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

$f(x)=ln(x)$
taylor expansion around a is
$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$
so for $a = 1$
$f'(x) = \frac{1}{x}$  $f''(x) = \frac{-1}{x^2}$ $f^{(3)}(x) = \frac{2}{x^3}$
$f^{(n)}(x) = \frac{(n-1)! (-1)^{n-1} }{x^n}$
$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!}(x-1)^n=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}(x-1)^n$
$ln(x)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}(x-1)^n$
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 $|x|<1$ 
