- #1

zenterix

- 524

- 74

- Homework Statement
- While reading a chapter from Simmons' "Differential Equations with Applications and Historical Notes" that reviews power series, I did not really understand a paragraph that tried to explain how to obtain the radius of convergence of a power series using the ratio test.

- Relevant Equations
- Consider the following power series in ##x##

$$\sum\limits_{n=0}^\infty a_nx^n$$

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##.

I'm a bit confused by this snippet.

Why does (1) determine the radius of convergence?

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

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?

Last edited: