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Show that $\ln\Big(|\frac{1+x}{1-x}|\Big)=2\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1},$ for $|x|<1$. The previous excercise (which was within my limited reach) was to show that $\frac{1}{1-x^2}=\sum_{n=0}^{\infty}x^{2n},$ for $|x|<1$. I suspect there is a (not overly subtle) connection here but, needless to say, I can't see it. I don't know why the absolute value signs are included, but it might be because the solution includes integrating some power series. The excercise is contained in a chapter on derivation and integration of (convergent) power series; one is also assumed to be familiar with multiplication of power series.

Very thankful for any help.

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up vote 4 down vote accepted


$1$. Write down the power series for $\ln(1+x)$.

$2$. Substitute $-x$ for $x$ to find the power series for $\ln(1-x)$.

If you don't know the power series for $\ln(1+x)$, but you probably do, use the power series for $\frac{1}{1+t}$, and the fact that $\ln(1+x)=\int_0^x\frac{dt}{1+t}$. Expand, and integrate term by term.

$3$. We have $\ln\left(\frac{1+x}{1-x}\right)=\ln(1+x)-\ln(1-x)$.

The absolute value signs surrounding $\frac{1+x}{1-x}$ that you asked about are pointless. Since we are only working with $|x|<1$, the quantity $\frac{1+x}{1-x}$ is positive, so the absolute value signs do nothing.

Some detail: It is standard that $\frac{1}{1-u}=1+u+u^2+u^3+\cdots$, with the series converging when $|u|<1$. Put $u=-t$. We obtain $$\frac{1}{1+t}=1-t+t^2-t^3+\cdots=\sum_{k=0}^\infty (-1)^kt^k.$$ It follows that (for $|x|<1$) $$\ln(1+x)=\int_0^x\frac{dt}{1+t}=\int_0^x\left(\sum_{k=0}^\infty (-1)^kt^k \right)dt=\sum_{k=0}^\infty(-1)^k\frac{x^{k+1}}{k+1}.$$ We have now carried out Hint $1$.
Now substitute $-x$ for $x$ in our expression for $\ln(1+x)$. We get $$\ln(1-x)=\sum_{k=0}^\infty(-1)^k\frac{(-x)^{k+1}}{k+1}=-\sum_{k=0}^\infty \frac{x^{k+1}}{k+1}.$$ (The terms $(-1)^k$ and $(-1)^{k+1}$ have product $(-1)^{2k+1}$, which is identically equal to $-1$.) We have now carried out Hint $2$.

Now use Hint $3$. We get $$\ln\left(\frac{1+x}{1-x}\right)=\sum_{k=0}^\infty \left[(-1)^k-(-1)\right]\frac{x^{k+1}}{k+1}.$$ Consider the coefficient $[(-1)^{k}-(-1)]$ above. This is $2$ if $k$ is even and $0$ if $k$ is odd. To put it another way, we can put $k=2n$, since odd $k$ make no contribution to the sum. Thus $$\ln\left(\frac{1+x}{1-x}\right)=\sum_{n=0}^\infty 2\frac{x^{2n+1}}{2n+1}.$$

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Well, but using the series for $ln(1+x)$ and $ln(1-x)$ together with $\ln(a/b)=\ln(a)-\ln(b)$, which I had already tried, doesn't yield the deisred identity. I am pretty sure one is supposed to use the sum for $\frac{1}{1-x^2}$ mentioned above. And the absolute values, pointless though they are, I think may be a hint that one is supposed to integrate both sides of something, since \integral 1/y = ln|y| + C. But I still have no idea how to solve this exercise. – Bart Patzer May 4 '12 at 11:08
@Bart Patzer: I wrote out the details. I even used summation symbols, which I don't like to do, since they make things less visually obvious. The hints (which you will note are essentially the same as those of David Mitra) give the desired expression. – André Nicolas May 4 '12 at 13:10

Hint: $\ln{a\over b}=\ln a-\ln b$.

Also, for $|x|<1$: $$ \ln(1+x)= \phantom{-}x-{x^2\over 2}+{x^3\over3}-{x^4\over 4}+\cdots, $$ and $$ \ln(1-x)= -x-{x^2\over 2}-{x^3\over3}-{x^4\over 4}+\cdots . $$

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