Problem solving a recurrence relation

Question

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?

Answer 1

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 $$J_n=J_0-\frac{\Gamma (n+1,-1)}{e \,\Gamma (n+1)}$$