We can use the fact that, for [tex]|x|<1[/tex],
[tex]\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n[/tex]
Notice that
[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1{1-x}\right]=\dfrac1{(1-x)^2}[/tex]
so that
[tex]f(x)=\displaystyle\frac5{(1-x)^2}=5\frac{\mathrm d}{\mathrm dx}\left[\sum_{n=0}^\infty x^n\right][/tex]
[tex]f(x)=\displaystyle5\sum_{n=0}^\infty nx^{n-1}[/tex]
[tex]f(x)=\displaystyle5\sum_{n=1}^\infty nx^{n-1}[/tex]
[tex]f(x)=\displaystyle5\sum_{n=0}^\infty(n+1)x^n[/tex]
By the ratio test, this series converges if
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{(n+2)x^{n+1}}{(n+1)x^n}\right|=|x|\lim_{n\to\infty}\frac{n+2}{n+1}=|x|<1[/tex]
so the series has radius of convergence [tex]R=1[/tex].
