1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ There's only one singularity on the unit circle: at $z=1$. $\endgroup$ – Lord Shark the Unknown Oct 28 '18 at 17:12
  • $\begingroup$ So why is, say, $z=-1$ not a singularity, even if the series diverges at that point? $\endgroup$ – P Collier Oct 28 '18 at 17:13
  • $\begingroup$ There's an analytic continuation there, as your formula indicates. $\endgroup$ – Lord Shark the Unknown Oct 28 '18 at 17:14
  • $\begingroup$ 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? $\endgroup$ – P Collier Oct 28 '18 at 17:20
1
$\begingroup$

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 $1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.