Timeline for Evaluating $\sum_{n=0}^\infty n^k n!$ in p-adics, and its connection to the summation of divergent series

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Jan 31, 2022 at 18:14	comment	added	Caleb Briggs		@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.
Jan 31, 2022 at 6:53	comment	added	KConrad		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 $\sum_{n \geq 1} 1/n^2 = \pi^2/6$.
Jan 31, 2022 at 5:35	comment	added	Fedor Petrov		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.
Jan 31, 2022 at 5:09	history	asked	Caleb Briggs	CC BY-SA 4.0