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

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert{a_n \left({x - \xi}\right)^n}\right\vert^{1/n}\)	\(<\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle \left\vert{a_n}\right\vert^{1/n} \left\vert{x - \xi}\right\vert\)	\(<\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle \left\vert{a_n}\right\vert^{1/n}\)	\(<\)	\(\displaystyle \frac 1 {\left\vert{x - \xi}\right\vert}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

The result follows from the definition of radius of convergence.

$\blacksquare$

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert{\frac {a_{n+1} \left({x - \xi}\right)^{n+1} }{a_n \left({x - \xi}\right)^n} }\right\vert\)	\(<\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle \left\vert{\frac {a_{n+1} }{a_n} }\right\vert \left\vert{x - \xi}\right\vert\)	\(<\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle \left\vert{\frac {a_{n+1} }{a_n} }\right\vert\)	\(<\)	\(\displaystyle \frac 1 {\left\vert{x - \xi}\right\vert}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

The result follows from the definition of radius of convergence.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 15.2$