I am supposed to find the Maclaurin expantion of

$ \log\left( \frac{1+x}{1-x} \right) $

So I noticed the obvious that $\log(1+x) - \log(1-x)$

Then Maclaurin polynomial of $\log (1+x)$ equals $\sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n}$

So by doing a quick quick change of variables (-x) I obtained the maclaurin expansion of $\log(1-x)$

In total, I obtained that the Maclaurin expansion of $ \log\left( \frac{1+x}{1-x} \right) $ equals.

$ \displaystyle P(x) = \sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n} - \sum_{n=0}^{\infty} (-1)^{n+1}\frac{(-x)^n}{n}$

But in my book the answer says $2 \sum_{n=0}^{\infty} \frac{1}{1+2n} x^{2n+1}$, I can see that this is correct, by writing out a few terms. But how do I show this by algebra?

My question is therefore, how do I show that

$\displaystyle \sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n} - \sum_{n=0}^{\infty} (-1)^{n+1}\frac{(-x)^n}{n} = 2 \sum_{n=0}^{\infty} \frac{1}{1+2n} x^{2n+1}$


  • 1
    $\begingroup$ Use $\log(1+x) = \log(1-(-x))$ and expand around $x=0$ just as you did for $\log(1-x)$. $\endgroup$ – Sasha Mar 6 '12 at 14:54
  • $\begingroup$ But I have already done this note the $x-2$ $\endgroup$ – N3buchadnezzar Mar 6 '12 at 14:55
  • $\begingroup$ $1+x$ becomes $1+x-2 = x-1 $ after replacing $x$ with $x-2$, not $1-x$ as required. To make $1+x$ become $1-x$, replace $x$ everywhere with $-x.$ $\endgroup$ – Ragib Zaman Mar 6 '12 at 14:58
  • $\begingroup$ Where did you get $x-2$ from? $\endgroup$ – Ilya Mar 6 '12 at 14:58
  • $\begingroup$ Silly mistake! I will fix it now $\endgroup$ – N3buchadnezzar Mar 6 '12 at 15:01

The expansion for $\log (1+x)$ is not $\sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n}$, but $$\begin{equation*} \log (1+x)=\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{n+1}% x^{n+1}=\sum_{n=1}^{\infty }\frac{\left( -1\right) ^{n+1}}{n}x^{n+1} \end{equation*}.$$


$$ \begin{eqnarray*} \log (1-x) &=&\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{n+1}\left( -x\right) ^{n+1}=\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}\left( -1\right) ^{n+1}}{n+1}x^{n+1} \\ &=&-\sum_{n=0}^{\infty }\frac{1}{n+1}x^{n+1}, \end{eqnarray*} $$

and $$ \begin{eqnarray*} \log \left( \frac{1+x}{1-x}\right) &=&\log (1+x)-\log (1-x) \\ &=&\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{n+1}x^{n+1}+\sum_{n=0}^{ \infty }\frac{1}{n+1}x^{n+1} \\ &=&\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}+1}{n+1}x^{n+1} \\ &=&\sum_{n=0}^{\infty }\frac{2}{2n+1}x^{2n+1}, \end{eqnarray*}$$

because $$ \begin{equation*} \left( -1\right) ^{n}+1=\left\{ \begin{array}{c} 2\quad \text{if }n\text{ even} \\ 0\quad \text{if }n\text{ odd}. \end{array} \right. \end{equation*}$$


It makes :

${\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{x^{n}}{n}}-(-1)^{n+1+n}\dfrac{x^{n}}{n}={\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{x^{n}}{n}}+\dfrac{x^{n}}{n}={\displaystyle \sum_{n=0}^{\infty}\dfrac{2x^{n}}{2n+1}}+{\displaystyle \sum_{n=1}^{\infty}-\dfrac{x^{n}}{2n}+\dfrac{x^{n}}{2n}}={\displaystyle \sum_{n=0}^{\infty}\dfrac{2x^{n}}{2n+1}}$

  • $\begingroup$ What did you do from 2 to 3 ? $\endgroup$ – N3buchadnezzar Mar 6 '12 at 15:28
  • $\begingroup$ I'm french, so I hope that it will be "understandable". I separated the even and odd terms. $\endgroup$ – matovitch Mar 6 '12 at 15:32

In fact you have $ (-1)^{n+1} (-x)^n = -(x)^n$. Thus if you look the term of the two sum you have: $x^n\frac{1 - (-1)^{n}}{n}$ and for the complete sum : $2\sum_{k .odd} \frac{x^k}{k}$ it's exactly that you want.


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