Does the following series have a closed-form expression:
$$\sum_{k=0}^{\infty} \frac{z^k}{(k!)^2}$$
I know that it must converge because:
$$\sum_{k=0}^{\infty} \frac{z^k}{k!} = e^z$$
and the $(k!)^2$ denominator obviously increases more quickly than the $k!$ denominator.
This problem came up in computing the probability of a draw in a football match with each team's goal scoring modeled as a Poisson process.
Thanks,
John
Denote
$$f(z)=\sum_{n=0}^\infty\frac{z^n}{(n!)^2}$$
By term-wise differentiation, we find that
$$f'(z)+zf''(z)=f(z)$$
A rather simple differential equation with the general solution
$$f(z)=c_1I_0(2\sqrt z)+c_2K_0(2\sqrt z)$$
where $I_n$ is a modified Bessel function of the first kind, and $K_n$ is a modified Bessel function of the second kind. By using
$$f(0)=f'(0)=1$$
we find that
$$f(z)=I_0(2\sqrt z)$$
Another approach: Consider this sequence as the direct product of the reciprocal of the factorial function with itself.
Given that the exponential function is the generating function of the reciprocal of the factorial $$e^z=\sum_{n\in\mathbb{N}} \frac{z^n}{n!},$$ take $$f(z)=\sum_{n\in\mathbb{N}}\frac{z^n}{(n!)^2}$$ as the generating function of the termwise product of the former sequence with itself and use the Hadamard product of two generating functions formula: $$f(z)= e^z\cdot e^z = \sum_{n\in\mathbb{N}} \frac{1}{n!}\frac{1}{n!}z^n = \frac{1}{2\pi i}\oint_{x=0} e^{x\sqrt{z}}e^{\frac{\sqrt{z}}{x}} \frac{dx}{x},$$ so $$\boxed{\sum_{n\in\mathbb{N}}\frac{z^n}{(n!)^2} = \frac{1}{2\pi i}\oint_{x=0} e^{\sqrt{z}(x+\frac{1}{x})} \frac{dx}{x}}$$ which indeed matches with the $0$-th Modified Bessel Function of the First Kind asserted in the previous answer.
Now the task is to find a closed form of this function or to prove that this modified Bessel function of the first kind has no closed expression in terms of the standard elementary functions.
