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Partial Sums#

Definition: Partial Sum

Let \(\{z_n\}_{n \in \mathcal{D}}\) be a complex sequence.

The \(k\)-th partial sum of \(\{z_n\}\) is the sum of its first \(k\) numbers:

\[ \sum_{\begin{aligned}j &\in \mathcal{D} \\ j &\le k \end{aligned}} a_j \]
Notation
\[ s_k \]

Complex Series#

Given a complex sequence \((z_n)_{n \in \mathcal{D}}\), it is apparent that the function which maps each \(k \in \mathcal{D}\) to the \(k\)-th partial sum of \((z_n)_{n \in \mathcal{D}}\) is itself a complex sequence \((s_n)_{n \in \mathcal{D}}\).

Definition: Complex Series

A complex series is the complex sequence of the partial sums of some other complex sequence.

Notation

If the sequence is \((z_n)_{n \in \mathcal{D}}\), then we denote the sequence of its partial sums by \(\displaystyle \sum_{n \in \mathcal{D}} z_n\).

When \(\mathcal{D} = \mathbb{N}_0\) or \(\mathcal{D} = \mathbb{N}\), we write \(\displaystyle \sum_{n = 0}^{\infty} z_n\) or \(\displaystyle \sum_{n = 1}^{\infty} z_n\), respectively.

In the case, when \(\mathcal{D} = \mathbb{N} \setminus \{0, 1, \dotsc, p - 1\}\) for some integer \(p\), we write \(\displaystyle \sum_{n = p}^{\infty} z_n\).

Convergence#

Since a complex series \(\displaystyle \sum_{n \in \mathcal{D}} z_n\) is just a complex sequence, the definitions and terminology for convergence of series is the same as those used for convergence of sequence,. What is different, however, is the notation used.

Notation: Convergence of Complex Series

If \(\displaystyle \sum_{n \in \mathcal{D}} z_n\) converges, to \(L\) we write

\[ \sum_{n \in \mathcal{D}} z_n = L \]

Instead of calling \(L\) the limit of \(\displaystyle \sum_{n \in \mathcal{D}} z_n\), we usually say it is its value.

Definition: Absolute Convergence of Complex Series

A complex series \(\displaystyle \sum_{n \in \mathcal{D}} z_n\) converges absolutely to \(L\) iff the series \(\displaystyle \sum_{n \in \mathcal{D}} |z_n|\) converges to \(L\).

Example: \(\sum_{n=1}^{\infty} \frac{2}{n^2}-\mathrm{i}\frac{(-1)^n}{n^2}\)

The complex series \(\sum_{n=1}^{\infty} \frac{2}{n^2}-\mathrm{i}\frac{(-1)^n}{n^2}\) is absolutely convergent, since

\[ \sum_{n=1}^{\infty} \left|\frac{2}{n^2}-\mathrm{i}\frac{(-1)^n}{n^2}\right| = \sum_{n=1}^{\infty} \sqrt{\frac{4}{n^4}+\frac{1}{n^4}} = \sum_{n=1}^{\infty}\frac{\sqrt{5}}{n^2}, \]

which is convergent.

Definition: Conditional Convergence

A complex series is conditionally convergent if it is convergent but not absolutely convergent.

Theorem: Absolute Convergence \(\implies\) Convergence

If a complex series is absolutely convergent towards \(L \in \mathbb{C}\), then it is also convergent towards \(L\).

Proof

TODO

Theorem: Convergence \(\implies\) Zero Sequence

Let \(\displaystyle \sum_{n \in \mathcal{D}} z_n\) be a complex series.

If \(\displaystyle \sum_{n \in \mathcal{D}} z_n\) converges, then \((z_n)_{n \in \mathcal{D}}\) converges to \(0\).

Proof

TODO

Theorem: Arithmetic with Convergent Series

If \(\sum_{n\in\mathcal{D}} z_n\) and \(\sum_{n \in \mathcal{D}}w_n\) converge, then

\[ \sum_{n \in \mathcal{D}} (\lambda z_n + \mu w_n) = \lambda \sum_{n \in \mathcal{D}} z_n + \mu \sum_{n \in \mathcal{D}} w_n \]

for all \(\lambda, \mu \in \mathbb{C}\).

Proof

TODO

Theorem: Majorant Convergence Test

Let \(\sum_{n \in \mathcal{D}} z_n\) be a complex series.

If there exist a convergent real series \(\sum_{n \in \mathcal{D}} b_n\) and some \(N \in \mathcal{D}\) such that

\[ |z_n| \le b_n \qquad \forall n \ge N, \]

then \(\sum_{n = 1}^{\infty} z_n\) is absolutely convergent.

Proof

TODO

Example: \(\sum_{n=1}^{\infty} z_n = \frac{(1+\mathrm{i})^n}{3^n n}\)

We examine the following complex series:

\[ \sum_{n=1}^{\infty} z_n = \frac{(1+\mathrm{i})^n}{3^n n} \]

We have:

\[ |z_n| = \frac{|1+\mathrm{i}|^n}{3^n n} = \frac{\sqrt{2}^n}{3^n n} = \left(\frac{\sqrt{2}}{3}\right)^n\frac{1}{n} \]

It is obvious that

\[ |z_n| = \left(\frac{\sqrt{2}}{3}\right)^n\frac{1}{n} \le \left(\frac{\sqrt{2}}{3}\right)^n \]

for all \(n \in \mathbb{N}\) and since \(\sum_{n = 1}^{\infty} \left(\frac{\sqrt{2}}{3}\right)^n\) converges as a geometric series with \(q \lt 1\), we know that \(\sum_{n=1}^{\infty} z_n\) is absolutely convergent.

Theorem: Ratio Convergence Test

Let \(\sum_{n \in \mathcal{D}} z_n\) be a complex Series.

  • (I) If there exists some \(N \in \mathcal{D}\) and some \(q \in (0;1)\) such that \(\left\vert \frac{z_{n+1}}{z_n}\right\vert \le q\) for all \(n \ge N\), then \(\sum_{n \in \mathcal{D}} z_n\) is absolutely convergent.

  • (II) If there exists some \(N \in \mathcal{D}\) such that \(\left\vert \frac{z_{n+1}}{z_n}\right\vert \ge 1\) for all \(n \ge N\), then \(\sum_{n \in \mathcal{D}} z_n\) is divergent.

Proof

TODO

Theorem: \(n\)-th Root Convergence Test

Let \(\sum_{n \in \mathcal{D}} z_n\) be a complex series.

If the real sequence \(\sqrt[n]{|z_n|}\) converges, then \(\sum_{n \in \mathcal{D}} z_n\) is:

\[ \displaystyle \lim_{n \to \infty} \sqrt[n]{|z_n|} \lt 1 \]
\[ \displaystyle \lim_{n \to \infty} \sqrt[n]{|z_n|} \gt 1 \]
Proof

TODO