Radius of Convergence from Limit of Sequence

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

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

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

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

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

Proof of First Result

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

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

Thus:

The result follows from the definition of radius of convergence.

Proof of Second Result

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

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

Thus:

The result follows from the definition of radius of convergence.

Also see

 * Radius of Convergence from Limit of Sequence/Complex Case