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!

  • 1
    $\begingroup$ 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. $\endgroup$
    – user779130
    Aug 10, 2021 at 2:42
  • $\begingroup$ 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. $\endgroup$ Aug 10, 2021 at 3:00
  • $\begingroup$ If you are interested not only in series but also integrals, these links may be useful: mathoverflow.net/questions/388628/… ; mathoverflow.net/questions/389694/… ; mathoverflow.net/questions/394326/… $\endgroup$
    – Anixx
    Aug 10, 2021 at 7:11

1 Answer 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:

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



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.

  • $\begingroup$ This is interesting! Do you have any examples with series that aren't made up of only positive terms? $\endgroup$ Aug 10, 2021 at 17:57
  • $\begingroup$ @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. $\endgroup$
    – Anixx
    Aug 10, 2021 at 18:04
  • $\begingroup$ @CalebBriggs also, examples here mathoverflow.net/questions/388628/… and this formula $\int_0^\infty (x^k-(k+1)!x^{-(k+2)})dx=0$ $\endgroup$
    – Anixx
    Aug 10, 2021 at 18:10
  • $\begingroup$ 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. $\endgroup$ Aug 29, 2022 at 23:08
  • 1
    $\begingroup$ It seems that consensus is the discrete sequence normalizes to 1. See the conversation in the comments here: mathoverflow.net/questions/429326/… $\endgroup$ Dec 13, 2022 at 19:50

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