Closed-form expression for a sum of reciprocals of factorials

Question

Is there a closed-form expression for the finite sum $$\sum_{s=1}^{2^{n-1}}\frac1{(s-1)!}$$ as a function of $n$?

Answer 1

Hint: The Taylor series for $e^x$ is $$\sum_{k = 0}^{\infty} \frac{1}{k!} x^k = 1 + x + \frac{1}{2!} x^2 + \cdots.$$

I guess "is there a closed form expression" is the correct terminology

I'm not sure whether you consider these closed-form, but here are two expressions for $$\sum_{k = 0}^N \frac{1}{k!}$$ that don't explicitly involve a sum:

Solution 1 $$\sum_{k = 0}^N \frac{1}{k!} = \frac{\Gamma(N + 1, 1)}{\Gamma(N)},$$ where $\Gamma(a)$ is the usual Gamma function, and $\Gamma(a, b)$ is the Incomplete Gamma Function---I'm not sure how satisfactory this is, since this identity basically amounts to unwinding definitions.

Solution 2 $$\sum_{k = 0}^N \frac{1}{k!} = \frac{1}{N!} \lfloor N! e \rfloor, \qquad N > 0;$$ this uses the fact that, after the zeroth, each $N$th partial sum of the series $\sum \frac{1}{k!}$ is largest multiple of $\frac{1}{N!}$ smaller than $e$, which is itself an easy consequence of the form of that series (and of course, that $e$ is its limit).

Taking $N = 2^{n - 1} - 1$ and reindexing gives expressions for your own sum.