Radius of Convergence from Limit of Sequence

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

Let $$S \left({x}\right) = \sum_{n=0}^\infty a_n \left({x - \xi}\right)^n$$ be a power series about $$\xi$$.

Then the radius of convergence $$R$$ of $$S \left({x}\right)$$ is given by:
 * $$\frac 1 R = \limsup_{n \to \infty} \left|{a_n}\right|^{1/n}$$;
 * $$\frac 1 R = \lim_{n \to \infty} \left|{\frac {a_{n+1}} {a_n}}\right|$$.

If either:
 * $$\frac 1 R = \limsup_{n \to \infty} \left|{a_n}\right|^{1/n} = 0$$; or
 * $$\frac 1 R = \lim_{n \to \infty} \left|{\frac {a_{n+1}} {a_n}}\right| = 0$$

then the radius of convergence is infinite and therefore the interval of convergence is $$\R$$.

Proof of First Result

 * $$\frac 1 R = \limsup_{n \to \infty} \left|{a_n}\right|^{1/n}$$:

From the $n$th root test, $$S \left({x}\right)$$ is convergent if $$\limsup_{n \to \infty} \left|{a_n \left({x - \xi}\right)^n}\right|^{1/n} < 1$$.

Thus:

$$ $$ $$

The result follows from the definition of radius of convergence.

Proof of Second Result

 * $$\frac 1 R = \lim_{n \to \infty} \left|{\frac {a_{n+1}} {a_n}}\right|$$.

From the ratio test, $$S \left({x}\right)$$ is convergent if $$\lim_{n \to \infty} \left|{\frac {a_{n+1} \left({x - \xi}\right)^{n+1}}{a_n \left({x - \xi}\right)^n}}\right| < 1$$.

Thus:

$$ $$ $$

The result follows from the definition of radius of convergence.