Convergence of power series beyond radius of convergence?

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 ?

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.

• Well I did not get your answer actually could you please explain? – Bijayan Ray Apr 22 at 15:57
• @BijayanRay: what part don't you understand ? – Yves Daoust Apr 22 at 16:00
• I am saying that the power series is converging beyond the radius of convergence my arguments are in the question where am I wrong? – Bijayan Ray Apr 22 at 16:02
• @BijayanRay: even more confuse than before, sorry. – Yves Daoust Apr 22 at 16:04
• Going through the proof of the theorem might make my question clearer. – Bijayan Ray Apr 22 at 16:04

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.

• In the inequality the one given in rudin, plug 1 in place of the mod z (after applying triangle inequality offcourse in the numerator) and hence the rhs of inequality remains unchanged, so of course the $A_n$ . Is my reasoning wrong anywhere in that case please point it and clarify it, thanks for your time and your answer. – Bijayan Ray Apr 22 at 18:08