A Taylor series of a function is an infinite sum of function derivatives calculated at that point.
Taylor series of f(x) => \[f(a) + \frac{f^{'}(a)}{1!}*(x-a) + \frac{f^{''}(a)}{2!}*(x-a)^{2} + \frac{f^{'''}(a)}{3!}*(x-a)^{3}...\] => \[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}a}{n!}*(x-a)^{n} \]
\(f^{'}(x) = \frac{1}{{(1-x)}^{2}}, and\space for\space x=0\space we\space get => \space 1\) \(f^{''}(x) = \frac{2}{{(1-x)}^{3}}, and\space for\space x=0\space we\space get => \space 2\) \(f^{'''}(x) = \frac{6}{{(1-x)}^{4}}, and\space for\space x=0\space we\space get => \space 6\)
Upon applying the same for x = 0, we get below, where limits for x are -1 < x < 1
\[f(x) = f(\frac{1}{(1-x)}) = 1 + x + x^2 + x^3 + ... = \sum_{n=0}^{\infty} x^n\]
\(f^{'}(x) = e^x, and\space for\space x=0\space we\space get => \space 1\) \(f^{''}(x) = e^x, and\space for\space x=0\space we\space get => \space 1\) \(f^{''}(x) = e^x, and\space for\space x=0\space we\space get => \space 1\)
Upon applying the same for x = 0, we get below \[f(x) = f(e^x) = 1 + x + \frac{1}{2!}*x^2 + \frac{1}{3!}*x^3 + ... = \sum_{n=0}^{\infty}\frac{1}{n!}*x^n\]
\(f(x) = ln(x+1), and\space for\space x=0\space we\space get => \space 0\) \(f^{'}(x) = \frac{1}{1+x}, and\space for\space x=0\space we\space get => \space 1\) \(f^{''}(x) = \frac{-1}{(1+x)^2}, and\space for\space x=0\space we\space get => \space -1\) \(f^{'''}(x) = \frac{2}{(1+x)^3}, and\space for\space x=0\space we\space get => \space 2\) \(f^{''''}(x) = \frac{-6}{(1+x)^4}, and\space for\space x=0\space we\space get => \space -6\)
Upon applying the same for x = 0, we get below \[f(x) = f(ln(1+x)) = 0 + \frac{1}{1!}*x - \frac{1}{2!}*x^2 + \frac{2}{3!}*x^3 - \frac{6}{4!}*x^4 + ... = \sum_{n=1}^{\infty}{-1}^{(n+1)} \frac{1}{n}*x^n\]