Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'd like to prove the following proposition:

The power series $\sum_{j=0}^{\infty} a_j(x - c)^j$ and the series $\sum_{j=0}^{\infty} \frac{a_j}{j+1}(x - c)^{j+1}$ obtained from term by term integration have the same radius of convergence, and the function $F$ defined by $F(x) = \sum_{j=0}^{\infty} \frac{a_j}{j+1}(x - c)^{j+1}$ on the common interval of convergence satisfies $F'(x) = \sum_{j=0}^{\infty} a_j(x - c)^j = f(x)$.

Does this occur as a result of $\sum_{j=0}^{\infty} a_j(x - c)^j$ and $\sum_{j=0}^{\infty} \frac{a_j}{j+1}(x - c)^{j+1}$ being real analytic? (Are they real analytic?)

share|improve this question
    
Cauchy-Hadamard theorem. –  davin Mar 16 '12 at 9:27

1 Answer 1

Recall that the radius of convergence $R$ of the series $\sum\limits_na_nx^n$ is characterized by the fact that, for every $|x|\lt R$, $a_nx^n\to0$ geometrically fast while, for every $|x|\gt R$, the sequence $(a_nx^n)_n$ is unbounded.

Introduce $b_n=a_n/(n+1)$ and let $R'$ denote the radius of convergence of the series $\sum\limits_nb_nx^n$.

For every $|x|\lt R$, $a_nx^n\to0$ hence $b_nx^n\to0$. This shows that $R'\geqslant R$. For every $|x|\gt R$, $|a_nx^n|\geqslant1$ infinitely often hence $|b_nx^n|\geqslant1/(n+1)$ infinitely often. In particular, the sequence $(b_nx^n)_n$ does not converge to $0$ geometrically fast, hence $R'\leqslant R$.

Hence $R'=R$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.