Recall that for |x| < 1, we have
[tex]\displaystyle \frac1{1-x} = \sum_{n=0}^\infty x^n[/tex]
Differentiating both sides gives
[tex]\displaystyle \frac1{(1-x)^2} = \sum_{n=0}^\infty nx^{n-1} = \sum_{n=0}^\infty (n+1)x^n[/tex]
and multiplying both sides by 8 gives the series for f(x) :
[tex]f(x)=\displaystyle \frac8{(1-x)^2} = \boxed{8\sum_{n=0}^\infty (n+1)x^n}[/tex]
and this converges over the same interval, |x| < 1, so that the radius of convergence is 1.
