Series of Power over Factorial Converges

Theorem
The series $\displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ converges for all real values of $x$.

Proof
If $x = 0$ the result is trivially true as:
 * $\forall n \ge 1: \dfrac {0^n} {n!} = 0$

If $x \ne 0$ we have:
 * $\left|{\dfrac{\left({\dfrac {x^{n+1}} {(n+1)!}}\right)}{\left({\dfrac {x^n}{n!}}\right)}}\right| = \dfrac {\left|{x}\right|} {n+1} \to 0$

as $n \to \infty$.

This follows from the results:
 * Sequence of Powers of Reciprocals is Null Sequence, where $\dfrac 1 n \to 0$ as $n \to \infty$
 * The Squeeze Theorem for Real Sequences, as $\dfrac 1 {n + 1} < \dfrac 1 n$
 * The Multiple Rule for Real Sequences, putting $\lambda = \left|{x}\right|$.

Hence by the Ratio Test: $\displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ converges.

Alternatively, the Comparison Test could be used but this is more cumbersome in this instance.

Another alternative is to view this as an example of Radius of Convergence of Power Series over Factorial setting $\xi = 0$.

Also see

 * Equivalence of Definitions of Exponential Function, where it is shown that this series converges to the exponential function.