Find a few derivatives, and calculate their values at a=0. Remember that there is a (-1) that comes from the derivative of (1-x)

\(f(x)=(1-x)^{-2}\)

\(f'(x)=-2(1-x)^{-3}(-1)\)

\(=2(1-x)^{-3}\)

\(f''(x)=(-3)2(1-x)^{-4}(-1)\)

\(=(3)(2)(1-x)^{-4}\)

\(f'''(x)=(-4)(3)2(1-x)^{-5}(-1)\)

\(=(4)(3)(2)(1-x)^{-5}\)

\(f^((4))(x)=(-5)(4)(3)(2)(1-x)^{-6}(-1)\)

\(=(5)(4)(3)(2)(1-x)^{-5}\)

\(f^n(x)=(n+1)!(1-x)^{-(n+2)}\)

Plug everything into Maclaurin general form:

\(f(x)=f(a)+\frac{f'(a)}{1!}x+\frac{f''(a)}{2!}x^2+\frac{f'''(a)}{3!}x^3+...\)

\(f(x)=1+2x+\frac{3!}{2!}x^2+\frac{4!}{3!}x^3+\frac{5!}{4!}x^4+...\)

\(=1+2x+3x^2+4x^3+5x^4+...\)

Find the pattern of the numbers to write in summation form

\(f(x)=\sum_{(n=0)}^\infty(n+1)x^n\) Use the Ratio Test \(|\frac{a_{n+1}}{a_n}|=|\frac{((n+1)+1)x^{n+1}}{(n+1)x^n}|=|\frac{(n+2)x^{n+1}}{(n+1)x^n}|=\frac{(n+2)|x|}{n+1}\)

\(\lim_{n\rightarrow\infty}\frac{(n+2)|x|}{(n+1)}\cdot\frac{\frac{1}{n}}{\frac{1}{n}}=\lim_{n\rightarrow\infty}\frac{(1+\frac{2}{n})|x|}{1+\frac{1}{n}}=\frac{(1+0)|x|}{(1+0)}=|x|\)

This converges when

\(|x|<1\)

The radius of convergence is half the width of this interval

\(R=\frac{1-(-1)}{2}=1\)

Result

\(f(x)=\sum_{n=0}^{\infty}(n+1)x^n,R=1\)