Cauchy-Hadamard confusion

Question

I was reading the proof here, but did not understand a few key points, so I'm asking here for the following clarifications:

(1) I don't see why $|c_n|\le (t+\varepsilon)^n$ for all but a finite number of $n$. Is there some theorem for this result in analysis?

(2) Why does (1) imply that $\sum c_nz^n$ converges for $|z|<1/(t+\varepsilon)$? We are not talking about absolute convergence here, I think, but the moduli seem to point to absolute convergence. There is probably a connection between the fact that the modulus of a complex sequence is less than one and its power series' convergence.

(3) Why does the fact that $|c_n|=(t+\varepsilon)^n$ imply that $\limsup\limits_{n\to\infty}|c_n|^{1/n}=1/R$?

(4) Why is $|c_n|\ge (t-\varepsilon)^n$ for infinitely many $n$?

(5) Finally, I've seen another statement of this theorem, with the difference that $R$ was defined there as $R:=\liminf\limits_{n\to\infty}|c_n|^{-1/n}$. Why are these statements equal?

Thank you very much!

Answer 1

Shaun is completely correct about 1. The $\limsup$ of a sequence $a_n$ is by definition the lowest value such that only finitely many of the $a_n$ are above it. Conversely, the $\liminf a_n$ is the largest value such that only finitely many values of $a_n$ are below it. Now suppose $f(x)$ is a strictly decreasing continuous function of $x$. Then $f(a_n)$ reverses every order relationship in $a_n$. If $L = \limsup a_n$, so only finitely many $a_n > L$, then only finitely many $f(a_n)$ will be below $f(L)$. For any value $Q$ less than $L$, there are infinitely many $a_n$ greater than $Q$. But that means that there are infinitely many $f(a_n)$ that are less than $f(Q)$. Which proves that $f(L) = \liminf f(a_n)$.

In particular $f(x) = \frac 1x$ is a continuous decreasing function on the positive reals. Therefore, when $\sqrt[n]{|c_n|}$ is a positive sequence, we have $$\liminf \frac 1 {\sqrt[n]{|c_n|}} = \frac 1{\limsup \sqrt[n]{|c_n|}}$$

The Wikipedia version is preferred for a simple reason: that "positive" sequence requirement. When infinitely many of the $c_n = 0$, the expression on the left requires the use of actual infinities for $\frac 10$ to define it. Of course, this can work and gives the same answer in the end because the $\liminf$ will only give $\infty$ if there is no lower limit point. But the $\limsup$ version does not require an actual infinity to work (the case when the $\limsup = 0$ can be treated as a conceptual infinity only).

As for (2), we have $|c_n| \le (t + \epsilon)^n$ and $|z| < 1/(t + \epsilon)$, so defining $k = |z|(t + \epsilon)$, this means $k < 1$. Therefore $$\sum |c_nz^n| = \sum |c_n||z^n| \le \sum (t + \epsilon)^n|z|^n \le \sum k^n$$ but since $k<1$, that geometric series is known to converge, and so $\sum c_nz^n$ is absolutely convergent for $|z| < 1/(t + \epsilon)$. This shows that the radius of convergence is at least $R$.

For (3), as Shaun says, you appear to have it backwards. Because $t = 1/R$ is the limit supremum, $t - \epsilon$ is a number below the limit supremum and therefore by definition, there are infinitely many $c_n$ with $\sqrt[n]{|c_n|} > t - \epsilon$ and therefore infinitely many $c_n > (t - \epsilon)^n$. Therefore when $z = 1/(t - \epsilon)$ then there are infinitely many $|c_nz^n| > (t - \epsilon)^n(t - \epsilon)^-n = 1$. So it cannot be that $\lim_n c_nz^n = 0$, wich is a requirement for the series to converge. Since any value greater than $R$ can be expressed as $1/(t - \epsilon)$ for some positive $\epsilon$, no value greater than $R$ can be the radius of convergence. So $R$ must be the radius of convergence.

Note that this proof does not itself show that the series diverges for every $|z| > R$, only that there is a least one such $z$ for any radius greater than $R$. However, it would take only a minor change to prove the fuller result.

Answer 2

1. $t$ is defined by $\limsup_{n\to\infty}|c_n|^\frac{1}{n}$. If it were true that $|c_n|^\frac{1}{n}\geq t+\epsilon$ for infinitely many $n$, then the lim sup would also be $\geq t+\epsilon$.

2. Let $N$ be such that $|c_n|^\frac{1}{n}N$, and assume that $|z|<1/(t+\epsilon)$. Then: $$\sum_{n} c_{n} z^n=\sum_{n =1}^Nc_{n}z^n+\sum_{n=N+1}^\infty c_nz^n<\sum_{n=1}^Nc_nz^n+\sum_{n=N+1}\frac{c_n}{(t+\epsilon)^n}<\sum_{n =1}^Nc_{n}z^n+\sum_{n>N}c_n$$ The sum of the last series is finite because both series converge; one is a finite sum, the other converges because $\sum_nc_n$ converges.

3. I'm a bit confused what you mean by this. The proof defines $\limsup_{n\to\infty}|c_n|^\frac{1}{n}=1/R$, not the other way around.

4. The reasoning is similar to 1.

5. Important not to get the notation from different proofs confused: in those proofs $R$ was the radius of convergence. In the Wikipedia article $1/R$ is the radius. Keeping that in mind see here.

Hope this helps. Let me know if you want further clarification.