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I know that $$ \ln(1+x)=\sum _{n=1}^{\infty }\:\left(-1\right)^{n-1}\frac{x^{n}}{n} $$

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Hint: $$\ln(5-x)=\ln\left[5\left(1-\frac{x}{5}\right)\right]=\ln 5+\ln\left(1-\frac{x}{5}\right). $$

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  • $\begingroup$ But then we use $\sum _{n=0}^{\infty }\:\frac{\left(\frac{x}{5}\right)^{n-1}}{n-1}\left(-1\right)^n$ instead of $\sum _{n=0}^{\infty }\:\frac{\left(x\right)^{n-1}}{n-1}\left(-1\right)^n$ ? $\endgroup$ – DDDD Jun 11 '14 at 14:43
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In your case it's more convenient to use $$ \log \bigg( \frac{1}{1-w} \bigg) = \sum_{k=1}^{n} \frac{w^k}{k} $$ just multiply by $-1$ and set $w=5x$.

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