Series Expansion for $\ln(x)$ I have a mathematics assignment, which requires me to proof that $$\ln\frac{2}{3} = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{2^{n}n}$$.
I know, I can solve this by proving $\ln x$ = $\sum_{n=1}^{\infty }\frac{1}{n}\left ( \frac{x-1}{x} \right )^{n}$, but I don't know how to prove this, so can anybody offer some help?
 A: Begin with what is probably the simplest, and thus for $\;|x|<1\;$ :
$$\frac1{1+x}=\sum_{n=0}^\infty (-1)^nx^n\stackrel{\text{we can integ. within the converg. radius}}\implies\log(1+x)=\sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n+1}=$$
$$=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}n$$
and now just substitute $\;x=\cfrac12\;$ :
$$\;\log\frac32=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{2^nn}$$ 
Final step: for you. Can you see what's the little, tiny step lacking?
A: $\displaystyle -\ln (1-x)=\sum\limits_{k=1}^\infty \frac{x^k}{k}$. 
( It comes from $\displaystyle \int\limits_0^x \frac{dt}{1-t}$ with $\displaystyle \frac{1}{1-x}=\sum\limits_{k=1}^\infty x^k$ for $|x|<1$. )
Substitute $x$ by $\displaystyle \frac{x-1}{x}$.
A: If $\dfrac{x-1}x=y$ then $x=\dfrac1{1-y}$.
By Taylor, you get the development
$$-\ln(1-y)=\sum_{k=1}^\infty\frac{y^k}k$$ hence the claim(s).
A: If you know the Taylor expansion for $\ln(1+t)$, that is,
$$
\ln(1+t)=\sum_{n\ge1}\frac{(-1)^{n+1}t^n}{n}\tag{*}
$$
which follows from integrating
$$
\frac{1}{1+x}=\sum_{n\ge0}(-1)^nx^n
$$
then it's easy: set $1+t=1/x$, so
$$
t=\frac{1}{x}-1=\frac{1-x}{x}
$$
and
$$
\ln x=-\ln\frac{1}{x}=
-\sum_{n\ge1}\frac{(-1)^{n+1}}{n}\left(\frac{1-x}{x}\right)^n
=\sum_{n\ge1}\frac{1}{n}\left(\frac{x-1}{x}\right)^n\tag{**}
$$
For $x=2/3$, we have
$$
\frac{2/3-1}{2/3}=1-\frac{3}{2}=-\frac{1}{2}
$$
Note that the series expansion (*) is valid for $-1<t\le1$, so
$$
0<1+t=\frac{1}{x}\le 2
$$
Therefore (**) holds for $x\ge\frac{1}{2}$
A: Using the definition of geometric series 
$$\frac{1}{1-x}=\sum_{k=0}^{\infty} x^{k} $$
Integrating both sides 
$$\int\frac{1}{1-x}=\sum_{k=0}^{\infty}\frac{(-1)^k x^{k+1}}{k+1} $$
So $$\ln (1-x)=-\sum_{k=1}^{\infty} \frac { x^k}{k} $$
$$\ln (y)= \sum_{k=1}^{\infty} \frac{1}{k} \left( \frac{y-1}{y} \right) ^{k}  $$
