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Next: 7.3 Null Sequences Up: 7. Complex Sequences Previous: 7.1 Some Examples.   Index

7.2 Convergence

7.6   Definition (Convergent sequence.) Let $f$ be a complex sequence, and let $L\in\mbox{{\bf C}}$. We will say $f$ converges to $L$ and write $f\to L$ if for every disc $D(L,r)$ there is a number $N\in\mbox{{\bf N}}$ such that

\begin{displaymath}\mbox{ for every } n\in\mbox{{\bf Z}}_{\geq N},\;\;\left( f(n)\in D(L,r)\right).\end{displaymath}

We say $f$ converges if there is some $L\in\mbox{{\bf C}}$ such that $f\to L$. We say $f$ diverges if and only if $f$ does not converge.

It appears from figure a 7.1 that for every disc $D(0,r)$ centered at $0$ the terms of the sequence $\displaystyle { \left\{ \left( {{1+i}\over 2}\right)^n\right\}}$ eventually get into $D(0,r)$; i.e., it appears that $\displaystyle { \left\{
\left( {{1+i}\over 2}\right)^n\right\}\to 0}$. Similarly, it appears that $\displaystyle { \left\{ \left( {{1+2i}\over 3}\right)^n\right\}\to 0}$.

From figure b, it appears that there are numbers $P,Q$ such that $\displaystyle {\left\{ \sum_{j=0}^n\left( {{1+i}\over 2}\right)^j\right\}\to P}$, and $\displaystyle {\left\{ \sum_{j=0}^n\left( {{1+2i}\over 3}\right)^j\right\}\to Q}$. You should be able to put your finger on $P$ and $Q$, and maybe to guess what their exact values are. We will return to these examples later.

Let $\displaystyle {w={{7+24i}\over {25}}}$. The figure below represents the sequence $\{w^n\}$. It appears from the figure that there is no number $L$ such that $\{w^n\}\mapsto L$. The following theorem shows that this is the case. (Note that $\displaystyle { \left\vert {{7+24i}\over{25}}\right\vert=\sqrt{ {{49+576}\over

${\left\{\Big({7+24i\over 25}\Big)^n\right\} }$

7.7   Theorem. Let $w\in\mbox{{\bf C}}$ satisfy $\vert w\vert\geq 1$ and $w\neq 1$. Then $\{w^n\}$ diverges.

Proof: Suppose that $\vert w\vert\geq 1$ and $w\neq 1$. Then for all $n\in\mbox{{\bf N}}$,

\vert w^n-w^{n+1}\vert=\vert w^n(1-w)\vert=\vert w\vert^n \vert 1-w\vert\geq \vert 1-w\vert>0.
\end{displaymath} (7.8)

Now suppose, to get a contradiction, that there is a number $L\in\mbox{{\bf C}}$ such that $\{w^n\}\to L$. Then corresponding to the disc $\displaystyle {D\left( L,{{\vert 1-w\vert}\over
2}\right)}$, there is a number $N\in\mbox{{\bf N}}$ such that

\begin{displaymath}n\in\mbox{{\bf Z}}_{\geq N}\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}w^n\in D\left(L,{{\vert 1-w\vert}\over 2}\right).\end{displaymath}

In particular,

\begin{displaymath}w^N\in D\left(L,{{\vert 1-w\vert}\over 2}\right) \mbox{ and } w^{N+1}\in
D\left(L,{{\vert 1-w\vert}\over 2}\right)\end{displaymath}


\begin{displaymath}\vert w^N-L\vert<{{\vert 1-w\vert}\over 2}\mbox{ and } \vert w^{N+1}-L\vert<{{\vert 1-w\vert}\over 2}.\end{displaymath}

By the triangle inequality,

\vert w^N-w^{N+1}\vert&=&\vert(w^N-L)+(L-w^{N+1})\vert \\
...vert 1-w\vert}\over 2}+{{\vert 1-w\vert}\over 2}=\vert 1-w\vert.

Combining this result with (7.8), we get

\begin{displaymath}\vert 1-w\vert\leq\vert w^N-w^{N+1}\vert<\vert 1-w\vert,\end{displaymath}

so $\vert 1-w\vert<\vert 1-w\vert$. This contradiction shows that $\{w^n\}$ diverges. $\mid\!\mid\!\mid$

We can also show that constant sequences converge.

7.9   Theorem. Let $\alpha\in\mbox{{\bf C}}$. Then the constant sequence $\tilde\alpha$ converges to $\alpha$.

Proof: Let $\alpha\in\mbox{{\bf C}}$. Let $D(\alpha,r)$ be a disc centered at $\alpha$. Then

\begin{displaymath}\tilde\alpha (n)=\alpha\in D(\alpha,r) \mbox{ for all }n\in\mbox{{\bf Z}}_{\geq 0},\end{displaymath}

Hence, $\tilde\alpha\to\alpha$. $\mid\!\mid\!\mid$

For purposes of calculation it is sometimes useful to rephrase the definition of convergence. Since the disc $D(\alpha,r)$ is determined by its radius $r$, and for all $z\in\mbox{{\bf C}}$, $z\in D(\alpha,r)\hspace{1ex}\Longleftrightarrow\hspace{1ex}\vert z-\alpha\vert<r$, we can reformulate definition 7.6 as

7.10   Definition (Convergence.) Let $f$ be a sequence in $\mbox{{\bf C}}$, and let $L\in\mbox{{\bf C}}$. Then $f\to L$ if and only if for every $r\in\mbox{${\mbox{{\bf R}}}^{+}$}$ there is some $N\in\mbox{{\bf N}}$ such that

\begin{displaymath}\mbox{ for every } n\in\mbox{{\bf Z}}_{\geq N},\;\; (\vert f(n)-L\vert<r).\end{displaymath}

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Next: 7.3 Null Sequences Up: 7. Complex Sequences Previous: 7.1 Some Examples.   Index