Approximations of $\sum_{n=0}^\infty\frac{x^n}{n!}$ and $\sum_{n=0}^\infty\frac{x^n}{n!^2}$ We know that
$$f(x)={\rm e}^x=\sum_{n=0}^\infty\frac{x^n}{(n!)\color{red}{^1}}\tag{1}.$$
We can approximate $f(x)$ by the function $$g(x)=b^x\,\,{\rm with}\,\, b\approx {\rm e}.\tag{2}$$
By letting  $b$ become closer to ${\rm e}$ we can adjust the goodness of the approximation and in the limit
$$\lim_{b\to {\rm e}}g(x)=\sum_{n=0}^\infty\frac{x^n}{(n!)\color{red}{^1}} \tag{3}$$
the approximation coincides with the infinite sum.
We slightly modify the infinite sum in eq.$(1)$ and get
$${\rm I}_0\left(2\sqrt{x}\right)=\sum_{n=0}^\infty\frac{x^n}{(n!)\color{red}{^2}}\tag{4},$$
where ${\rm I}_0$ is the modified Bessel function of first kind and zeroth order. Is it possible to approximate eq.$(4)$  in the way as $g(x)$ approximates $f(x)$? There we have the property, that the less the constants $b$ and ${\rm e}$ differ, the better is the approximation for the infinite sum.
As eq.$(1)$ and eq.$(4)$ differ only by $n!$ I am motivated to find for eq.$(4)$ an approximative function $h$ with the structure $$h(x,c_1,c_2,\ldots,c_k)\approx{\rm I}_0\left(2\sqrt{x}\right),\tag{5}$$ where $c_i$ are a finite set of constants. Analogous to eq.$(3)$ the approximation shall become exact in the limit
$$\lim_{c_i\to d_i \forall i} h(x,c_1,c_2,\ldots,c_k)={\rm I}_0\left(2\sqrt{x}\right),\tag{6}$$
where $d_i$ are unknown (transcendental) constants.
What I tried:
For $x=1$ we get the constant ${\rm e}$ from eq.$(1)$
$$\sum_{n=0}^\infty\frac{1}{(n!)\color{red}{^1}}={\rm e}\approx 2.71828\tag{7}$$
and analogous we get another less known constant from eq.$(4)$
$$\sum_{n=0}^\infty\frac{1}{(n!)\color{red}{^2}}={\rm I}_0(2)\approx 2.27959\tag{8}.$$
Now one could use an approximate value of ${\rm I}_0(2)$ (together with some other constants) to get an approximation for ${\rm I}_0(2\sqrt{x})$ without using infinite sums.
 A: Quoting from Wikipedia: In mathematics, an elementary function is a function of a single variable (typically real or complex) that is defined as taking sums, products, and compositions of finitely many polynomial, rational, trigonometric, hyperbolic, and exponential functions, including possibly their inverse functions (e.g., $\arcsin$, $\log$, or $x^{1/n}$).
As you can see, the exponential function is on the list. Now, the modified Bessel function $I_{0}$ is a non-elementary function. You can express it in terms of elementary functions but such an expression will always involve an infinite process, e.g., infinite series, integration, etc. It may be hard to accept that adding an extra factorial to the sum changes things so dramatically, but that is the case here. You have to accept that what you are asking for is not possible, not because mathematicians are incompetent but because of the nature of mathematics.
It was shown by Hankel that
$$
I_0 (2\sqrt x ) \sim\frac{{\mathrm{e}^{2\sqrt x } }}{{2\sqrt{\pi} x^{1/4} }}\left( {1 + \frac{1}{{16\sqrt x }} + \frac{9}{{512x}} +  \cdots } \right),
$$
as $x\to +\infty$. This is a divergent asymptotic expansion in the sense of Poincaré. For the full expansion, see $(10.40.1)$ in the NIST DLMF. The series is divergent for all $x>0$ but any truncated version of it yields a good approximation for $I_0$ for large positive values of $x$. This asymptotics also shows you how the function deviates from $\mathrm{e}^x$.
When $x<0$, your sum can be written in terms of the $J_0$ Bessel function as $J_0 (2\sqrt { - x} )$. The behaviour for large negative $x$ is given by another result of Hankel and it shows an oscillatory behaviour. At leading order
$$
J_0 (2\sqrt { - x} ) \approx \frac{1}{{\sqrt \pi  ( - x)^{1/4} }}\cos \left( {2\sqrt { - x}  - \frac{\pi }{4}} \right),
$$
when $x$ is large and negative.
The above two asymptotics shows that the behaviour of your series is rather different when $x>0$ or $x<0$ and may also convince you that it is not possible to capture two such different behaviours by a single approximation.
