Interval of convergence of power series from ratio test

[zenterix]

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 $L>1$.

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]

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]

I guess you've answered your own question.