Complex Sequences#

Definition: Complex Sequence

A complex sequence is a sequence of complex numbers.

Theorem: Equality of Complex Sequences

Let $(a_n)_{n \in \mathcal{D}}$ and $(b_n)_{n \in \mathcal{D}}$ be infinite complex sequences.

If there exist complex power series $\sum_{n \in \mathcal{D}} a_n (x - c)^n$ and $\sum_{n \in \mathcal{D}} b_n (x - c)^n$ which have the same center $c \in \mathbb{C}$ and for which there is some $r \gt 0$ such that both series converge to the same number, i.e.

\[ \sum_{n \in \mathcal{D}} a_n (x - c)^n = \sum_{n \in \mathcal{D}} b_n (x - c)^n, \]

when when $|x - c| \lt r$, then $a_n = b_n$ for all $n \in \mathcal{D}$.

Proof

TODO

Convergence#

Definition: Convergence

A complex sequence $(z_n)_{n \in \mathcal{D}}$ converges to the limit $L \in \mathbb{C}$ if and only if for each $\varepsilon \gt 0$, there exists some $N \in \mathcal{D}$ such that

\[ |z_n - L| \lt \varepsilon \qquad \forall n \ge N. \]

Notation

The most common notation is

\[ \lim_{n \to \infty} z_n = L \]

In text, one also writes "$z_n \to L$ as $n \to \infty$" or just "$z_n \to L$". Sometimes, one might also encounter $z_n \underset{n \to \infty}{\longrightarrow} L$ and $z_n \overset{n \to \infty}{\longrightarrow} L$.

Theorem: Uniqueness of the Limit

The limit of a convergent complex sequence is unique - if $\{z_n\}$ converges to both $L$ and $M$, then $L = M$.

Proof

Pick some arbitrary $\varepsilon \gt 0$.

Since $\{a_n\} \to L$, by definition, there exists some integer $N_L$ such that

\[ |a_n - L| \lt \varepsilon \qquad \forall n \ge N_L. \]

Similarly, since $\{a_n\} \to M$, there exists some integer $N_M$ such that

\[ |a_n - M| \lt \varepsilon \qquad \forall n \ge N_M. \]

Now, let $N = \max\{N_L, N_M\}$. For all $n \ge N$, both of the aforementioned inequalities hold. Therefore, for all $n \ge N$, we have

\[ |L - M| = |L - a_n + a_n - M| \le |L - a_n| + |a_n - M| \lt \varepsilon + \varepsilon = 2\varepsilon. \]

Therefore,

\[ \frac{1}{2}|L - M| \lt \varepsilon \]

So far, the argument does not actually depend on the particular choice of $\varepsilon$ and is thus true for every $\varepsilon \gt 0$. This means that $\frac{1}{2}|L - M|$ is smaller that every positive real number. This is only possible if $\frac{1}{2}|L - M|$ is zero which is in turn only possible if $|L - M| = 0$. Therefore, $L = M$.

Theorem: Complex Convergence via Real Convergence

A complex sequence $(z_n)_{n \in \mathcal{D}}$ converges to $L \in \mathbb{C}$ the sequence $|z_n - L|$ converges to $0$:

\[ \lim_{n \to \infty} z_n = L \iff \lim_{n \to \infty} |z_n - L| = 0 \]

Proof

TODO

Example: $z_n = \frac{1}{2n} + \frac{n}{n+1}\mathrm{i}$

We examine the following complex sequence $(z_n)_{n \in \mathbb{N}}$:

\[ z_n = \frac{1}{2n} + \frac{n}{n+1}\mathrm{i} \]

Since $\frac{1}{2n} \to 0$ and $\frac{n}{n+1} \to 1$, we see that $(z_n)$ probably converges to $\mathrm{i}$. To show this, we examine $z_n - \mathrm{i}$:

\[ z_n - \mathrm{i} = \frac{1}{2n} + \left(\frac{n}{n+1} -1\right)\mathrm{i} = \frac{1}{2n} + \frac{-1}{n+1}\mathrm{i} \]

We now examine the limit of the absolute values:

\[ \lim_{n\to\infty} |z_n - \mathrm{i}| = \sqrt{\frac{1}{4n^2}+\frac{1}{(n+1)}} = 0 \]

Therefore, $(z_n)$ converges to $\mathrm{i}$.

