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Let $\displaystyle \sum_{n=2}^\infty a_nx^n$ be a power series with radius of convergence $R>0$ prove, $\displaystyle \int_0^x \left (\sum_{n=0}^\infty a_nt^n \right ) \ dt = \sum_{n=0}^\infty a_n \dfrac{x^{n+1}}{n+1} $

The question gives a hint to use the mean value theorem, I have given it an attempt,

define $f(x) = \displaystyle \sum_{n=0}^\infty a_nx^n$,

$g(x) = \displaystyle \int_0^x f(t) \ dt - \sum_{n=0}^\infty a_n \dfrac{x^{n+1}}{n+1}$ then the derivative $g'(x) = 0$, so $g(x)$ is constant,

how would I use the mean value theorem from here? I suppose I could say there exists a point $c \in (-R,R)$ s.t. $g'(c) = \dfrac{g(-R) - g(R)}{-2R} = 0$ but what information does that give me?

Warz
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  • Please avoid using display style in titles. It makes them too big. – Cameron Buie Mar 03 '14 at 20:18
  • I don't see what use the mean value theorem could be here. For $\lvert x\rvert < R$, the power series converges uniformly on the interval of integration, so we can interchange summation and integration. Or, we can, like you did, see that $g$ is constant, and since evidently $g(0) = 0$, the result follows. No MVT needed anywhere. – Daniel Fischer Mar 03 '14 at 20:20
  • @DanielFischer you need MVT to conclude that there exists $c$ such that $g(c)=0$ – gt6989b Mar 03 '14 at 20:22
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    @gt6989b How so? $g(0) = 0$ is immediate. – Daniel Fischer Mar 03 '14 at 20:30
  • @DanielFischer Thanks, that does work, but really curious why the hint was for the MVT – Warz Mar 03 '14 at 20:30
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    I have absolutely no idea. – Daniel Fischer Mar 03 '14 at 20:32
  • @DanielFischer sorry, just being dumb – gt6989b Mar 03 '14 at 20:32
  • I'm realy curious to know how we can use the MVT to prove the equality. Notice that the common way to have the result is to prove the uniform convergence and then integrate term by term. –  Mar 03 '14 at 20:59

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