Sum the power series,
$$\sum_{n=0}^\infty\frac{nx^n}{n+9}$$
I found my radius of convergence to be 1 and my interval was $-1<x<1$. For end points I checked and I wasn't too sure if they are included in the interval
If they are would the values for which series converges absolutely include the -1 and 1?
For a positive integer $m$ ($m=9$ in your case)
$\begin{array}\\ s_m(x) &=\sum_{n=0}^\infty\frac{nx^n}{n+m}\\ &=\sum_{n=m}^\infty\frac{(n-m)x^{n-m}}{n}\\ &=\sum_{n=m}^\infty\frac{nx^{n-m}}{n}-\sum_{n=m}^\infty\frac{mx^{n-m}}{n}\\ &=\sum_{n=m}^\infty x^{n-m}-mx^{-m}\sum_{n=m}^\infty\frac{x^{n}}{n}\\ &=\sum_{n=0}^\infty x^{n}-mx^{-m}\sum_{n=m}^\infty\frac{x^{n}}{n}\\ &=\dfrac1{1-x}-mx^{-m}(\sum_{n=1}^\infty\frac{x^{n}}{n}-\sum_{n=1}^{m-1}\frac{x^{n}}{n})\\ &=\dfrac1{1-x}-mx^{-m}(-\ln(1-x)-\sum_{n=1}^{m-1}\frac{x^{n}}{n})\\ \end{array} $