Radius of Convergence of Power Series over Factorial/Real Case

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Theorem

Let $\xi \in \R$ be a real number.

Let $\displaystyle f \left({x}\right) = \sum_{n \mathop = 0}^\infty \frac {\left({x - \xi}\right)^n} {n!}$.


Then $f \left({x}\right)$ converges for all $x \in \R$.


That is, the interval of convergence of the power series $\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left({x - \xi}\right)^n} {n!}$ is $\R$.


Proof

This is a power series in the form $\displaystyle \sum_{n \mathop= 0}^\infty a_n \left({x - \xi}\right)^n$ where $\left \langle {a_n} \right \rangle = \left \langle {\dfrac 1 {n!}} \right \rangle$.

Applying Radius of Convergence from Limit of Sequence, we find that:

\(\displaystyle \lim_{n \mathop \to \infty} \left\vert{\dfrac {a_{n+1} } {a_n} }\right\vert\) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \left\vert{\dfrac {\dfrac 1 {\left({n + 1}\right)!} } {\dfrac 1 {n!} } }\right\vert\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \left\vert{\dfrac {n!} {\left({n + 1}\right)!} }\right\vert\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \left\vert{\dfrac 1 {n+1} }\right\vert\)
\(\displaystyle \) \(=\) \(\displaystyle 0\) Sequence of Powers of Reciprocals is Null Sequence

Hence the result.

$\blacksquare$


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