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