Value of $\sum_{n=0}^{+\infty}\frac{x^n}{n!}$ for $x=0$

Question

Why the value of $\sum_{n=0}^{+\infty}\frac{x^n}{n!}$ for $x=0$, is 1 and not 0?

Answer 1

It's intuitive that the sum of no numbers is defined as zero. Less intuitive is that the product of no numbers is defined as one.

Symbolically:

$$\sum_{\text{false}} x = 0$$

$$\prod_{\text{false}} x = 1$$

No matter what number $x$ is, even zero.

One reason $0^0 = 1$ by definition is because the cardinality (how many elements there are in a set, denoted $\vert \cdot \vert$) of the number of functions from a set $X$ to a set $Y$, denoted $Y^X$, is given by:

$$\left\vert Y^X \right\vert = |Y|^{|X|}$$

and there is one and only one function from the empty set to itself:

$$1 = \left\vert \varnothing^\varnothing \right\vert = |\varnothing|^{|\varnothing|} = 0^0$$

See https://proofwiki.org/wiki/Zero_to_the_Power_of_Zero for more reasons.

Edit: In the context of taking limits, we refer to $0^0$ as an indeterminate form because you cannot conclude $x^y \to 1$ just because $x \to 0$ and $y \to 0$.

Answer 2

We have by convention $$\left.\sum_{n=0}^{+\infty}\frac{x^n}{n!}\right|_{x=0}=\left.\frac{x^0}{0!}\right|_{x=0}+0+0+0+\dots=1+0+0+\dots=1.$$

Answer 3

Because when dealing with power series it is assumed that $0^0=1$.