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It's more of a confusion for me know: in what cases sums of infinite series can't be integrated and differentiated? I understand that e.g. $$ \sum_{k=0}^{\infty}z^k = \frac{1}{1-z} $$

has the remainder term of for $n+1, n+2...$ equal to $\frac{z^{n+1}}{1-z}$, so it does not converge uniformly for $|z|<1$. But nevertheless I can manipulate for $z<1$: $$ \sum_{k=0}^{\infty}k z^k = z\sum_{k=0}^{\infty} \frac{\partial}{\partial z} (z^k) $$

and so on. What do I miss here?

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3 Answers

up vote 1 down vote accepted

What you are missing is that, even though the series does not converge uniformly for $|z|<1$, it does converge uniformly for $|z|<r$, for any $r<1$. And that is all you need for the usual manipulations.

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I see. What is the difference between $|z|<1$ and $|z|<r<1$? –  user19821 Feb 21 '12 at 6:54
    
@user19821: It doesn’t converge uniformly on $(-1,1)$, but it converges uniformly on $(-r,r)$ for any positive $r<1$. –  Brian M. Scott Feb 21 '12 at 7:14
    
@BrianM.Scott: yes, I see that. The question is, why? –  user19821 Feb 21 '12 at 7:23
    
@user19821: I thought that you wanted to know what the difference is, so that’s the question that I answered. I’m not sure what you’re asking now. Are you asking why it converges uniformly on $(-r,r)$ if $0<r<1$? –  Brian M. Scott Feb 21 '12 at 7:37
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The convergence can not be uniformly for $|z|<1$ because the limit function is unbounded on that domain. –  Dirk Feb 21 '12 at 8:28
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In your example, the differentiation works because the series does converge uniformly on any interval $(-1+\delta,1-\delta)$, for any $\delta$ between $0$ and $1$. And also the series of the derivatives converges uniformly on those intervals.

The conclusion is that for your series you can differentiate and integrate term by term for any $z$ in $(-1,1)$.

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In addition to the other answers: The right notion here is locally uniform convergence of a sequence of analytic functions.

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