Finding the area of convergence of a complex power series

Question

I'm trying to find the radius of convergence and the area of convergence of the following power series :

$$\sum_{n=0}^{\infty } (n+1/2)z^n$$

The radius of convergence is easily found to be 1 by the ratio test. So inside the unit circle we have convergence.

But how about the entire area of convergence? It's not like in the real case where you can just plug in the interval end points and thereby know if these are in the area of convergence or not. There seems to be a infinite amounts of points on the unit circle to deal with in this complex case? Is there some trick involved here?

In general, the behavior on the circle of convergence can be extremely complicated. — saulspatz, Jun 08 '21 at 15:59
Of course, strictly outside the radius of convergence it diverges. — GEdgar, Jun 08 '21 at 16:01
@saulspatz: the behavior on the circle of convergence can be extremely complicated --- For those wanting more details about this, see the mathoverflow question Behaviour of power series on their circle of convergence and the Mathematics Stack Exchange question Power series which diverges precisely at the roots of unity, converges elsewhere and #6 in this answer. — Dave L. Renfro, Jun 08 '21 at 17:12

score 3 · Accepted Answer · answered Jun 08 '21 at 15:56

3

The series cannot possibly converge when $|z|\geq1$, since its terms don't go to zero.

answered Jun 08 '21 at 15:56

Martin Argerami

205,756

Well that was easy! – Filip Johansen Jun 08 '21 at 15:58
1

Now do $\sum z^n/(n+1/2)$. – GEdgar Jun 08 '21 at 16:00
@GEdgar I get the same radius of convergence, but now the test for divergence doesn't give me anything to work with? – Filip Johansen Jun 08 '21 at 16:08
1

@FilipJohansen: you should be able to do something like this. – Martin Argerami Jun 08 '21 at 18:38

score 2 · Answer 2 · answered Jun 08 '21 at 17:13

if we define: $$f(z)=\lim_{N\to\infty}S_N(z)\qquad S_N(z)=\sum_{n=1}^N(n+1/2)z^n$$ now notice that we can split up this summation into two: $$\underbrace{\sum_1^Nnz^n}_{\mathcal{S}}+\frac12\sum_1^Nz^n$$ We need both of these to converge in order for our overall sum to converge, and by looking at it, it is the left sum that we care about since $n>\frac12$ so we can say the radius of converge of $S_N$ is the same as this left sum, which I will call $\mathcal{S}$.

For $\mathcal{S}$ to converge we need the ratio of terms to converge, so if we take: $$a_n=nz^n$$ then: $$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{(n+1)z^{n+1}}{nz^n}\right|=\left(1+\frac1n\right)|z|$$ taking the limit we get: $$L=|z|$$ and so for $|z|<1$ the sum converges

Finding the area of convergence of a complex power series

2 Answers2