I am currently studying undergraduate calculus and I've encountered this specific problem:
Prove that if a power series converges absolutely at one endpoint of its interval of convergence, then the power series is absolutely convergent at each endpoint.
So we need to prove that if $\sum_{n=0}^\infty C_n(x-a)^n$ is absolutely convergent at one endpoint, then it is absolutely convergent at the other endpoint
So far I can only do a proof for the case when $a=0$ and I want to ask if it is correct
Given the infinite series $f(x)=\sum_{n=0}^\infty C_nx^n$ which is absolutely convergent at $-R<x<R$
- consider the case when $f(R)$ is absolutely convergent. Then $\sum_{n=0}^\infty \vert C_nR^n\vert$ is convergent.
- Now, we check $f(-R)$ for absolute convergence. $f(-R)=\sum_{n=0}^\infty \vert C_n(-R)^n \vert =\sum_{n=0}^\infty \vert C_n(R)^n(-1)^n \vert =\sum_{n=0}^\infty \vert C_n(R)^n \vert $
- Since the absolute value of the power series at $x=-R$ is equal to the absolute value of the power series at $x=R$, which is convergent, therefore if $f(R)$ is absolutely convergent, then $f(-R)$ is also absolutely convergent.
- now consider the case when $f(-R)$ is absolutely convergent. Then $\sum_{n=0}^\infty \vert C_n(-R)^n \vert$ is absolutely convergent.
- Now, we wish to check if the absolute value of the power series $f(R)=\sum_{n=0}^\infty \vert C_n(R)^n \vert$ is also absolutely convergent.
- Further evaluation of the absolute value of the power series $f(-R)$ gives us: $f(-R)= \sum_{n=0}^\infty \vert C_n(-R)^n \vert =\sum_{n=0}^\infty \vert C_n(R)^n(-1)^n \vert =\sum_{n=0}^\infty \vert C_n(R)^n \vert = f(R)$
- Since the absolute value of the power series evaluated at $x=R$ is equal to the absolute value of the power series evaluated at $x=-R$ which is convergent, we can conclude that the power series $f(R)$ is absolutely convergent. Therefore if the $f(-R)$ is absolutely convergent, then $f(R)$ is also absolutely convergent. end of proof
I am only a beginner in doing proofs and that is why I will be grateful to hear any corrections on my proof.
Lastly, I wish to know how to prove the more general case when $a\neq0$ which is $\sum_{n=0}^\infty C_n(x-a)^n$
The idea is correct. However, the proof should have ended after the fourth $\bullet$. In fact, since you define $f(x)$ as $\sum_{n=0}^\infty C_nx^n$, there is no reason for you to assert that $f(R)=\sum_{n=0}^\infty\left\lvert C_nR^n\right\rvert$. And you are assuming that the series $\sum_{n=0}^\infty C_nR^n$ is absolutely convergent, which means that the series $\sum_{n=0}^\infty\left\lvert C_nR^n\right\rvert$ converges.