Alternative argument to show that function diverges everywhere

Consider the function:

$$f(x) = x^{\frac{1}{2}} + \frac{1}{2}x^{-\frac{1}{2}} +\frac{2}{3}x^{\frac{3}{2}} - \frac{1}{4}x^{-\frac{3}{2}} +\frac{1}{15}x^{\frac{5}{2}} + \cdots$$

which is constructed such that $f$ is (or at least looks like) a solution to $f' = f$.

Then, $f(x) = Ae^x$ for some constant $A$. But $f$ diverges at $x=0$, so "$A=\pm\infty$" and $f$ diverges everywhere.

I appreciate that this is hardly rigorous, and that a rigorous proof could be given through analytical methods. But is the concept of the argument flawed in any way, and could it be made rigorous?

Solution 1:

Clearly, $$ f(x)=\sum_{n=0}^\infty \frac{4^n n!}{(2n+1)!}x^{(2n+1)/2} +\sum_{n=0}^\infty \frac{(2n-1)!}{4^n (n-1)!}x^{-(2n+1)/2} $$ The first term converges when $0<x$ while the radius of convergence of the second term is $R=0$.

Hence the function is nowhere definable!