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Let $u=(u_n)_{n\ge0}$ be a strictly increasing sequence of natural integers.

Are there known (and reasonably explicit) formulas for some series of the form :

$$S_u=\sum_{n=0}^\infty\frac{1}{(u_n)!}$$

?

Of course, I put aside the well known case where : $\exists p\in\mathbb{N}^\star;\forall n\in\mathbb{N},u_n=pn$.

For example, can we expect some formulas for :

$$\sigma=\sum_{n=0}^\infty\frac{1}{(n^2)!}$$ $$\tau=\sum_{n=0}^\infty\frac{1}{(n!)!}$$

?

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    The simpliest of the sequences $\{u_n\}$, namely, $u_n=n$, yields a trascendental number. So... I don't think so. – 2017-01-14
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    For the second one, see https://math.stackexchange.com/questions/1850150 – 2017-01-14
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    It can be shown using Taylor series that if $(u_n)$ is an arithmetic sequence of the form $u_n = kn$ with $k \in \mathbb{N}^*$ then $S_u = (e^{\omega_1} + e^{\omega_2} + \cdots + e^{\omega_k})/k$ where $\omega_i$ are the $k$th roots of unity. In the general case, I agree with @ajotatxe that it is highly unlikely that a simple formula should exist, seen as some very simple cases (such as $u_n = n^2$) are already very difficult if not impossible. – 2017-01-14

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You did not yet formally rule out all the easy cases, like $$ \sum_{n=1}^\infty \frac{1}{(2n+1)!} = \frac{e-e^{-1}}{2} $$

So to get a hard case, you must rule out a subset $A$ of $\mathbb N$ that is eventually periodic.

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    Absolutely right. – 2017-01-14