Infinite product include summation

Question

I would like to find an infinite product of$$\prod _{n=2}^{\infty} \left(1+\frac{(-1)^{n-1}}{a_n}\right)$$

where $a_n = \sum_{k=1}^{n-1} \frac{n!(-1)^{k-1}}{k!} $

I tried to compute $a_2 , a_3 ,...$, but nothing pattern can see.

Any ideas?

Answer 1

Observe that, because

$$\frac{a_{n+1}}{(n+1)!} = \frac{a_n}{n!} + \frac{(-1)^{n-1}}{n!}$$

the given sequence satisfies

$$a_{n+1} = (n+1) (a_n + (-1)^{n-1}) $$

Then

$$1+\frac{(-1)^{n-1}}{a_n} = \frac{a_n+(-1)^{n-1}}{a_n} = \frac1{n+1} \frac{a_{n+1}}{a_n}$$

Then

$$\prod_{n=2}^N \left (1+\frac{(-1)^{n-1}}{a_n} \right ) = \frac{2}{(N+1)!} \frac{a_{N+1}}{a_2} = \frac{a_{N+1}}{(N+1)!}$$

The product sought is just the limit of this ratio as $N \to \infty$, which is just

$$\lim_{N \to \infty} \sum_{k=1}^{N} \frac{(-1)^{k-1}}{k!} = 1-\frac1{e} $$