2

One way to understand how the sum of all factorials $$ S_{\mathbb{N}!} = 1! + 2! + 3! + \ldots $$ can be regularized comes from an observation that the function $$ f(x) = \text{P.V.}\int \limits_0^\infty \dfrac{\operatorname{e}^{-t}}{1-t\,x} \operatorname{d}t $$ is finite for $x>0$, while its Taylor expansion is $$ f(x) = 0! + 1!x + 2!x^2 + 3!x^3 + \ldots $$ So, we can assign $S_{\mathbb{N}!} = f(1)$.

I'm wondering if same can be done for the sum of all integers $$ S_{\mathbb{N}} = 1 + 2 + 3 + \ldots $$ Clearly, we cannot use the function $\dfrac{x}{(x-1)^2}$, since it is not defined at $x=1$. Still, I'm wondering if it's possible to come up with smth similar to the case of factorials.

This is actually a closely related question.

mavzolej
  • 1,442
  • Let $F(z) = \int_0^\infty \frac{e^{-t}}{1-tz}dt$ which is analytic on $\mathbb{C} \setminus [0,\infty)$ and $\lim_{z \to 0} \frac{F^{(k)}(z)}{k!} = \int_0^\infty t^k e^{-t}dt = k!$. Then what ? – reuns Nov 30 '17 at 20:54
  • In physics we often encounter asymototic series with $a_k\propto k!$ (tunneling processes in QM and QFT). In these situations we 'regularize' the result by evaluating $F(z)$ instead of $\sum a_nz^n$ (which is divergent). We reconstruct $F(z)$ from the series using Borel summation or other more involved techniques. So my question is how to do same with a series whose coefficients grow linearly, not factorially. – mavzolej Dec 01 '17 at 01:02
  • So you interpret $F(z) = \int_0^\infty \frac{e^{-t}}{1-tz}dt, z \in \mathbb{C}\setminus [0,\infty)$ as a regularized version of $\sum_{n=0}^\infty n! z^n$ ? With $a_n=n$ then $G(z)=\sum_{n=0}^\infty a_n z^n$ does converge for $|z| < 1$. Then what ? – reuns Dec 01 '17 at 01:21
  • The expansion of $F(z)$ is divergent for any $z>0$. The point of regularization is that it allows me to assign a finite value to the series $\sum n!z^n$ at the points where it diverges. I would like to do the same thing with the second series and evaluate it at $z=1$. – mavzolej Dec 01 '17 at 01:49
  • 1
    $G(z)= \frac{z}{(z-1)^2}$ has a double pole at $z=1$ $$\lim_{\epsilon \to 0}\frac{G(1+\epsilon)+G(1+i\epsilon)+G(1-\epsilon)+G(1-i\epsilon)}{4}=0$$ is finite. Note this is quite similar to what you do with $F(z)$ as $$\text{pv}.\int_0^\infty \frac{e^{-t}}{1-tx}dt = \lim_{\epsilon \to 0} \frac{F(x+i\epsilon)+F(x-i\epsilon)}{2}$$ – reuns Dec 01 '17 at 02:18
  • Thanks!! This really helps. – mavzolej Dec 01 '17 at 02:20
  • I'm surprised though the answer is not $-1/12$, as given by other 'regularizations'. – mavzolej Dec 01 '17 at 04:43

0 Answers0