My task is to show that the Maclaurin series of $\ln\frac{x+1}{x-1}$ is given as $\sum_{n=1}^{\infty} (\frac{2}{2n-1})x^{2n-1}$

My approach is using the standard expansion of $\ln(x+1)$ to get $x-\frac{x^2}{2}+\frac{x^3}{3}$

I got $$\ln(x-1) = -x-\frac{x^2}{2}-\frac{x^3}{3}$$

After this point I'm a little confused as to how to proceed. My guess would be to subtract $\ln(x-1)$ from $\ln(x+1)$ but I still don't understand how to get this form or what exactly it's asking.

Any help would be greatly appreciated.

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    $\begingroup$ So far so good. When you subtract the two series, you'll see that the remaining terms form a pattern. Match that pattern to the series you are supposed to show is equal to it. $\endgroup$ – Matthew Leingang Aug 29 '18 at 3:11
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    $\begingroup$ I suppose that the question is for $\log \left(\frac{x+1}{1-x}\right)$ and not for what you wrote. Otherwise, some $i \pi$ would appear in addition to the expression. $\endgroup$ – Claude Leibovici Aug 29 '18 at 3:25
  • $\begingroup$ It's a show that so I presume that it's correct and we have to try finding out why, is ln just log to the base 10 from my understanding no? $\endgroup$ – Chris Latter Aug 29 '18 at 3:38
  • $\begingroup$ $\ln x$ is the logarithm to base e, although $\log x$ is often used instead. $\log x$ also denotes a base-10 logarithm when they're first introduced, but that fades out quickly unless you become an engineer. $\endgroup$ – Bladewood Aug 29 '18 at 3:54
  • $\begingroup$ I see, so how does this fact change the answer to this question? $\endgroup$ – Chris Latter Aug 29 '18 at 3:59

You have $$ \ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+...$$

and $$\ln (x-1) = -x-\frac{x^2}{2}-\frac{x^3}{3}+...$$

Thus $$\ln\frac{x+1}{x-1}= \ln (1+x)-\ln (x-1)$$


$$= 2x+(2/3) x^3 + 2/5 (x^5) +(2/7)x^7 +..... $$

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

  • $\begingroup$ This is not correct. $\endgroup$ – Claude Leibovici Aug 29 '18 at 3:26
  • $\begingroup$ why is that not correct ?? $\endgroup$ – Mohammad Riazi-Kermani Aug 29 '18 at 3:34
  • $\begingroup$ Have a look to my comment. $\endgroup$ – Claude Leibovici Aug 29 '18 at 3:39
  • $\begingroup$ Which part is incorrect? $\endgroup$ – Chris Latter Aug 29 '18 at 3:40
  • $\begingroup$ That second expansion belongs to $\log(1 - x)$, which up to a constant term is the same as $\log(x - 1)$, but it's an invalid assertion over the real numbers (mainly for $x \le 1$). If we're dealing with complex numbers, it should have a constant term of $i(\pi + 2\pi k)$ (or as a tauist, $i(\tau/2 + \tau k)$). $\endgroup$ – Bladewood Aug 29 '18 at 3:57

With $$\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots$$ $$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots$$ gives $$\ln(1-x)-\ln(1+x)=-2x-\frac{2x^3}{3}-\frac{2x^5}{5}-\cdots$$ or $$\ln\dfrac{1-x}{1+x}=\sum_{n=0}^\infty\dfrac{-2x^{2n+1}}{2n+1}$$ valid for $|x|<1$.

  • $\begingroup$ This is not correct. $\endgroup$ – Claude Leibovici Aug 29 '18 at 3:27
  • $\begingroup$ @ClaudeLeibovici yes. I just rewrote OP's results. $\endgroup$ – Nosrati Aug 29 '18 at 3:34
  • $\begingroup$ I'm getting a little confused now haha, how is this wrong exactly? $\endgroup$ – Chris Latter Aug 29 '18 at 3:37
  • $\begingroup$ The problem is convergence radius. the convergence radius for both primary series is $|x|<1$. $\endgroup$ – Nosrati Aug 29 '18 at 3:40
  • $\begingroup$ Why is it ln(1-x) - ln(1+x) ? isn't it supposed to be the other way around? $\endgroup$ – Chris Latter Aug 29 '18 at 4:12

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