Taylor expansion of exponential function and singularities

Stackexchange community,

Lets consider the function $f(z)=\exp(-z)$.

This function can be expanded as a taylor series:

$$f(z)=1-z+\frac{z^2}{2!}-\frac{z^3}{3!}\pm$$

As this is "some sort of polynomial" it will never have a singularity on every finite domain of $z$.

But if I express $f(z)=\exp(-z)=\frac{1}{\exp(z)}$ and then apply the taylor series on $\exp(z)$ I get:

$$f(z)=\frac{1}{1+z+\frac{z^2}{2!}+\frac{z^3}{3!}\pm}$$

Now here comes my confusion, the new representation allows singularities of $f(z)$, as it is "some kind of polynomial" in the denominator. What is the problem here? Is it because of the radius of convergence?

EDIT: Ok it is because the fundamental theorem of calculus isn't true for infinite sums. Is there some intuitive way of understanding this phenomenon?

• But what is the problem with second representation? – qwenty Nov 9 '15 at 17:22
• The second representation allows singularities by the fundamental theorem of algebra. – MrYouMath Nov 9 '15 at 17:39
• "As this is a polynomial" -- no it is not a polynomial. – zhw. Nov 9 '15 at 18:31
• Ok it is power sum, but why exactly does the fundamental theorem of algebra not hold for infinite polynomials? – MrYouMath Nov 9 '15 at 19:56
• See math.stackexchange.com/questions/109360/… for the behaviour of the roots of the partial sums of the exponential function. Also math.stackexchange.com/questions/51586/… – Lutz Lehmann Jul 3 '17 at 18:05

Consider the polynomials $\sum_{k=0}^n x^k$ for any positive integer $k$ these are defined for all real (and complex) $x$.
But, if we consider the infinite power series $\sum_{k=0}^{\infty} x^k$, this diverges for $|x| > 1$ and converges for $|x| < 1$.