Convergence of power series beyond radius of convergence?

Question

In rudin analysis text book, 3.44 theorem , it says about the convergence on the boundary of the circle of the power series, the theorem is roughly:-

Suppose the radius of convergence of $\Sigma c_nz^n $ and suppose the $c_n$ series is monotonically decreasing$c_n->0 $ as n-> infinity. Then $\Sigma c_nz^n $ converges at every point of mod z = 1 except possibly at z=1.

My question is :-

Why would not the same proof applies under the given hypothesis for when the radius of convergence is $<1$ ?

But if the proof is true for that case too then is not it a contradiction to 3.39 theorem (as there would be points mod z = 1 may not be z=1 such that the series converges that is the series is converging beyond the radius of convergence)(so where am I mistaking?)(saying about the existence of a radius of convergence) ?

Well and in 3.42 theorem ('partial sums of series of $a_n$ is bounded' and ' $ b_n$ decreases monotonically' and $b_n->0$ ' the $\Sigma a_nb_n$ converges.) saying about convergence of series of $a_nb_n$ do the series $a_n$ needs to converge ?

Answer 1

In an entire series, say

$$\sum c_n z^n,$$ if the radius of convergence is $1$, the modified series

$$\sum c_n\left(\frac zr\right)^n=\sum\frac{c_n}{r^n}z^n$$ has the radius of convergence $r$. Hence any proof can be easily adapted to non-unit radii.

Answer 2

In the proof of Theorem 3.44, Rudin appeals to Theorem 3.42. He says,"the hypotheses of Theorem 3.42 are satisfied." This is not true if the radius of convergence is less than $1$. The first hypothesis of theorem 3.42 is, "the partial sums form a bounded sequence." If the radius of convergence is less than $1$ and $|z|=1$ then the partial sums are certainly not bounded.