Answer
$2$
Work Step by Step
Write the Taylor series for the function as: $(1-x)^{-2}=1+2x+3x^2+4x^3+....$
Now, we have:
$\Sigma_{k=1}^{\infty} k (\dfrac{1}{2})^k=\Sigma_{k=1}^{\infty} \dfrac{k}{2^k}\\=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+......\\=\dfrac{1}{2}(1+2(\dfrac{1}{2})+3(\dfrac{1}{2})^2+4(\dfrac{1}{2})^3+......)$
Therefore, we have:
$\Sigma_{k=1}^{\infty} k (\dfrac{1}{2})^k=\dfrac{1}{2}[\dfrac{1}{1-(\dfrac{1}{2})^2}]\\=2$
