I have to compute this series :
$$\sum_{k=0}^\infty (k+1)x^{2k}$$
First, I have $$|x|<1$$ but then I don't know how to begin ...
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4$x^{2k}=(x^2)^k$. Now, see this. – David Mitra Dec 17 '14 at 09:04
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1We usually say series, even when its just one. – MathMajor Dec 17 '14 at 09:04
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Use the fact that this is a power series with radius of convergence $r=1$. Thus, for $\lvert y\rvert < r$, $$ \sum_{k=0}^\infty (k+1)y^k = \frac{d}{dy} \sum_{k=0}^\infty y^{k+1} = \frac{d}{dy} \sum_{k=1}^\infty y^{k} $$ and $$ \sum_{k=1}^\infty y^{k} = \frac{y}{1-y}. $$ Now, you can compute the derivative of $\frac{y}{1-y}$, and evaluate it on $x^2$ (which has $x^2 < r$ indeed).
Clement C.
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Thanks ! But is the first equality trivial ? I mean getting $d/dy$ out of the sum ? Or is it because the series converge for $|y|<1$ ? – dcholleton Dec 17 '14 at 09:14
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It is not trivial -- it holds because power series (in particular) uniformly converge in their disk of convergence. – Clement C. Dec 17 '14 at 09:15