset{x\to0^+}{\operatorname{germ}} \frac2{x^3}\\ \pi^2\delta(0)^2-\pi\delta(0)+1/4&\omega_-^2&\frac16&2 \int_0^{\infty } \left(x-\frac{1}{2}\right) \, dx+\frac{1}{6}&\underset{x\to\infty}{\operatorname{germ}}B_2(x)&e^{-2\gamma}\\ \pi^2\delta(0)^2+\pi\delta(0)+1/4&\omega_+^2&\frac16&2 \int_0^{\infty } \left(x+\frac{1}{2}\right) \, dx+\frac{1}{6}&\underset{x\to\infty}{\operatorname{germ}}B_2(x+1)&e^{-2\gamma}\\ \pi^2\delta(0)^2&\tau^2&-\frac1{12}&\int_{-\infty}^{\infty } |x| \, dx-\frac{1}{12}&\underset{x\to\infty}{\operatorname{germ}}B_2(x+1/2)&\frac{e^{-2\gamma}}{16} \\ &\ln \omega_++\gamma&0&\int_1^\infty \frac{dx}x;\sum_{k=1}^\infty \frac1x -\gamma&\underset{x\to\infty}{\operatorname{germ}}\ln x\\ -3\pi\delta''(0)-\frac14 \pi\delta(0);\pi^3\delta(0)^3&\tau^3&0&\int_0^\infty \left(3x^2-\frac1{4}\right)dx&\underset{x\to\infty}{\operatorname{germ}}B_3(x+1/2)&\frac{e^{-3\gamma}}{64} \\ \frac{2\pi\delta(i)+1}{e-1}&e^{\omega_-}&\frac1{e-1}&\frac1{e-1}+\frac1{e-1}\int_{-\infty}^\infty e^x dx&\underset{x\to\infty}{\operatorname{germ}} \frac{e^x+1}{e-1}&\frac1{\sqrt{e}}\\ \frac{2\pi\delta(i)+1}{1-e^{-1}}&e^{\omega_+}&\frac1{1-e^{-1}}&\frac1{1-e^{-1}}+\frac1{1-e^{-1}}\int_{-\infty}^\infty e^x dx&\underset{x\to\infty}{\operatorname{germ}} \frac{e^x+1}{1-e^{-1}}&\sqrt{e}\\ &(-1)^\tau&\frac\pi{2}&&&1\\ \end{array} $$

Missing from the table above:

$(1)=e^{\psi _e(\ln (e-1))}$

$(2)=e^{\psi\left(\frac12+\frac{i}{2\sqrt{3}}\right)+\psi\left(\frac12-\frac{i}{2\sqrt{3}}\right)}$

$(3)=e^{\psi\left(\frac12+\frac{1}{2\sqrt{3}}\right)+\psi\left(\frac12-\frac{1}{2\sqrt{3}}\right)}$

For more formulas and the methods used you can refer to this page (a MathML capable browser is needed, such as Firefox or PaleMoon).

Another MathOverflow post on this topic.

Anixx
  • 9,302
  • This is interesting! Do you have any examples with series that aren't made up of only positive terms? – Caleb Briggs Aug 10 '21 at 17:57
  • @CalebBriggs in those cases usually Abel or Borel regularization works. One of examples, $\int_0^\infty \tan x dx=\ln 2$, but here Cesaro is enough anyway. – Anixx Aug 10 '21 at 18:04
  • @CalebBriggs also, examples here https://mathoverflow.net/questions/388628/a-set-of-divergent-integrals-that-i-think-equal-to-gamma and this formula $\int_0^\infty (x^k-(k+1)!x^{-(k+2)})dx=0$ – Anixx Aug 10 '21 at 18:10
  • Is there a regularization of $e^{e^x}$ as $x$ goes to infinity? I have some reason to believe that this is equal to 1, but im not sure how to verify this fact. – Sidharth Ghoshal Aug 29 '22 at 23:08
  • @SidharthGhoshal as a discrete sequence or as a smooth function? Notice that in these cases the regularization will be different (in one case we would have to regularize series, in the other - an integral). – Anixx Aug 30 '22 at 05:29
  • As a discrete sequence $e^{e^0} \rightarrow e^{e^1} \rightarrow e^{e^2} \rightarrow ... $ – Sidharth Ghoshal Aug 30 '22 at 15:34
  • @SidharthGhoshal I think there is no closed form. – Anixx Sep 02 '22 at 21:26
  • 1
    It seems that consensus is the discrete sequence normalizes to 1. See the conversation in the comments here: https://mathoverflow.net/questions/429326/how-to-correctly-renormalize-this-function-at-the-pole-x-1-evaluating-sum – Sidharth Ghoshal Dec 13 '22 at 19:50