Easy way to remember Taylor Series for log(1+x)? Assuming $|x|<1$, if
one can easily remember that 
$$
\dfrac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}
$$
then one can derive the following
\begin{eqnarray*}
\mbox{log}(1-x) & = & -\sum_{n=1}^{\infty}\dfrac{x^{n}}{n}\qquad\qquad\mbox{(integrate)}\\
\dfrac{1}{(1-x)^{2}} & = & \sum_{n=1}^{\infty}nx^{n-1}\qquad\qquad\mbox{(differentiate)}\\
\dfrac{x}{(1-x)^{2}} & = & \sum_{n=1}^{\infty}nx^{n}\qquad\qquad\mbox{(multiply by x)}\\
\dfrac{2}{(1-x)^{3}} & = & \sum_{n=1}^{\infty}(n-1)nx^{n-2}.
\end{eqnarray*}
Is there a similar handy way to remember
\begin{eqnarray*}
log(1+x) & = & \sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{x^{n}}{n}\\
 & = & x-\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}-\dfrac{x^{4}}{4}+\cdots
\end{eqnarray*}
using either a similar (or a different) approach?

$$\frac{1}{1+x}=\frac{1}{1-(-x)}=\sum\limits_{n=1}^{\infty}(-x)^n$$
Can you take it from here? :)
