Recall that for [tex]|x|<1[/tex], we have the convergent geometric series
[tex]\displaystyle \sum_{n=0}^\infty x^n = \frac1{1-x}[/tex]
Now, for [tex]\left|\frac x5\right| < 1[/tex], we have
[tex]\dfrac1{5 - x} = \dfrac15 \cdot \dfrac1{1 - \frac x5} = \dfrac15 \displaystyle \sum_{n=0}^\infty \left(\frac x5\right)^n = \sum_{n=0}^\infty \frac{x^n}{5^{n+1}}[/tex]
Integrating both sides gives
[tex]\displaystyle \int \frac{dx}{5-x} = C + \int \sum_{n=0}^\infty \frac{x^n}{5^{n+1}} \, dx[/tex]
[tex]\displaystyle -\ln(5-x) = C + \sum_{n=0}^\infty \frac{x^{n+1}}{5^{n+1}(n+1)}[/tex]
If we let [tex]x=0[/tex], the sum on the right side drops out and we're left with [tex]C=-\ln(5)[/tex].
It follows that
[tex]\displaystyle \ln(5-x) = \ln(5) - \sum_{n=0}^\infty \frac{x^{n+1}}{5^{n+1}(n+1)}[/tex]
or
[tex]\displaystyle \ln(5-x) = \boxed{\ln(5) - \sum_{n=1}^\infty \frac{x^n}{5^n n}}[/tex]
