# Divergence of Series vs Dense Singularities

I am preparing for a Complex Analysis exam and have a question where I have to show that $$\sum_{n=1}^\infty nz^n=\frac{z}{(1-z)^2}$$ which I did using a geometric series of $$\frac{1}{1-z}$$. The next part of the question asks me to find all points where the original series diverges. My initial guess is that the series diverges for all $$z$$ such that $$|z|\ge1$$.

However, it then seems that the series has a dense set of singularities on the unit circle and so cannot be analytically continued, which clearly isn't true. Is it that these points are not actually singularities or am I missing something else?

Additionally, I rely on uniform convergence when using the geometric series for $$|z|<1$$ but uniform convergence of power series is technically defined on compact sets. Is it incorrect to use uniform convergence as justification when the actual domain is open rather than compact?

• There's only one singularity on the unit circle: at $z=1$. – Lord Shark the Unknown Oct 28 '18 at 17:12
• So why is, say, $z=-1$ not a singularity, even if the series diverges at that point? – P Collier Oct 28 '18 at 17:13
• There's an analytic continuation there, as your formula indicates. – Lord Shark the Unknown Oct 28 '18 at 17:14
• Thanks, but I still don't quite see what makes this series different to $\sum_{n=0}^\infty z^{2^n}$, which cannot be analytically extended? both series diverge on the unit circle, so what causes the difference? – P Collier Oct 28 '18 at 17:20

Your guess is right (and easy to justify): that series diverges of and only if $$\lvert z\rvert\geqslant1$$. Obviously, the sum of the series has one and only one analytic continuation to $$\mathbb{C}\setminus\{1\}$$ and that analytic continuation has one an only one singularity, located at $$1$$.