Behavior on the Circle of Convergence using Dirichlet's Test

Question

Let $O$ be a circle of convergence with center $z_o$ and radius $R = \vert z^*-z_o\vert = 1.$ The goal is to find where the following sums converge inside $O$.

$$A.\:\: \sum\frac{z^n}{n^2}$$

$$B.\:\: \sum\frac{z^n}{n}$$

$$C.\:\: \sum z^n$$ Let $z^*-z_o = z$ be a complex number. For $A$, $$\biggl|\frac{z^n}{n^2}\biggr| = \biggl|\frac{r^n(\cos(n\theta)+i\sin(n\theta))}{n^2}\biggr|\leq\frac{1}{n^2}$$ hence, $$\biggl|\sum\frac{z^n}{n^2}\biggr|\leq\sum\biggl|\frac{z^n}{n^2}\biggr|\leq\sum\frac{1}{n^2}$$ As $n$ approaches $\infty$, the rightmost part of the last inequality converges since this is a sum of infinite geometric series. Hence $A$ converges absolutely for all $|z|\leq1$ and the sum also converges in an ordinary sense. Using the same process, it follows that $C$ converges everywhere inside $O$ except at $z = 1$.

Finally for $B$, two notable cases $z=1$ and $z=-1$ immediately came to my mind, $$\biggl|\sum\frac{z^n}{n}\biggr|=\sum\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}...\qquad \text{when}\quad z=1$$ $$\biggl|\sum\frac{z^n}{n}\biggr|=\left|\sum\frac{(-1)^n}{n}\right|=\left|-1+\frac{1}{2}-\frac{1}{3}...\right|\qquad \text{when}\quad z=-1$$ Although absolute convergence don't exist for both sums, the latter will still converge (sum of alternating harmonic series converges) Other than this, I don't know where in $O$ the sum converge absolutely. Luckily in here Determining precisely where $\sum_{n=1}^\infty\frac{z^n}{n}$ converges? ,one user used Dirichlet's test (I'm not so familiar with this probably I'll ask something about this too later.) The most crucial part of his answer is this :$$\left|\sum_{n=1}^Nz^{ n }\right|=\left|\frac{z-z^{N+1}}{1-z}\right|\leq\frac{2}{|1-z|}$$ and concluded that only when $z\neq1$ all the conditions of Dirichlet's Test are satisfied.
But why is it that $$\sum_{n=1}^N z^{n} = \frac{z-z^{N+1}}{1-z}\:\:?$$ If $z$ is a real then it will become the sum of geometric series but is this true for complex numbers as well?
Finally if I apply Dirichlet's test to $A$ using the same process with $B$ done by the user from the link, clearly with ${a_n}=\frac{1}{n^2},\, a_1\geq a_2\geq\cdots$ and $\,\lim_{n\to\infty}a_n=0$. But then again from $B$, $\left|\sum_{n=1}^Nz^{ n }\right|$ is only convergent when $z\neq1$.

To summarize it all, my observations are the following:

• So far, we looked for values of $z$ for which the sums are absolutely convergent. If they are absolutely convergent to those values of $z$, then they are convergent in the ordinary sense for those values as well. Absolute convergence implies convergence. However, the converse is not true.

• In the case $B$, we saw that if $|z|=1$ may it be $+1,-1$ or even $|z^*-z_o|=1$ like $$ \sqrt\frac{1}{2}+\sqrt\frac{1}{2}i, $$ the sum doesn't converge absolutely. But this does not imply that the series doesn't strictly converge in an ordinary sense for these values except at $z=1$(as demonstrated by Dirichlet's Test).

For my questions:

1. What happens if there are sums for which absolute convergence don't exist but ordinary convergence exist? How can we determine the values for which they are convergent?

2. Is $\sum_{n=1}^N z^{n} = \frac{z-z^{N+1}}{1-z}$ true for complex numbers?

3. Why Dirichlet's Test failed in finding values of $z$ for which $A$ is convergent. That is $z=1$ should be excluded contradictory to $|z|\leq1$ (again absolute convergence implies ordinary convergence).

I apologize for making this too long. These sums were taken from the book Advanced Engineering Mathematics By Kreyzig page 682. IMO, the author did not properly elaborate how these sums converge/diverge nor did he give helpful illustrations. This is hard for me since I don't have proper knowledge about complex analysis/real analysis so I decided to put this here to ease the pain in my head. I expect many errors above and I also hope I will get clear answers from my questions. Thank you for reading.

