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Prove that $\sum x^j$ is differentiable on (-1,1), and $$\frac{d}{dx} \sum x^j = \sum (j+1) x^j$$

I am able to prove that $\sum x^j$ converges uniformly to $\frac{1}{1+x}$. However, how do I get this derivative? It doesn't seem to follow traditional derivative rules.

Matthew Towers
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kiwifruit
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  • I don't know how formal of a proof you want, but as far as getting the derivative from typical rules: $$\sum_{j\ge0}(j)x^{j-1} = \sum_{j\ge1}(j)x^{j-1} = \sum_{j\ge0}(j+1)x^{j}$$ – apnorton Apr 19 '14 at 16:40
  • But how does this derivative come by differentiating $\sum x^j$? – kiwifruit Apr 19 '14 at 16:41
  • The derivative of a sum is the sum of the derivatives. This extends to series as well. – apnorton Apr 19 '14 at 16:42
  • @kiwifruit since a serie converges absolutely on its open domain of convergence, you can make the limit that comes from the derivative and the one that comes from the infinite sum commute to show that the derivatives of the infinite sum is the infinite sum of the derivative of each term. – yago Apr 19 '14 at 16:45
  • @kiwi Sorry, I misled you, its absolutely convergent on any intervals $(a,b)$ with $-1 < a < b < 1$ but not on the whole $(-1,1)$. However, this is sufficient to show our claim on derivative – yago Apr 19 '14 at 16:57

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It does follow the traditional derivative rules. But we need to treat the case $j=0$ separately, since the traditional rule for $x^j$ does not apply to it. $$\frac d {dx} \sum_{j=0}^\infty x^j = \frac d {dx}\left(x^0 + \sum_{j=1}^\infty x^j\right) = \frac d {dx}\left(1 + \sum_{j=0}^\infty x^{j+1}\right) = \sum_{j=0}^\infty (j+1)x^{j}$$

Klaas van Aarsen
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