Recall that for |x| < 1, we have
[tex]\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n[/tex]
Replace x with 2x, multiply 4, and call this function f :
[tex]f(x)=\dfrac4{1-2x}=\displaystyle4\sum_{n=0}^\infty(2x)^n[/tex]
Take the derivative:
[tex]f'(x)=\dfrac8{(1-2x)^2}=\displaystyle8\sum_{n=0}^\infty n(2x)^{n-1}=\boxed{8\sum_{n=0}^\infty (n+1)(2x)^n}[/tex]
By the ratio test, the series converges for
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{(n+2)(2x)^{n+1}}{(n+1)(2x)^n}\right|=|2x|\lim_{n\to\infty}\frac{n+2}{n+1}=|2x|<1[/tex]
or |x| < 1/2, so the radius of convergence is 1/2.
