Singular point of a series in $\mathbb{C}$

I'm given the following problem. Let $f(x)=\sum c_nz^n$ in the unit disk, and furthermore $c_n\in \mathbb{R}, \ c_n >0$. Then the point $z_0=1$ is singular.

I don't seem to understand this, because if we let $c_n=2^{-n}$, then the sum $\sum 2^{-n}$ converges, so that would be a counterexample.

What am I doing wrong here?

-
What are the bounds of summation? And are you sure you're not overlooking some condition on $c_n$? I agree that $\sum_{n=0}^{\infty} 2^{-n}$ converges. – Dimitrije Kostic Nov 25 '11 at 18:43

I believe the proof by contradiction goes as follows: if $f$ is analytic at $z=1$, then it is analytic in a small disk centered at $z=1$. It follows that the power series of $f$ at $z=1/2$ converges in a disk of radius slightly greater than $1/2$. Choose a point $z_1>1$ at which this last series converges, and then rewrite the coefficients of that series (which are derivatives of $f$ at $z=1/2$) in terms of the original series $\sum c_n z^n$ at $z=0$. When you rearrange the resulting double sum (this is where the nonnegativity hypothesis is used!), you conclude that the original series converges at $z_1$, which contradicts the assumption that the radius of convergence is exactly 1.