convergence divergence - What is the $I_{0}(x)$ function?

While trying to calculate the following infinite sum:

$$\sum_{k=0}^{\infty} \frac {4^k}{(k!)^{2}}$$

I got the result: $I_{0}(4) = 11.301...$

I've never encountered this function before ($I_{0}(x)$), can someone please describe it and explain why the above infinite sum converges to an output of this function?

I expected something having to do with the exponential function since $$\sum_{k=0}^{\infty} \frac {\mu^k}{k!} = e^\mu$$

Answer

The modified Bessel function of the first kind has a power series expansion
$$I_{\alpha}(x)=\sum_{k=0}^{\infty}\frac{1}{k!\Gamma(k+\alpha+1)}\Big(\frac{x}{2}\Big)^{2k+\alpha}$$

Taking $\alpha=0$ and using $\Gamma(k+1)=k!$, and then setting $x=4$, we get
$$I_0(4)=\sum_{k=0}^{\infty}\frac{4^k}{(k!)^2}$$
which is your sum.