Sumfunction of the series $\sum_{n=0}^{\infty}(2n-1)(2x)^{n}$ for all points in the interval of convergence.

The question is about to find the sumfunction of the series $\sum_{n=0}^{\infty}(2n-1)(2x)^{n}$ for all points in the interval of convergence.

Before I show you what I have tried, the earlier question was about to find the formula for the series $\sum (an+b)x^{n}$ by using two the formulas $\sum_{n=0}^{\infty} (n+1)x^{n}=\frac{1}{(1-x)^{2}}$ and $\sum_{n=0}^{\infty} x^{n}\frac{1}{1-x}$ . I got; \begin{align*} \sum_{n=0}^{\infty} (an+b)x^{n}&=\sum_{n=0}^{\infty} \left (a(n+1)+(b-a) \right )x^{n} \\ &=a\sum (n+1)x^{n} + (b-a)\sum x^{n} \\ &=a\frac{1}{(1-x)^{2}}+(b-a)\frac{1}{1-x} \\ &=\frac{b+x(a-b)}{(1-x)^{2}} \end{align*} Its interval of convergence is $|x|<1$. So the convergence of interval of the series $\sum_{n=0}^{\infty}(2n-1)(2x)^{n}$ is $|x|<\frac{1}{2}$ (by using the ratio test). Now I am trying to find the sumfunction of the series with the abovementioned formula \begin{equation*}f(x)=\sum_{n=0}^{\infty}(2n-1)(2x)^{n}=\frac{-1+2x(2-(-1))}{(1-2x)^{2}}=\frac{-1+6x}{(1-2x)^{2}}\end{equation*} Then $f(-\frac{1}{2})=-1$ and $f(\frac{1}{2})=\infty$. So the sumfunction would be between $]-1, \infty[$. Is it correct or is there a better way to answer? Thanks.

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It looks fine to me, though I wouldn't write $\,f(.)=\infty\,$ . Also, the series diverges at $\,-\frac12\,$ so I'd rather write $$\lim_{x\to-\frac12^+}f(x)=-1$$ –  DonAntonio Jun 2 '13 at 12:58