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I have to find an expression for $J_n$ given the following recurrence relation (that is, solving it): $$J_n=J_{n-1}+\frac{(-1)^{n+1}}{n!}$$ I know that $$J_n=J_{n-1}+\frac{(-1)^{n+1}}{n!}=J_{n-2}+\frac{(-1)^{n}}{(n-1)!}+\frac{(-1)^{n+1}}{n!}=\ldots=\sum_{k=1}^{n}\frac{(-1)^{k+1}}{k!}$$ However, I am not very happy with this result. I would like to find a formula for $J_n$ that doesn't include a $\sum$. Can anyone tell me how to solve this problem in a different way?

Bernard
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kubo
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    What's the first term $J_0$ (or $J_1$)? It seems to be derangement permutation: https://en.wikipedia.org/wiki/Derangement So we have something like $J_n = D_n/n! \simeq 1/e$ (maybe it's required some adjustment). – Sávio Jun 15 '20 at 23:30
  • @Savio Yes I though about the $D_n/n!$ expression but it didn't occur to me that I could express it as $1/e$, thanks!. How did you deduce that $D_n/n!=1/e$? – kubo Jun 15 '20 at 23:36
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    Actually, we have $\lim_n D_n/n! = 1/e$ by Taylor (or Maclaurin) series, but $D_n/n! \neq 1/e$ for every $n$ since $e$ is irrational. Some closed formulas are in the content "Counting derangements" at Wikipedia (although they depend on floor function). – Sávio Jun 15 '20 at 23:45

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From what you wrote $$J_n=J_{n-1}+\frac{(-1)^{n+1}}{n!}=\sum_{k=1}^{n}\frac{(-1)^{k+1}}{k!}=1-\frac{\Gamma (n+1,-1)}{e \,\Gamma (n+1)}$$ But, since we do not know the value assigned to $J_0$, use $$J_n=J_0-\frac{\Gamma (n+1,-1)}{e \,\Gamma (n+1)}$$

Claude Leibovici
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