Use the differentiation property of power series to find the derivative of $$f(x) =\dfrac{1}{1-x}=\sum\limits_{k\,=\,0}^{\infty }x^{k}$$

Solution The function $f(x) =\dfrac{1}{1-x}$, defined on the open interval $(-1,\,1)$, is represented by the power series $$f(x) =\frac{1}{1-x}=1+x+x^{2}+ \cdots +x^{n}+\cdots = \sum\limits_{k\,=\,0}^{\infty }x^{k}$$

Using the differentiation property, we find that $$f^{\prime} (x) =\frac{1}{(1-x)^{2}}=1+2x+3x^{2}+\cdots +nx^{n-1}+\cdots =\sum\limits_{k\,=1}^{\infty }kx^{k-1}$$

whose radius of convergence is $1$.