[Math] Ratio test for radius of convergence

calculuspower seriesreal-analysissequences-and-series

Consider the power series $\sum c_nz^n$. Here is a theorem-definition from baby Rudin:
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Now, Theorem 3.37 implies that $Q:=\frac{1}{\lim \sup |c_{n+1}/c_n|}\le R$. So if $Q$ is infinite, then it is equal to $R$. But if $Q$ is finite, it need not be equal to $R$. Thus, in general, $Q\ne R$.

However, I don't understand why one can't mimic the proof of 3.39 to prove actually that $Q=R$ always. Namely,

$$\lim \sup |\frac{c_{n+1}z^{n+1}}{c_nz^n}|=|z|\lim\sup |\frac{c_{n+1}}{c_n}|,$$

and if the latter is $< 1$, then the series converges (in that case $|z|<Q)$; if the latter is $> 1$, then the series diverges (and $|z|> Q$ in that case).

So it seems we can define the radius of convergence as $Q$. But, as noted above, this is not true. Where is a mistake in my reasoning?

Best Answer

It's not true that if $|z|\limsup|c_{n+1}/c_n|>1$ then the series diverges.

In fact there are big problems here if $c_n=0$ for some $n$. But that's not the only problem. Consider the sequence

$$(c_n)=(1,2,1/4,2/4,1/9,2/9\dots),$$

or $c_{2k}=1/(k+1)^2$, $c_{2k+1}=2/(k+1)^2$. Let $z=1$. Then $|z|\limsup|c_{n+1}/c_n|=2$, but $\sum c_nz^n$ converges.

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