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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 always viewed this fact as an interesting property of geometric series, rather than suggesting any deep connection between the summation of p-adic series and divergent series summation. This consideration is further supported by the fact that there are many series that converge in the p-adics, but don't converge in the reals, or vice-versa.

However, I started to reconsider this position upon seeing that in the p-adics: $$\sum_{n=1}^\infty n \cdot n! = -1$$ This very same result can be obtained in the reals using divergent series methods! For example, the easiest method is to replace $n!$ by its integral representation to obtain: $$\sum_{n=0}^\infty n! x^n = \int_0^\infty \frac{e^{-t}}{1-xt}dt$$ This has a singularity for $x>0$, but that singularity cancels with itself, so $$\sum_{n=0}^\infty n! x^n = \int_0^{\frac{1}{x} - \varepsilon} \frac{e^{-t}}{1-xt}dt + \int_{\frac{1}{x} + \varepsilon}^\infty \frac{e^{-t}}{1-xt}dt$$ Evaluating the derivative at x=1(which produces the extra $n$ term) gives $-1$. This is more than just a coincidence with this one method. For example, Euler's derivation for the sum of the series $\sum_{n=0}^\infty (-1)^n n! x^{k+1}$ involving find a solution to the differential equation $\frac{ds}{dx} = \frac{x-s}{x^2}$, leads to the same value of $-1$.

As a second test, this paper also gives that $$\sum_{n=1}^\infty n^2 \cdot n! = - \sum_{n=1}^\infty n!$$ This agrees with the case of divergent series, where, at x=1 $$\frac{d}{dx} x\frac{d}{dx} \int_0^\infty \frac{e^{-t}}{1-xt} = 1-\int_0^\infty \frac{e^{-t}}{1-xt}$$ (the extra 1 comes from the fact that the integral represents a series starting at n=0, intead of n=1).

Unfortunately, I am not well versed in p-adic numbers, so I am unsure about how to numerically check that $\sum_{n=1}^\infty n!$ in the p-adics agrees with the integral. Can it be shown that the sum $\sum_{n=1}^\infty n^k \cdot n!$ in the p-adics agrees with the divergent series summation in the real numbers? In the case they do agree-- is there a reason for this connection? Is there more generally an agreement between p-adic summation and summation in the reals under certain conditions?

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  • $\begingroup$ The sums $\sum n\cdot n!$ and $\sum n!(n^2+1)$ are simply telescoping (as $n\cdot n!=(n+1)!-n!$, $n! (n^2+1)=(n+1)!n-n!(n-1)$). For non-telescoping sums like $\sum n! $ evaluation even modulo $p$ looks hopeless. $\endgroup$
    – Fedor Petrov
    Commented Jan 31, 2022 at 5:35
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    $\begingroup$ It is unclear what it means "to numerically check that $\sum_{n \geq 1} n!$ in the $p$-adics agrees with the integral." There is no integral "in the $p$-adics" here, and that $p$-adic sum of $n!$ over all $n$ has no known simple formula. It's perhaps transcendental over $\mathbf Q$. Compare the simple $\sum_{n \geq 1} 1/n(n+1) = 1$ and the much more complicated $\sum_{n \geq 1} 1/n^2 = \pi^2/6$. $\endgroup$
    – KConrad
    Commented Jan 31, 2022 at 6:53
  • $\begingroup$ @KConrad I had in mind that because the p-adic series converged, and the divergent series integral converges, one could simply approximate both and check if they converge to the same value. However, with more thought, this is obviously wrong. Being close in their respective norms to the limit gives us nothing about whether they converge to the same value. Perhaps I need to rethink the question a bit. $\endgroup$
    – Caleb Briggs
    Commented Jan 31, 2022 at 18:14

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