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$$R=\dfrac{1}{\limsup_{n\to\infty}|a_n|^{\frac{1}{n}}} $$

I've been reading this paper regarding asymptotic growth, and I stumbled upon this relation between the radius of convergence and the root test. From my knowledge, the root test shows if the series converge of diverge with conditions $$ \limsup_{n\to\infty}|a_n|^{\frac{1}{n}}=L $$ If $L<1$, series converge, if $L>1$ series diverge, and if $L=1$ the series may be divergent, conditionally convergent, or absolutely convergent.

How have they connected the root test to the radius of convergence of the series?

Sine of the Time
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Mihailo
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1 Answers1

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Root test.
Consider a series $\sum_{n=1}^\infty a_n$. Let $$ L := \limsup_{n\to\infty} |a_n|^{1/n} $$ Here $L \in [0,+\infty]$. Then
$\bullet $ If $L<1$, the series $\sum a_n$ converges absolutely [by comparison with a geometric series].
$\bullet $ If $L>1$, the series $\sum a_n$ diverges [since $a_n$ does not converge to $0$].
$\bullet $ If $L=1$, the test is inconclusive.

Now consider a power series $\sum c_n x^n$. Let $$ R:=\dfrac{1}{\limsup_{n\to\infty}|c_n|^{1/n}} . $$ Fix a value of $x$ with $0 < |x| < +\infty$. Let $a_n = c_n x^n$. Then $$ L = \limsup_{n\to\infty} |a_n|^{1/n} = \limsup_{n\to\infty} |c_n x^n|^{1/n} = \limsup_{n \to \infty} |c_n|^{1/n} |x| = \frac{|x|}{R} . $$ [we use the conventions $\frac{|x|}{0} = +\infty$ and $\frac{|x|}{+\infty} = 0$.]

So:
$\bullet $ If $|x|<R$, then $L<1$ so $\sum c_n x^n$ converges absolutely.
$\bullet $ If $|x|>R$, then $L>1$ so $\sum c_n x^n$ diverges.

These two hold for all $x$ with $0<|x|<+\infty$. So the "radius of convergence" of the series $\sum c_n x^n$ is $R$.

GEdgar
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