On the ratio test for power series

  • #1
schniefen
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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:

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.
 
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  • #2
It is fairly easy to show from the definition of the limit of a sequence and the definition of a sequence diverging to [itex]+\infty[/itex] that [tex]
\lim_{n \to \infty} |b_n| = \infty\quad\Rightarrow\quad\lim_{n \to \infty} \frac{1}{|b_n|} = 0[/tex] and [tex]
\lim_{n \to \infty} |b_n| = 0 \quad\Rightarrow\quad \lim_{n \to \infty} \frac{1}{|b_n|} = \infty.[/tex]
 
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  • #3
pasmith said:
It is fairly easy to show from the definition of the limit of a sequence and the definition of a sequence diverging to [itex]+\infty[/itex] that [tex]
\lim_{n \to \infty} |b_n| = \infty\quad\Rightarrow\quad\lim_{n \to \infty} \frac{1}{|b_n|} = 0[/tex] and [tex]
\lim_{n \to \infty} |b_n| = 0 \quad\Rightarrow\quad \lim_{n \to \infty} \frac{1}{|b_n|} = \infty.[/tex]
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!
 

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