Answer 1

It seems to me that the question need first an analysis of the Cauchy-Hadamard theorem on the meaning of the circle of convergence and the radius of convergence of a power series: therefore I will state the theorem below but prove it only for the parts on the absolute and uniform convergence of power series inside their circle of convergence and their divergence outside it.

Cauchy-Hadamard theorem (see for example [1], chapter III §2 pp. 53-54). Let $\{ c_n \}_{n\in \Bbb N}$ a sequence of complex numbers, $z_o\in \Bbb C$ and consider the power series $$ \sum\limits_{n=0}^\infty c_n (z-z_o)^n.\label{1}\tag{1} $$ Then there exists an extended real number $R\in [0,\infty]$ such that

1. The series \eqref{1} converges absolutely for all $z\in\Bbb C$ such that $|z-z_o|<R$ and uniformly in every disk $B(z_o,\rho)$ centered in $z_o$ with radius $\rho < R$.
2. The terms of the series \eqref{1} are unbounded (and hence the series diverges) for every $z\in \Bbb C$
3. The extended real number $R$ is the reciprocal of the extended real number $\limsup_{n\to\infty} \sqrt[n]{|c_n|}$ and is called the radius of convergence of the series \eqref{1}.

As told above I will prove only statement 1 and 2, assuming known $R=1$: let's start.
Proof of 1 and 2. Let $R=1$ be the least upper bound in $[0,\infty]$ of the set $$ \{r\in[0,\infty]\,|\,|c_n| r^n\text{ is a bounded sequence}\}. $$ By definition, if $|z-z_o|>R$ the sequence $\{ |c_n (z-z_o)|\}_{n\in\Bbb N}$ is unbounded and the series \eqref{1} diverges since $$ |s_m-s_{m-1}|=\left|\sum\limits_{n=0}^m c_n (z-z_o)^n-\sum\limits_{n=0}^{m-1} c_n (z-z_o)^n\right|=|c_m(z-z_o)^m|\underset{m\to\infty}{\nrightarrow}0. \label{2}\tag{D} $$ Statement 2 above is proved. Now, assuming $0<\rho<R=1$ and choosing $r$ such that $\rho < r < 1$, the sequence $\{ |c_n (z-z_o)|\}_{n\in\Bbb N}$ is bounded is bounded by definition, say $0\le|c_n (z-z_o)|\le|c_n|r^n\le M<\infty$ for all $n\in\Bbb N$. Then, for all $z\in B(z_o,\rho)$ $$ |c_n(z-z_o)^n|=|c_n|r^n\cdot\bigg(\frac{|z-z_o|}{r}\bigg)^{\!n}\le M\cdot\left(\frac{\rho}{r}\right)^n $$ thus $$ 0\le\left|\sum\limits_{n=0}^\infty c_n (z-z_o)^n\right|\le \sum\limits_{n=0}^\infty |c_n (z-z_o)^n|\le M\sum\limits_{n=0}^\infty \left(\frac{\rho}{r}\right)^n= M \frac{r}{r-\rho}<\infty, $$ and the statement 1 is finally proven. $\blacksquare$

The theorem of Cauchy-Hadamard implies that a power series at a point $z_o$ converges absolutely on each point inside its convergence disk and furthermore the convergence is uniform on each disk centered in $z_o$ and entirely contained in it.

Having clarified the framework, we can answer the three instances of the question:

1. What happens if there are sums for which absolute convergence don't exist but ordinary convergence exist? How can we determine the values for which they are convergent?

To my knowledge, general methods (read necessary and sufficient conditions) to deal with such problems are Cauchy's convergence criterion and Tauber's second theorem: in this answer I described some facts related to this problem. Furthermore Cauchy-Hadamard solves completely the question for power series only inside their convergence disk, but leaves completely aside what happens on its the boundary, i.e. for $|z|=1$ in our case. There can be convergence in ordinary (not absolute) sense, divergence and oscillating behavior: again, in those cases, only the two criterions cited above seem to be available.

