On the ratio test for power series

[schniefen]

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:

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.

 

[pasmith]

It is fairly easy to show from the definition of the limit of a sequence and the definition of a sequence diverging to $+\infty$ that $$\lim_{n \to \infty} |b_n| = \infty\quad\Rightarrow\quad\lim_{n \to \infty} \frac{1}{|b_n|} = 0$$ and $$\lim_{n \to \infty} |b_n| = 0 \quad\Rightarrow\quad \lim_{n \to \infty} \frac{1}{|b_n|} = \infty.$$

 

[schniefen]

It is fairly easy to show from the definition of the limit of a sequence and the definition of a sequence diverging to $+\infty$ that $$\lim_{n \to \infty} |b_n| = \infty\quad\Rightarrow\quad\lim_{n \to \infty} \frac{1}{|b_n|} = 0$$ and $$\lim_{n \to \infty} |b_n| = 0 \quad\Rightarrow\quad \lim_{n \to \infty} \frac{1}{|b_n|} = \infty.$$

Thank you for replying.

Ok, I guess in computing $r$ the author assumed $x\neq c$. Because if $R=0$, then, as you write, $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim_{n\to\infty}\frac1{\left|\frac{a_{n}}{a_{n+1}}\right|}=\infty.$$ And so $r=[|x-c|\cdot\infty]=\infty$, i.e. we have divergence of the power series for $R=0$ for all values of $x\neq c$. Thanks!