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- Homework Statement
- I have a question about a proof of the ratio test for power series.

- Relevant Equations
- The following definition is relevant. If the power series ##\sum_{n=0}^\infty a_n (x-c)^n## converges for ##|x-c|<R## and diverges for ##|x-c|>R##, then ##0\leq R\leq \infty## is called the radius of convergence of the power series.

In these lecture notes, there is the following theorem and proof:

I'm confused about "...the power series converges if ##0\leq r<1##, or ##|x-c|<R##...". In other words, why is ##|x-c|<R## equivalent to ##0\leq r<1##?

I guess the author reasons as follows. If $$R=\lim _{n\to \infty }\left|\frac{a_n}{a_{n+1}}\right|,$$ then for ##0<R<\infty##, we have $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim_{n\to\infty}\frac1{\left|\frac{a_{n}}{a_{n+1}}\right|}=\frac1{R}.\tag1$$ So $$r=|x-c|\frac{1}{R}$$ and then clearly ##0\leq r<1## is equivalent to ##|x-c|<R##. But what about when ##R=0## and ##R=\infty##? Then ##(1)## is not defined/not valid. This confuses me and I'd be grateful for a comment or two.

Theorem 10.5.Suppose that ##a_n\neq0## for all sufficiently large ##n## and the limit $$R=\lim _{n\to \infty }\left|\frac{a_n}{a_{n+1}}\right|$$ exists or diverges to infinity. Then the power series ##\sum_{n=0}^\infty a_n (x-c)^n## has radius of convergence ##R##.

Proof.Let $$r=\lim_{n\to\infty}\left|\frac{a_{n+1}(x-c)^{n+1}}{a_n(x-c)^n}\right|=|x-c|\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|.$$ By the ratio test, the power series converges if ##0\leq r<1##, or ##|x-c|<R##, and diverges if ##1<r\leq\infty##, or ##|x-c|>R##, which proves the result.

I'm confused about "...the power series converges if ##0\leq r<1##, or ##|x-c|<R##...". In other words, why is ##|x-c|<R## equivalent to ##0\leq r<1##?

I guess the author reasons as follows. If $$R=\lim _{n\to \infty }\left|\frac{a_n}{a_{n+1}}\right|,$$ then for ##0<R<\infty##, we have $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim_{n\to\infty}\frac1{\left|\frac{a_{n}}{a_{n+1}}\right|}=\frac1{R}.\tag1$$ So $$r=|x-c|\frac{1}{R}$$ and then clearly ##0\leq r<1## is equivalent to ##|x-c|<R##. But what about when ##R=0## and ##R=\infty##? Then ##(1)## is not defined/not valid. This confuses me and I'd be grateful for a comment or two.