Theorem: Component-Wise Convergence

A complex sequence $(z_n)_{n \in \mathcal{D}}$ converges to $L \in \mathbb{C}$ if and only if the sequences of its real part and its imaginary part converge to the real part and imaginary part of $L$, respectively.

\[ \lim_{n \to \infty} z_n = L \iff \lim_{n \to \infty} \operatorname{Re}(z_n) = \operatorname{Re}(L) \qquad \text{and} \qquad \lim_{n \to \infty} \operatorname{Im}(z_n) = \operatorname{Im}(L) \]

Proof

TODO

Example: $z_n = \left(1 - \frac{3}{n}\right)^n + \mathrm{i}\frac{2n^2+3n+5}{n^2+1}$

We examine the following complex sequence $(z_n)_{n\in\mathbb{N}}$:

\[ z_n = \left(1 - \frac{3}{n}\right)^n + \mathrm{i}\frac{2n^2+3n+5}{n^2+1} \]

Since $\lim_{n\to\infty} \left(1 - \frac{3}{n}\right)^n = \mathrm{e}^{-3}$ and $\lim_{n\to\infty} \frac{2n^2+3n+5}{n^2+1} = 2$, we know that $(z_n)_{n\to\mathcal{N}}$ converges to $\mathrm{e}^{-3} + 2\mathrm{i}$:

\[ \lim_{n \to \infty} = \left(1 - \frac{3}{n}\right)^n + \mathrm{i}\frac{2n^2+3n+5}{n^2+1} = \mathrm{e}^{-3}+2\mathrm{i} \]

Example: $z_n = \frac{5}{n} + (-1)^n\mathrm{i}$

We examine the following complex sequence $(z_n)_{n\in\mathbb{N}}$:

\[ z_n = \frac{5}{n} +(-1)^n \mathrm{i} \]

Since $\Im(z_n) = (-1)^n$ diverges, we know that $(z_n)_{n \to\mathbb{N}}$ must also diverge.

Theorem: Cauchy Sequences

A complex sequence $(z_n)_{n \in \mathcal{D}}$ is convergent if and only if for each $\varepsilon \gt 0$ there exists some $N \in \mathcal{D}$ such that

\[ \lim_{n \to \infty} |z_n - z_m| = 0 \qquad \forall n,m \ge N \]

Proof

TODO

Note

Sequences for which the above holds, i.e. convergent sequences, are also known as Cauchy sequences.

Theorem: Boundedness of Convergent Sequences

Every convergent complex sequence is bounded.

Proof

Suppose that $\{a_n\}$ converges to some $L \in \mathbb{C}$. Then, by definition, for each $\varepsilon \gt 0$, there exists some integer $N$ such that

\[ |a_n - L| \lt \varepsilon \qquad \forall n \ge N. \]

Choose $\varepsilon = 1$. The actual choice is irrelevant, it will just result in a different bound. Then,

\[ |a_n - L| \lt 1 \qquad \forall n \ge N. \]

Let's look at the absolute value of $a_n$:

\[ |a_n| = |a_n - L + L| \]

Using the triangle inequality, we get

\[ |a_n| = |a_n - L + L| \le |a_n - L| + |L|. \]

For all $n \ge N$,

\[ |a_n - L| + |L| \lt 1 + |L| \]

and so $|a_n| \lt 1 + |L|$ for all $n \ge N$. This means that the modulus of all sequence terms from the $N$-th one onwards is less than $1 + |L|$. Amongst the first $N-1$ terms of the sequence, choose the one whose modulus $M$ is greatest. The moduli of the first $N-1$ terms are thus all less than or equal to $M$. Essentially, we have

$|a_n| \le M$ for every $n \lt N$;
$|a_n| \lt 1 + |L|$ for every $n \ge N$.

