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?
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$\begingroup$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?
$\endgroup$ 3 $\begingroup$$\displaystyle -\ln (1-x)=\sum\limits_{k=1}^\infty \frac{x^k}{k}$. ( It comes from $\displaystyle \int\limits_0^x \frac{dt}{1-t}$ with $\displaystyle \frac{1}{1-x}=\sum\limits_{k=1}^\infty x^k$ for $|x|<1$. )
Substitute $x$ by $\displaystyle \frac{x-1}{x}$.
$\endgroup$ $\begingroup$If $\dfrac{x-1}x=y$ then $x=\dfrac1{1-y}$.
By Taylor, you get the development
$$-\ln(1-y)=\sum_{k=1}^\infty\frac{y^k}k$$ hence the claim(s).
$\endgroup$ $\begingroup$If you know the Taylor expansion for $\ln(1+t)$, that is, $$ \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 $$ Therefore (**) holds for $x\ge\frac{1}{2}$
$\endgroup$ $\begingroup$Using the definition of geometric series $$\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} $$
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