\(f(x) = ln(1+x)\) and \(c=0\)
\(f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n\)
Derivatives:
\(f'(x) = \frac{1}{x+1}\)
\(f''(x) = \frac{-1}{(x+1)^2}\)
\(f'''(x) = \frac{2}{(x+1)^3}\)
\(f^{(4)}(x) = \frac{-6}{(x+1)^4}\)
\((c = 0, n = 0) \rightarrow \frac{ln(1+0)}{0!}(x-0)^0 = ln(1) = 0\)
\((c = 0, n = 1) \rightarrow \frac{\frac{1}{0+1}}{1!}(x-0)^1 = x\)
\((c = 0, n = 2) \rightarrow \frac{\frac{-1}{(0+1)^2}}{2!}(x-0)^2 = \frac{-1}{2}x^2\)
\((c = 0, n = 3) \rightarrow \frac {\frac{2}{(0+1)^3}}{3!}(x-0)^3 = \frac{1}{3}x^3\)
\((c = 0, n = 4) \rightarrow \frac {\frac{-6}{(0+1)^4}}{4!} (x-0)^4 = \frac{-1}{4}x^4\)
In summation form:
\(\sum_{n=1}^{\infty} \frac {X^n}{n}(-1)^{(n+1)}\)
I had to start the summation at n=1 since it does not work when n=0.