Let $B = \max\{M, 1 + |L|\}$. Therefore, $|a_n| \le B$ for every integer $n$ and so $a_n$ is bounded.

Theorem: Convergence to Zero

A complex sequence $(z_n)_{n \in \mathcal{D}}$ converges to $0$ if and only if the real sequence $(|z_n|)_{n \in \mathcal{D}}$ of its absolute values converges to $0$.

\[ \lim_{n \to \infty} z_n = 0 \iff \lim_{n \to \infty} |z_n| = 0 \]

Proof

TODO

Theorem

Let $\{a_n\}$ and $\{b_n\}$ be complex sequences.

If $\{a_n\}$ converges to zero and there exists some integer $N$ such that $|b_n| \le |a_n|$ for all $n \ge N$, then $\{b_n\}$ also converges to zero.

Proof

TODO

Theorem: Limit Arithmetic

If $(a_n)_{n\in\mathcal{D}}$ and $(b_n)_{n\in\mathcal{D}}$ are both convergent complex sequences, then

\[\begin{aligned}& \lim_{n \to \infty} (\alpha a_n + \beta b_n) = \alpha \lim_{n \to \infty} a_n + \beta \lim_{n \to \infty} b_n \qquad \forall \alpha, \beta \in \mathbb{C} \\ & \lim_{n \to \infty} (a_n \cdot b_n) = \lim_{n \to \infty} a_n \cdot \lim_{n \to \infty} b_n \\ & \lim_{n \to \infty} \frac{a_n}{b_n} = \frac{\displaystyle \lim_{n \to \infty} a_n}{\displaystyle \lim_{n \to \infty} b_n}, \qquad \lim_{n \to \infty} b_n \ne 0 \\ & \lim_{n \to \infty} \overline{a_n} = \overline{\lim_{n \to \infty} a_n} \\ & \lim_{n \to \infty} |a_n| = \left|\lim_{n \to \infty} a_n\right|\end{aligned}\]

Proof

Let $A = \displaystyle \lim_{n \to \infty} a_n$ and $B = \displaystyle \lim_{n \to \infty} b_n$.

Proof of (1):

We have to prove that for each $\varepsilon \gt 0$, there exists some integer $N$ such that

\[ |\alpha a_n + \beta b_n - (\alpha A + \beta B)| \lt \varepsilon \qquad \forall n \ge N. \]

The case of $\alpha = \beta = 0$ is trivial, since then $|\alpha a_n + \beta b_n - (\alpha A + \beta B)|$ is always zero and is thus smaller than all $\varepsilon \gt 0$. As for the existence of $N$ - not only does such an integer $N$ exist, but there are actually infinitely many such integers, since the inequality holds irrespective of $N$.

If $\alpha$ and $\beta$ are not zero, choose some arbitrary $\varepsilon' \gt 0$. Since $a_n \to A$ and $b_n \to B$, there exist integers $N_A$ and $N_B$ such that

$$
\begin{aligned}

|a_n - A| \lt \varepsilon' \qquad \forall n \ge N_A \

|b_n - B| \lt \varepsilon' \qquad \forall n \ge N_B

\end{aligned}
$$

Let $N = \max \{N_A, N_B\}$. Then both inequalities hold for every $n \ge N$. Now, we look at

\[ |\alpha a_n + \beta b_n - (\alpha A + \beta B)| = |\alpha (a_n - A) + \beta (b_n - B)|. \]

By the triangle inequality, we obtain

\[ |\alpha (a_n - A) + \beta (b_n - B)| \le |\alpha (a_n - A)| + |\beta (b_n - B)| = |\alpha| |a_n - A| + |\beta| |b_n - B|. \]