2. Is $\sum_{n=1}^N z^{n} = \frac{z-z^{N+1}}{1-z}$ true for complex numbers?

This identity is a consequence of the following algebraic ones valid for real or complex $z$, which can be proved by using mathematical induction: $$ \begin{split} 1-z^2 & =(1-z)(z+1)\\ 1-z^3 & =(1-z)(1+z+z^2)\\ & \vdots\\ 1-z^N & =(1-z)\sum\limits_{n=0}^{N-1} z^n \end{split} \implies \sum\limits_{n=1}^{N} z^n=z\cdot\sum\limits_{n=0}^{N-1} z^n=z\frac{1-z^{N}}{1-z} $$

3. Why Dirichlet's Test failed in finding values of $z$ for which $A$ is convergent. That is $z=1$ should be excluded contradictory to $|z|\leq1$ (again absolute convergence implies ordinary convergence).

Dirichlet's test (see for example [2], chapter XII, §5 p. 592 and §6 p. 599) applies to a series of the form $\sum a_nb_n$ if the the sequence of partial sums $\big\{\sum^m a_n\big\}_{m\in\Bbb N}$ of the sequence $\{a_n\}_{n\in\Bbb N}$ is bounded and the sequence $\{b_n\}_{n\in\Bbb N}$ converges monotonically to $0$: in the example $A$ of the question, $a_n=1/n^2$ satisfies the hypothesis for the applicability, but $b_n=z^n$ does not if $|z|=1$, thus the test is not applicable in this specific case.

A note on the behavior of $|c_m(z-z_o)^m|$
Following the comment of the OP, I had to reconsider relation \eqref{2} and figured that, in order to be clearer, I should point out the following facts.

• The unboundedness of the set $\{ |c_n (z-z_o)|\}_{n\in\Bbb N}$ does not imply that $$ \lim_{m\to\infty}|c_m(z-z_o)^m|=\infty $$ but only that $$ \limsup_{m\to\infty}|c_m(z-z_o)^m|=\infty, $$ thus I have changed the final step in \eqref{2} from $\underset{m\to\infty}{\longrightarrow}\infty$ to $\underset{m\to\infty}{\nrightarrow}0$. However, the conclusion is the same, i.e. the power series \eqref{1} diverges for $|z-z_o|>R$ as a consequence of a standard result in analysis which states that if $\sum a_n$ converges, then $a_n\to 0$ (see for example [2], chapter IV, §1, theorem 1.1 pp. 153-154).

• The reasoning of the preceding point is true also when $c_m\to 0$ as $n\to\infty$, assuming that $0<R<+\infty$: to see this explicitly we should recall another fundamental theorem on the limit superior of a given sequence $\{a_m\}_{m\in\Bbb N}$:
  Theorem ([1], chapter III, §6, theorem 6.1(c) pp. 110-111) If $-\infty<\limsup_{n\to\infty} x_n$ and $-\infty<L$ is an extended real number, then $L \le \limsup x_n$ if and only if for each $B<L$ we have $x_n > B$ for infinitely many $n$'s.
  From this theorem the have that $$ \limsup_{n\to\infty} \sqrt[m]{|c_m|}=\frac{1}{R}\implies\sqrt[m]{|c_m|}>\frac{1}{R+\varepsilon}\iff|c_m|>\frac{1}{(R+\varepsilon)^m} $$ for infinitely many $m$'s. Since $|z-z_o|>R$, we can choose $\varepsilon$ in the following way: $$ 0<\varepsilon=\frac{|z-z_o|-R}{2}\iff |z-z_o|=R+2\varepsilon $$ The above choice implies the relation $$ |c_n(z-z_o)^m|=|c_m||z-z_o|^m>\frac{|z-z_o|^m}{(R+\varepsilon)^m}=\left(\frac{R+2\varepsilon}{R+\varepsilon}\right)^{\!m}\:\text{ for infinitely many}m\in\Bbb N $$ which in turn implies $$ \limsup_{m\to\infty}|c_m(z-z_o)^m|=\infty $$ and the consequences seen in the preceding point.

Bibliography

[1] Robert B. Burckel (1979), "An Introduction to Classical Complex Analysis", Vol. 1, Lehrbücher und Monographien aus dem Gebiete der exakten Wissenschaften. Mathematische Reihe, Band 64, Basel–Stuttgart: Birkhäuser Verlag, p. 570, ISBN 3-7643-0989-X, DOI:10.1007/978-3-0348-9374-9, MR0555733, Zbl 0434.30001.

[2] Emanuel Fischer (1983), "Intermediate Real Analysis", Undergraduate Texts in Mathematics, Berlin-Heidelberg-New York: Springer-Verlag, xiv+770, ISBN 0-387-90721-1, DOI: 10.1007/978-1-4613-9481-5, MR681692 (84e:26004), Zbl 0506.26002.