Consider a complex power series \(\displaystyle \sum_{k=0}^\infty a_k z^k\) with radius of convergence \(0 \lt R \lt \infty\) and suppose that for every \(w\) with \(\vert w \vert = R\), \(\displaystyle \sum_{k=0}^\infty a_k w^k\) converges.

We provide an example where the power expansion at the origin \[
\displaystyle f(z) = \sum_{k=0}^\infty a_k z^k\] is discontinuous on the closed disk \(\vert z \vert \le R \).

The function \(f\) is constructed as an infinite sum \[
\displaystyle f(z) = \sum_{n=1}^\infty f_n(z)\] with \(f_n(z) = \frac{\delta_n}{a_n-z}\) where \((\delta_n)_{n \in \mathbb N}\) is a sequence of positive real numbers and \((a_n)\) a sequence of complex numbers of modulus larger than one and converging to one. Let \(f_n^{(r)}(z)\) denote the sum of the first \(r\) terms in the power series expansion of \(f_n(z)\) and \(\displaystyle f^{(r)}(z) \equiv \sum_{n=1}^\infty f_n^{(r)}(z)\).

We’ll prove that:

1. If \(\sum_n \delta_n \lt \infty\) then \(\sum_{n=1}^\infty f_n^{(r)}(z)\) converges and \(f(z) = \lim\limits_{r \to \infty} \sum_{n=1}^\infty f_n^{(r)}(z)\) for \(\vert z \vert \le 1\) and \(z \neq 1\).
2. If \(a_n=1+i \epsilon_n\) and \(\sum_n \delta_n/\epsilon_n < \infty\) then \(\sum_{n=1}^\infty f_n^{(r)}(1)\) converges and \(f(1) = \lim\limits_{r \to \infty} \sum_{n=1}^\infty f_n^{(r)}(1)\)
3. If \(\delta_n/\epsilon_n^2 \to \infty\) then \(f(z)\) is unbounded on the disk \(\vert z \vert \le 1\).

First, let’s recall this corollary of Lebesgue’s dominated convergence theorem:

Let \((u_{n,i})_{(n,i) \in \mathbb N \times \mathbb N}\) be a double sequence of complex numbers. Suppose that \(u_{n,i} \to v_i\) for all \(i\) as \(n \to \infty\), and that \(\vert u_{n,i} \vert \le w_i\) for all \(n\) with \(\sum_i w_i < \infty\). Then for all \(n\) the series \(\sum_i u_{n,i}\) is absolutely convergent and \(\lim_n \sum_i u_{n,i} = \sum_i v_i\).
Continue reading A power series converging everywhere on its circle of convergence defining a non-continuous function →

The root test is a test for the convergence of a series \[
\sum_{n=1}^\infty a_n \] where each term is a real or complex number. The root test was developed first by Augustin-Louis Cauchy.

We denote \[l = \limsup\limits_{n \to \infty} \sqrt[n]{\vert a_n \vert}.\] \(l\) is a non-negative real number or is possibly equal to \(\infty\). The root test states that:

• if \(l < 1\) then the series converges absolutely;
• if \(l > 1\) then the series diverges.

The root test is inconclusive when \(l = 1\).

A case where \(l=1\) and the series diverges

The harmonic series \(\displaystyle \sum_{n=1}^\infty \frac{1}{n}\) is divergent. However \[\sqrt[n]{\frac{1}{n}} = \frac{1}{n^{\frac{1}{n}}}=e^{- \frac{1}{n} \ln n} \] and \(\limsup\limits_{n \to \infty} \sqrt[n]{\frac{1}{n}} = 1\) as \(\lim\limits_{n \to \infty} \frac{\ln n}{n} = 0\).

A case where \(l=1\) and the series converges

Consider the series \(\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}\). We have \[\sqrt[n]{\frac{1}{n^2}} = \frac{1}{n^{\frac{2}{n}}}=e^{- \frac{2}{n} \ln n} \] Therefore \(\limsup\limits_{n \to \infty} \sqrt[n]{\frac{1}{n^2}} = 1\), while the series \(\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}\) is convergent as we have seen in the ratio test article. Continue reading Root test →

The ratio test is a test for the convergence of a series \[
\sum_{n=1}^\infty a_n \] where each term is a real or complex number and is nonzero when \(n\) is large. The test is sometimes known as d’Alembert’s ratio test.

Suppose that \[\lim\limits_{n \to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert = l\] The ratio test states that:

• if \(l < 1\) then the series converges absolutely;
• if \(l > 1\) then the series diverges.

What if \(l = 1\)? One cannot conclude in that case.

Cases where \(l=1\) and the series diverges

Consider the harmonic series \(\displaystyle \sum_{n=1}^\infty \frac{1}{n}\). We have \(\lim\limits_{n \to \infty} \frac{n+1}{n} = 1\). It is well know that the harmonic series diverges. Recall that one proof uses the Cauchy’s convergence test based for \(k \ge 1\) on the inequalities: \[
\sum_{n=2^k+1}^{2^{k+1}} \frac{1}{n} \ge \sum_{n=2^k+1}^{2^{k+1}} \frac{1}{2^{k+1}} = \frac{2^{k+1}-2^k}{2^{k+1}} \ge \frac{1}{2}\]

An even simpler case is the series \(\displaystyle \sum_{n=1}^\infty 1\).

Cases where \(l=1\) and the series converges

We also have \(\lim\limits_{n \to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert = 1\) for the infinite series \(\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}\). The series is however convergent as for \(n \ge 1\) we have:\[
0 \le \frac{1}{(n+1)^2} \le \frac{1}{n(n+1)} = \frac{1}{n} – \frac{1}{n+1}\] and the series \(\displaystyle \sum_{n=1}^\infty \left(\frac{1}{n} – \frac{1}{n+1} \right)\) obviously converges.

Another example is the alternating series \(\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n}\).