Using this theorem:

Let $(a_k)$ be a sequence in $\Bbb R$ and let $x_0\in \Bbb R$:

  • The power series: $$\sum_{k=0}^\infty a_k(x-x_0)^k \qquad \text{and} \qquad \sum_{k=0}^\infty ka_k(x-x_0)^{k-1} $$ converge uniformly in $[x_0-r,x_0+r]$ for all $r\in (0,R)$ where $R$ is the radius of convergence of $\sum_{k=0}^\infty a_k(x-x_0)^k $.

  • The function $$f(x):= \sum_{k=0}^\infty a_k(x-x_0)^k $$ is continously diferentiable in $(x_0-R,x_0+R)$, and its derivative is given by: $$f'(x)=\sum_{k=0}^\infty ka_k(x-x_0)^{k-1} $$

calculate the following: $$\tag {1} \sum_{k=0}^\infty \frac {k}{2^{k-1}}$$

So, I'm confused on what do I have to do. I'm gessing that we have to find the $x_0$ of this: $$\sum_{k=0}^\infty \frac {k}{2^{k-1}} (x-x_0)^k$$ I know that $\mathbf {(1)}$ converges to $4$, but I didn't used the theorem, and using the definition of radius of convergence, (superficially) I found that $R:=\sup\{r\in(-2,2) \mid \sum_{k=0}^\infty \frac {k}{2^{k-1}}r^k<\infty \text{ (the sum converges in $\Bbb R$)} \}$. I don't see what they're asking me to do, and why/how to use the theorem.

  • 1
    $\begingroup$ The question is asking "to what value does $(1)$ converge?" But, how do you know that it converges to $4$? The formula you probably used is the derivative of the geometric series formula... And, derivatives are in that theorem... $\endgroup$ – apnorton Sep 29 '13 at 19:55

Hint: Take $x = 1/2$ and $x_0 = 0$. $\displaystyle \sum_{k=0}^\infty k (x - x_0)^{k-1}$ is the derivative of ...

  • $\begingroup$ Yesterday I put a comment, but now is not important and I don't think it contributes anything, that's why I deleted it. Thanks for the hint! $\endgroup$ – Ana Galois Oct 1 '13 at 3:28

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