For every $n \ge N$, we know that $|a_n - A| \lt \varepsilon'$ and $|b_n - B| \lt \varepsilon'$ and so we get

\[ |\alpha| |a_n - A| + |\beta| |b_n - B| \lt |\alpha| \varepsilon' + |\beta| \varepsilon'. \]

Since $|\alpha (a_n - A) + \beta (b_n - B)| \le |\alpha| |a_n - A| + |\beta| |b_n - B|$ and $|\alpha a_n + \beta b_n - (\alpha A + \beta B)| = |\alpha (a_n - A) + \beta (b_n - B)|$, we get

So far, the argument does not depend on the choice of $\varepsilon'$ which means that it must be true for all $\varepsilon' \gt 0$. Thus, we have proven that for each $\varepsilon = |\alpha| \varepsilon' + |\beta| \varepsilon'$, there exists some integer, namely $N$, such that

\[ |\alpha a_n + \beta b_n - (\alpha A + \beta B)| \lt \varepsilon \qquad \forall n \ge N, \]

which is what we set out to prove.

Proof of (2):

We have to prove that for each $\varepsilon \gt 0$, there exists some integer $N$ such that

\[ |a_n \cdot b_n - A \cdot B| \lt \varepsilon \qquad \forall n \ge N. \]

Choose some arbitrary $\varepsilon' \gt 0$. Since $a_n \to A$ and $b_n \to B$, there exist integers $N_A$ and $N_B$ such that

$$
\begin{aligned}

&|a_n - A| \lt \varepsilon' \qquad \forall n \ge N_A \

&|b_n - B| \lt \varepsilon' \qquad \forall n \ge N_B.

\end{aligned}
$$

Let $N = \max \{N_A, N_B\}$. Then both inequalities hold for all $n \ge N$. We transform the expression $|a_n \cdot b_n - A \cdot B|$ a bit:

\[ |a_n \cdot b_n - A \cdot B| = |a_n \cdot b_n - A \cdot b_n + A \cdot b_n - A \cdot B|. \]

Using the triangle inequality, we obtain

\[ |a_n \cdot b_n - A \cdot b_n + A \cdot b_n - A \cdot B| \le |a_n \cdot b_n - A \cdot b_n| + |A \cdot b_n - A \cdot B|, \]

which is equivalent to

\[ |a_n \cdot b_n - A \cdot B| \le |b_n| \cdot |a_n - A| + |A| \cdot |b_n - B|. \]

For all $n \ge N$,

\[ |b_n| \cdot |a_n - A| + |A| \cdot |b_n - B| \lt |b_n| \cdot \varepsilon' + |A| \cdot \varepsilon', \]

which in turn means that, for all $n \ge N$,

\[ |a_n \cdot b_n - A \cdot B| \lt |b_n| \cdot \varepsilon' + |A| \cdot \varepsilon' \]

Recall that every convergent sequence is bounded. This means that there exists some $B \gt 0$ and some integer $\mathcal{N}$ such that $|b_n| \le B$ for all $n \ge \mathcal{N}$. Therefore,

\[ |a_n \cdot b_n - A \cdot B| \lt B \cdot \varepsilon' + |A| \cdot \varepsilon' \qquad \forall n \ge \mathcal{N}. \]

Set $\varepsilon = B \cdot \varepsilon' + |A| \cdot \varepsilon'$. Therefore,

\[ |a_n \cdot b_n - A \cdot B| \lt \varepsilon \qquad \forall n \ge \mathcal{N}. \]

It is obvious that $\varepsilon$ can be any positive real number. Moreover, since the argument so far does not depend on any particular choice of $\varepsilon$, we know the argument holds for all $\varepsilon \gt 0$. We have thus proven that for each $\varepsilon \gt 0$, there exists some integer, namely $\mathcal{N}$, such that

\[ |a_n \cdot b_n - A \cdot B| \lt \varepsilon \qquad \forall n \ge \mathcal{N}, \]

which is what we set out to show.

Proof of (3):

TODO

Proof of (4):

TODO

Proof of (5):

TODO