Theorem 8.2 of Walter Rudin's Principles of mathematical analysis is stated as :
Suppose $\Sigma c_n$ converges. Put $f(x)=\Sigma_{n=0}^{\infty}c_nx^n \ \ \ \ \ \ (-1<x<1)$. Then $lim_{x\to 1} f(x)= \Sigma_{n=0}^{\infty} c_n$.
I do not understand why does the power series of $x$ have to converge at all.
We cannot apply the comparison test for series here as the terms may be negative for the power series.
Applying Cauchy condition I get :
$|c_nx^n+….+c_mx^n| \leq |c_nx^n|+…+|c_mx^m| \leq |c_n|+…+|c_m|$ for $|x|<1$.
We are not given that $\Sigma c_n$ converges absolutely to bound the first term in the above inequality.
The power series must converge because in an application of the theorem, the following is stated :
If $\Sigma a_n, \Sigma b_n, \Sigma c_n $ converge to A,B and C resp. and if $c_n=a_0b_n +…+a_nb_0$ then C=AB. Let $f(x) = \Sigma a_nx^n,g(x) = \Sigma b_nx^n,h(x) = \Sigma c_nx^n$ for $0\leq x\leq 1$. For $x<1$ series converges absolutely.
This application is also not clear to me. Please help.
Best Answer
For any bounded sequence $(c_n)$ the series $\sum c_nx^{n}$ converges for $|x|<1$ (by comparison with the geomertic series $\sum |x|^{n}$). If $\sum c_n$ converges then $c_n\to 0$ which makes $(c_n)$ bounded.