## Deceptively easy product

The problem is to evaluate

$$\prod_{n=2}^{\infty} \left (1+\frac{(-1)^{n-1}}{a_n} \right )$$

where

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

This is one of those cases where trying out a bunch of numbers really isn’t going to help all that much. The $a_n$ look kind of like $n!$, except off by some. Even so, I can find no closed form to the product by subbing $a_n$ with $n!$.

However, we make the following simple observation:

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

Accordingly, 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

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

This is a very deceptively simple result given the awkward pattern of numbers one gets from computing the sequence $a_n$.