The sum $\sum \frac{1}{(i!)^2}$

Question

We know the sum:

$$\sum\limits_{i=0}^\infty \frac{1}{(i!)} = e.$$

This makes me wonder about the sum:

$$S = \sum\limits_{i=0}^\infty \frac{1}{(i!)^2}.$$

It's clear that it converges and also that it's less than $e$ since all the terms are less than the first summation. But is there a closed form for it?

Numerically, it seems to be around $2.279$.

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In general, one can define a function similar to the Riemann Zeta:

$$S(u) = \sum\limits_{i=0}^\infty \frac{1}{(i!)^u}$$

This will converge for all $u>0$. We know the closed form for $u=1$.

For large $u$, it'll just become $2$.

But I wonder if any other $u$ has a closed form. And if we wanted to extend it to the complex numbers and analytically continue it, what would we get for:

$$S(-1) = \sum\limits_{i=0}^\infty (i!)$$

Answer 1

Specifically for your case of $$S = \sum\limits_{i=0}^\infty \frac{1}{(i!)^2}$$ this admits a closed form in terms of a modified bessel function of the first kind where we have $$I_v(z) = \left(\frac z2 \right)^v \sum_{i=0}^\infty \frac{\left(\frac {z^2}4\right)^i}{(i!)\,\Gamma(v+i+1)} \iff{\color{red} {I_0(2)}} = \sum\limits_{i=0}^\infty \frac{1}{(i!)^2}$$

In general, the "closed form" of $$S(u) = \sum\limits_{i=0}^\infty \frac{1}{(i!)^u}$$ for $u\in \mathbb Z^+$ can be found by looking at the general series definition of the hypergeometric function, and can be expressed in a sense as $$S(u) = {}_0F_u(; \underbrace{1, 1, \dots, 1;1}_{\text{there are }u\text{ 1's}})$$

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In regards to your limit of $$\lim_{u\to0^+}S(u)$$ I don't think your numerical evaluation was correct as Mathematica on my end claims it diverges

Answer 2

As far as I know, this constant does not have a closed-form expression in terms of elementary functions. That said, we can express it in closed form in terms of the Modified Bessel Function of the First Kind, $I_\alpha(x)$, which is defined as $$I_\alpha(x) := i^{-\alpha} J_\alpha(i x) \sum_{m = 0}^\infty \frac{1}{m! \Gamma(m + \alpha + 1)} \left(\frac{x}2\right)^{2 m + \alpha} .$$ Here, $J_\alpha(x)$ is the (unmodified) Bessel Function of the First Kind, which satisfies a similar differential equation.

Fixing $\alpha = 0$ gives $$I_0(x) = J_0(i x) = \sum_{n = 0}^\infty \frac{1}{m!^2} \left(\frac{x}2\right)^{2 m},$$ and setting $x = 2$ gives that our sum is $$\sum_{m = 0}^\infty \frac{1}{m!^2} = I_0(2) = J_0(-2 i) = 2.279585302\ldots .$$

More generally, substituting $u = \frac{x^2}{4}$ in our expression for $I_0(x)$ gives that the series in @KamalSaleh's comment has value $$\sum_{m = 0}^\infty \frac{u^i}{i!^2} = I_0(2 \sqrt u)$$ for suitable $u$.

Alternatively, it follows from the definition of hypergeometric function that we can write $$\sum_{m = 0}^\infty \frac{1}{i!^2} = {}_0 F_1(1; 1) ,$$ and more generally that $$\sum_{m = 0}^\infty \frac{u^m}{m!^2} = {}_0 F_1(1; u) .$$

(Even) more generally, for positive integers $k$, $$\sum_{m = 0}^\infty \frac{u^m}{m!^k} = {}_0 F_{k - 1}(\underbrace{1, \ldots, 1}_{k - 1}; u) .$$

Where the summands are defined, where $|u| < 1$ and as $k \searrow 0$, linearizing the series in $u$ gives the expansion $$\sum_{m = 0}^\infty \frac{u^m}{m!^k} \sim \frac{1}{1 - u} - \sum_{m = 0}^\infty u^m \log (m!) \cdot k + R(k) ,$$ where $R(k) \in O(k^2)$.

Answer 3

The Bessel Clifford function is defined as (for $n=0$) $$C_0(z)=\sum_{k=0}^\infty\frac{z^k}{k!^2}$$So your sum is equivalent to the integral$$C_0(1)=\frac1{2\pi}\int_0^{2\pi}e^{2\cos\theta}d\theta=\frac1{\pi}\int_0^\pi e^{2\cos\theta}d\theta$$Which doesn't seem to be solvable.

Credit to @TymaGaidash for making me aware of this function.