For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion.
\(f(x) = \frac{1}{(1-x)}\)
with generating derivatives,
first derivative: \(f'(x) = \frac{1}{(1-x)^2}\)
second derivative: \(f''(x) = \frac{2}{(1-x)^3}\)
third derivative: \(f'''(x) = \frac{6}{(1-x)^4}\)
fourth derivative: \(f''''(x) = \frac{24}{(1-x)^5}\)
when x = 0, \(f^n(0)=f(0)+f'(0)+f''(0)+f'''(0)+f''''(0)+...\)
then \(f^n(0)=1+1+2+6+24+..\)
if substitute back to Taylor Series, \[f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}0}{n!}x^n\]
\(=1+x+\frac{2x^2}{2!}+\frac{6x^3}{3!}+\frac{24x^4}{4!}+..\)
\(=1+x+x^2+x^3+x^4+...\)
Therefore, the Tylor Series expension is \[f(x)=\sum_{n=0}^{\infty}x^n\]
\(f(x)=e^x\)
since \(f(x)=e^x\), so the derivatives are \(f(x)=f'(x)=f''(x)=f'''(x) = f^n(x)=e^x\)
when x = 0, \(f(0)=f'(0)=f''(0)=f'''(0)=f^n(0)=e^0=1\)
Therefore, the Taylor Series expansion is \[\sum_{n=0}^{\infty}\frac{1}{n!}x^n=\sum_{n=0}^{\infty}\frac{x^n}{n!}\]
\(f(x)=ln(1+x)\)
first derivative:
\(f'(x)=\frac{1}{1+x}=(1+x)^-1\)
second derivative:
\(f''(x)=-1(1+x)^{-2}=\frac{-1}{(1+x)^2}\)
third derivative:
\(f'''(x)=2(1+x)^{-3}=\frac{2}{(1+x)^3}\)
fourth derivative:
\(f''''(x)=-6(1+x)^{-4}=\frac{-6}{(1+x)^4}\)
when x=0, \(f(0)=ln(1+0)=ln(1)=0\)
\(f'(0)=(1+0)^{-1}=1\)
\(f''(0)=-1(1+0)^{-2}=-1\)
\(f'''(0)=2(1+0)^{-3}=2\)
\(f''''(0)=-6(1+0)^{-4}=-6\)
since \(f^{(n)}(0)=-1^{(n+1)(n-1)!}\) when \(n\ge 1\)
Therefore
\(ln(1+x)\)
\(=\sum_{n=0}^{\infty}\frac{f^{(n)}0}{n!}(x)^n\)
\(=0+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(n-1)!}{n!}x^n\)
the Taylor Series expansion is
\(\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}\)