Looking for Taylor series expansion of $\ln(x)$

Question

We know that the expansion of $$\sin(x) $$ is $$x/1!-x^3/3!\cdots$$

Without using Wolfram alpha, please help me find the expansion of $\ln(x)$.

I have my way of doing it, but am checking myself with this program because I am unsure of my method.

Answer 1

$$f(x)=\ln(x).$$

The Taylor expansion around $a$ is $$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n,$$

so for $a = 1$,

$$f'(x) = \frac{1}{x},\; f''(x) = \frac{-1}{x^2},\; f^{(3)}(x) = \frac{2}{x^3},\ldots,\; f^{(n)}(x) = \frac{(-1)^{n-1}(n-1)!}{x^n}.$$

$$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!}(x-1)^n=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}(x-1)^n$$

$$\ln(x)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}(x-1)^n$$

The more popular formula is obtained when we substitute $x$ with $1+x$

$$\ln(1+x)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}x^n$$ or $$\ln(1+x)=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}x^n$$ This is for $|x|<1$.