Interval of convergence of power series from ratio test

zenterix · Mar 13, 2024

If ##a_n\neq 0## for all ##n##, consider the limit

$$\lim\limits_{n\to\infty} \left | \frac{a_{n+1}x^{n+1}}{a_nx^n} \right | = \lim\limits_{n\to\infty}\left |\frac{a_{n+1}}{a_n}\right | |x|=L$$

The ratio test asserts that this series converges if ##L<1## and diverges if ##L>1##.

These considerations yield the formula

$$R=\lim\limits_{n\to\infty}\left |\frac{a_{n+1}}{a_n}\right |\tag{1}$$

if this limit exists (we put ##R=\infty## if ##|a_n/a_{n+1}|\to\infty##).

Regardless of whether this formula can be used or not, it is known that ##R## always exists; and if ##R## is finite and nonzero, then it determines an interval of convergence ##-R<x<R## such that inside the interval the series converges and outside the interval it diverges.

I'm a bit confused by this snippet.

Why does (1) determine the radius of convergence?

zenterix · Mar 13, 2024

I guess it is because of the following

$$\lim\limits_{n\to\infty}\left |\frac{a_{n+1}}{a_n}\right | |x| \tag{1}$$

$$=\frac{1}{\lim\limits_{n\to\infty} \left |\frac{a_n}{a_{n+1}}\right |} |x|\tag{2}$$

$$=\frac{|x|}{R}\tag{3}$$

Then, for ##-R<x<R## we see that the expression in (3) is less than 1 and so the power series converges.

PeroK · Mar 13, 2024

I guess you've answered your own question.

Interval of convergence of power series from ratio test

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