Powerseries of a natural logarithm

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SUMMARY

The discussion focuses on proving the equation \(\frac{1}{x} \ln \frac{x+1}{x-1} = \sum_0^\infty \frac{2x^{2n}}{2n+1}\). Participants suggest using the derivative relation \(\frac{1}{1+x} = \frac{d}{dx}\ln (1+x)\) and expanding it as a geometric series, although initial attempts led to an alternating series. A key insight involves utilizing the logarithmic property \(\log\left(\frac{a}{b}\right) = \log a - \log b\) to simplify the series and eliminate the alternation through term cancellation.

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Homework Statement


Show that

\frac{1}{x} \ln \frac{x+1}{x-1} = \sum_0^\infty \frac{2x^{2n}}{2n+1}.


2. The attempt at a solution
I tried to use the relation

\frac{1}{1+x} = \frac{d}{dx}\ln (1+x)

and expand as a geometric series, but this did not lead anywhere since I then ended up with an alternating series.

Anyone got any ideas of where to start?
 
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HINT:

\log\left(\frac{a}{b}\right) = \log a - \log b
 
Yes, so I could subtract the two resulting series... and I see now that it will actually lead trough. Ofcourse I knew about the property, but I did not see how it would get rid of the 'alternation'. However I do see that now because of the cancellation of terms.

Thank you!
 

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