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 $r>0$ 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 $|x-c|<r$, then $a_n=b_n$ for all $n\in \mathcal{D}$.

PROOF

TODO

Convergence

Definition: Convergence

Let $\{a_n\}$ be a complex sequence.

A complex sequence $\{z_n\}$ converges to the limit $L\in \mathbb{C}$ if and only if for each $\epsilon >0$, there exists some integer $N$ such that

$$|a_n-L|<\epsilon \forall n\ge N$$

NOTATION

The most common notation is

$$\underset{n\to \infty }{\lim }{a}_n=L$$

In text, one also writes ”$a_n\to L$ as $n\to \infty$” or just "$a_n\to L$". Sometimes, one might also encounter $a_n\underset{n\to \infty }{⟶}L$ and $a_n\stackrel{n\to \infty }{⟶}L$.

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 $\epsilon >0$.

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

$$|a_n-L|<\epsilon \forall n\ge N_L.$$

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

$$|a_n-M|<\epsilon \forall n\ge N_M.$$

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

$$|L-M|=|L-a_n+a_n-M|\le |L-a_n|+|a_n-M|<\epsilon +\epsilon =2\epsilon .$$

Therefore,

$$\frac{1}{2}|L-M|<\epsilon$$

So far, the argument does not actually depend on the particular choice of $\epsilon$ and is thus true for every $\epsilon >0$. 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: Component-Wise Convergence

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

$$\underset{n\to \infty }{\lim }{z}_n=L⟺\underset{n\to \infty }{\lim }\mathrm{Re}(z_n)=\mathrm{Re}(L)\text{and}\underset{n\to \infty }{\lim }\mathrm{Im}(z_n)=\mathrm{Im}(L)$$

PROOF

TODO

Theorem: Cauchy Sequences

A complex sequence $\{z_n\}$ is convergent if and only if for each $\epsilon >0$ there exists some integer $N$ such that

$$\underset{n\to \infty }{\lim }|z_n-z_m|=0\forall n,m\ge N$$

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 $\epsilon >0$, there exists some integer $N$ such that

$$|a_n-L|<\epsilon \forall n\ge N.$$

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

$$|a_n-L|<1\forall n\ge N.$$

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 $n\ge N$,

$$|a_n-L|+|L|<1+|L|$$

and so $|a_n|<1+|L|$ 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 $n<N$;
• $|a_n|<1+|L|$ 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 $\{a_n\}$ converges to zero if and only if $\{|a_n|\}$ converges to zero.

$$\underset{n\to \infty }{\lim }{a}_n=0⟺\underset{n\to \infty }{\lim }|a_n|=0$$

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 $\{a_n\}$ and $\{b_n\}$ are both convergent complex sequences, then

$$\begin{array}{rl}& \underset{n\to \infty }{\lim }(\alpha a_n+\beta b_n)=\alpha \underset{n\to \infty }{\lim }{a}_n+\beta \underset{n\to \infty }{\lim }{b}_n\forall \alpha ,\beta \in \mathbb{C}\\ & \\ & \underset{n\to \infty }{\lim }(a_n\cdot b_n)=\underset{n\to \infty }{\lim }{a}_n\cdot \underset{n\to \infty }{\lim }{b}_n\\ & \\ & \underset{n\to \infty }{\lim }\frac{a_n}{b_n}=\frac{\underset{n\to \infty }{\lim }{a}_n}{\underset{n\to \infty }{\lim }{b}_n},\underset{n\to \infty }{\lim }{b}_n\ne 0\\ & \\ & \underset{n\to \infty }{\lim }\overline{a_n}=\overline{\underset{n\to \infty }{\lim }{a}_n}\\ & \\ & \underset{n\to \infty }{\lim }|a_n|=\left|\underset{n\to \infty }{\lim }{a}_n\right|\end{array}$$

PROOF

Let $A=\underset{n\to \infty }{\lim }{a}_n$ and $B=\underset{n\to \infty }{\lim }{b}_n$.

Proof of (1):

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

$$|\alpha a_n+\beta b_n-(\alpha A+\beta B)|<\epsilon \forall n\ge N.$$

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 $\epsilon >0$. 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 ${\epsilon }^{\prime }>0$. Since $a_n\to A$ and $b_n\to B$, there exist integers $N_A$ and $N_B$ such that

$$\begin{array}{r}|a_n-A|<{\epsilon }^{\prime }\forall n\ge N_A\\ |b_n-B|<{\epsilon }^{\prime }\forall n\ge N_B\end{array}$$

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 $|a_n-A|<{\epsilon }^{\prime }$ and $|b_n-B|<{\epsilon }^{\prime }$ and so we get

$$|\alpha ||a_n-A|+|\beta ||b_n-B|<|\alpha |{\epsilon }^{\prime }+|\beta |{\epsilon }^{\prime }.$$

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

$$|\alpha a_n+\beta b_n-(\alpha A+\beta B)|<|\alpha |{\epsilon }^{\prime }+|\beta |{\epsilon }^{\prime }\forall n\ge N$$

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

$$|\alpha a_n+\beta b_n-(\alpha A+\beta B)|<\epsilon \forall n\ge N,$$

which is what we set out to prove.

Proof of (2):

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

$$|a_n\cdot b_n-A\cdot B|<\epsilon \forall n\ge N.$$

Choose some arbitrary ${\epsilon }^{\prime }>0$. Since $a_n\to A$ and $b_n\to B$, there exist integers $N_A$ and $N_B$ such that

$$\begin{array}{rl}& |a_n-A|<{\epsilon }^{\prime }\forall n\ge N_A\\ & |b_n-B|<{\epsilon }^{\prime }\forall n\ge N_B.\end{array}$$

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 $n\ge N$,

$$|b_n|\cdot |a_n-A|+|A|\cdot |b_n-B|<|b_n|\cdot {\epsilon }^{\prime }+|A|\cdot {\epsilon }^{\prime },$$

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

$$|a_n\cdot b_n-A\cdot B|<|b_n|\cdot {\epsilon }^{\prime }+|A|\cdot {\epsilon }^{\prime }$$

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

$$|a_n\cdot b_n-A\cdot B|<B\cdot {\epsilon }^{\prime }+|A|\cdot {\epsilon }^{\prime }\forall n\ge \mathcal{N}.$$

Set $\epsilon =B\cdot {\epsilon }^{\prime }+|A|\cdot {\epsilon }^{\prime }$. Therefore,

$$|a_n\cdot b_n-A\cdot B|<\epsilon \forall n\ge \mathcal{N}.$$

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

$$|a_n\cdot b_n-A\cdot B|<\epsilon \forall n\ge \mathcal{N},$$

which is what we set out to show.

Proof of (3):

TODO

Proof of (4):

TODO

Proof of (5):

TODO