For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion.
Taylor Series Formula: \[ \begin{aligned} f(x) &= \sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n \end{aligned} \]
\[ \begin{aligned} f(x) &= \frac{1}{1-x} \end{aligned} \] The Taylor Series around a = 0; \[ \begin{aligned} \text{First derivative},\: f'(x) &= \frac{1}{(1-x)^2}, \:\text{evaluated at } x = 0 \rightarrow 1 \\ \text{Second derivative},\: f''(x) &= \frac{2}{(1-x)^3}, \:\text{evaluated at } x = 0 \rightarrow 2 \\ \text{Third derivative},\: f'''(x) &= \frac{6}{(1-x)^4}, \:\text{evaluated at } x = 0 \rightarrow 6 \\ \text{nth derivative},\: f^n(x) &= \frac{n!}{(1-x)^{(n+1)}}, \:\text{evaluated at } x = 0 \rightarrow n! \\ \end{aligned} \] Therefore, the full expansion of \(\frac{1}{1-x}\) as a Taylor series around a = 0, valid for \(|x| < 1\); \[ \begin{aligned} f(x) = \sum_{n = 0}^{\infty}\frac{n!}{n!}x^n &= \sum_{n = 0}^{\infty}x^n \end{aligned} \]
\[ \begin{aligned} f(x) &= e^x \end{aligned} \] The Taylor Series around a = 0; \[ \begin{aligned} \text{First derivative},\: f'(x) &= e^x, \:\text{evaluated at } x = 0 \rightarrow 1 \\ \text{Second derivative},\: f''(x) &= e^x, \:\text{evaluated at } x = 0 \rightarrow 1 \\ \text{nth derivative},\: f^n(x) &= e^x, \:\text{evaluated at } x = 0 \rightarrow 1 \\ \end{aligned} \] Therefore, the full expansion of \(e^x\) as a Taylor series around a = 0, valid for \(\forall x\); \[ \begin{aligned} f(x) = \sum_{n = 0}^{\infty}\frac{1}{n!}x^n &= \sum_{n = 0}^{\infty}\frac{x^n}{n!} \end{aligned} \]
\[ \begin{aligned} f(x) &= \ln(1+x) \end{aligned} \] The Taylor Series around a = 0; \[ \begin{aligned} \text{First derivative},\: f'(x) &= \frac{1}{(1+x)}, \:\text{evaluated at } x = 0 \rightarrow 1 \\ \text{Second derivative},\: f''(x) &= -\frac{1}{(1+x)^2}, \:\text{evaluated at } x = 0 \rightarrow -1 \\ \text{Third derivative},\: f'''(x) &= \frac{2}{(1+x)^3}, \:\text{evaluated at } x = 0 \rightarrow 2 \\ \text{Fourth derivative},\: f^{4}(x) &= -\frac{6}{(1+x)^4}, \:\text{evaluated at } x = 0 \rightarrow -6 \\ \text{nth derivative},\: f^n(x) &= (-1)^{n-1}\frac{(n-1)!}{(1+x)^{n}}, \:\text{evaluated at } x = 0 \rightarrow (-1)^{n-1}(n-1)! \\ \end{aligned} \] Therefore, the full expansion of \(\ln(1+x)\) as a Taylor series around a = 0, valid for \(x > -1\); \[ \begin{aligned} f(x) = \sum_{n = 0}^{\infty}\frac{(-1)^{n-1}(n-1)!}{n!}x^n &= \sum_{n = 0}^{\infty}\frac{(-1)^{n-1}x^n}{n} \end{aligned} \]
\[ \begin{aligned} f(x) &= x^{1/2} \end{aligned} \] Using the Binomial Theorem; \[ \begin{aligned} (1+u)^k &= \sum_{n = 0}^{\infty}\binom{k}{n}u^n \end{aligned} \] Substituting \((u + 1) = x\) and \(k = \frac{1}{2}\); \[ \begin{aligned} x^{\frac{1}{2}} &= \sum_{n = 0}^{\infty}\binom{\frac{1}{2}}{n}(x-1)^n \end{aligned} \] Valid for \(|x| < 1\), where \[ \begin{aligned} \binom{\frac{1}{2}}{n} &= \frac{(\frac{1}{2})(\frac{1}{2}-1)(\frac{1}{2}-2)...(\frac{1}{2}-n+1)}{n!} \end{aligned} \]