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11.4 Absolute Convergence

11.39   Definition (Absolute Convergence.) Let $f$ be a complex sequence. We say that $f$ is absolutely summable if and only if $\vert f\vert$ is summable; i.e., if and only if $\displaystyle {\{\sum_{j=0}^n\vert f(j)\vert\}}$ converges. In this case, we also say that the series $\sum f$ is absolutely convergent.

11.40   Example. $\displaystyle {\sum\left\{{{(-1)^n}\over n}\right\}_{n\geq 1}}$ is convergent, but is not absolutely convergent.

11.41   Theorem. Let $f$ be a complex sequence. If $\sum f$ is absolutely convergent, then $\sum f$ is convergent.


Case 1:
Suppose $f(n)$ is real for all $n\in\mbox{{\bf N}}$, and that $\sum\vert f\vert$ converges. Then

\begin{displaymath}0\leq f(n)+\vert f(n)\vert\leq\vert f(n)\vert+\vert f(n)\vert=2\vert f(n)\vert\end{displaymath}

for all $n\in\mbox{{\bf N}}$, so by the comparison test, $\sum(f+\vert f\vert)$ converges. Then $\sum(f+\vert f\vert)-\sum\vert f\vert$, being the difference of two convergent sequences, is convergent; i.e., $\sum f$ converges.

Case 2:
Suppose $f$ is an arbitrary absolutely convergent complex series. We know that for all $n\in\mbox{{\bf N}}$,

\begin{displaymath}0\leq \vert \mbox{\rm Re}(f)(n)\vert\leq \vert f(n)\vert\end{displaymath}


\begin{displaymath}0\leq \vert \mbox{\rm Im}(f)(n)\vert\leq \vert f(n)\vert,\end{displaymath}

so by the comparison test, $\sum\vert\mbox{\rm Re}(f)\vert$ and $\sum\vert\mbox{\rm Im}(f)\vert$ are convergent, and by Case 1, $\sum(\mbox{\rm Re}(f))$ and $\sum(\mbox{\rm Im}(f))$ are convergent. It follows that $\sum(\mbox{\rm Re}(f))+i\sum(\mbox{\rm Im}(f))=\sum f$ is convergent. $\mid\!\mid\!\mid$

11.42   Example. Let $z$ be a non-zero complex number. Let

\begin{displaymath}\{C_n(z)\}=\sum\{c_n(z)\}=\left\{\sum_{j=0}^n{{z^{2j}(-1)^j}\over {(2j)!}}\right\}.\end{displaymath}

I claim $ \sum\{c_n\}$ is absolutely convergent (and hence convergent). We have

\begin{displaymath}\vert c_n(z)\vert={{\vert z\vert^{2n}}\over {(2n)!}}.\end{displaymath}

We have

\begin{displaymath}\left\{ {{\vert c_{n+1}(z)\vert}\over {\vert c_n(z)\vert}}\ri...
...}=\left\{ {{\vert z\vert^2}\over
{(2n+1)(2n+2)}}\right\}\to 0<1\end{displaymath}

so by the ratio test, $\sum\{\vert c_n(z)\vert\}$ converges. Hence $\sum\{c_n(z)\}$ is absolutely convergent, and hence it is convergent. Clearly $\{C_n(0)\}\to 1$, so $\{C_n(z)\}$ converges for all $z\in\mbox{{\bf C}}$. In the exercises you will show that $\displaystyle {\sum\left\{ {{(-1)^jz^{2j+1}}\over {(2j+1)!}}\right\}}$ is also convergent for all $z\in\mbox{{\bf C}}$.

Motivated by the results of section 10.3, we make the following definitions:

11.43   Definition ($\sin$ and $\cos$.) For all $z\in\mbox{{\bf C}}$, we define

\cos(z)&=&\sum_{j=0}^\infty {{(-1)^jz^{2j}}\over {(2j)!}}.\\
\sin(z)&=&\sum_{j=0}^\infty {{(-1)^jz^{2j+1}}\over {(2j+1)!}}.

11.44   Remark. It is clear from the definition that

&\;&\sin(0)=0\mbox{ and }\cos(0)=1. \\
...}. \\
&\;&\cos(-z)=\cos(z)\mbox{ for all } z\in \mbox{{\bf C}}.

For all $n\in\mbox{{\bf N}}$, $z\in\mbox{{\bf C}}$, let

C_n(z)&=&\sum_{j=0}^n {{(-1)^jz^{2j}}\over {(2j)!}}, \cr
S_n(z)&=&\sum_{j=0}^n {{(-1)^jz^{2j+1}}\over {(2j+1)!}}.


S_n'(z)&=&\sum_{j=0}^n{{(-1)^j(2j+1)z^{2j}}\over {(2j+1)!}} \\
&=&\sum_{j=0}^n {{(-1)^jz^{2j}}\over {(2j)!}}=C_n(z).

I would now like to be able to say that for all $z\in\mbox{{\bf C}}$,

\{S_n(z)\}\to S(z)&\mbox{$\Longrightarrow$}&\{S_n'(z)\}\to S'(...
...\mbox{$\Longrightarrow$}&S'(z)=C(z)\mbox{ (since }\{C_n\}\to C);

i.e., I would like to have a theorem that says

\begin{displaymath}\{f_n(z)\}\to f(z)\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\{f_n'(z)\}\to f'(z).\end{displaymath}

However, the next example shows that this hoped for theorem is not true.

11.45   Example. Let $\displaystyle { f_n(z)={z\over {1+nz^2}}}$ for all $z\in\mbox{{\bf C}}$, $n\in\mbox{{\bf Z}}_{\geq 1}$. Then for all $z\in\mbox{{\bf C}}\backslash\{0\}$,

\begin{displaymath}\{f_n(z)\}={z\over n}{1\over {\left({1\over n}+z^2\right)}}\to0 \cdot {1\over {0+z^2}}=0,\end{displaymath}


\begin{displaymath}\{f_n(0)\}=\{0\}\to 0,\end{displaymath}


\begin{displaymath}f_n(z)\to\tilde 0(z)\mbox{ for all }z\in\mbox{{\bf C}}.\end{displaymath}

Now $\displaystyle {f_n'(z)={{(1+nz^2)-2nz^2}\over {(1+nz^2)^2}}={{1-nz^2}\over
{(1+nz^2)^2}}}$. So $f_n'(0)=1$ for all $n$, and thus $\{f_n'(0)\}\to 1\neq\tilde
0'(0)$. Eventually we will show that $\sin '=\cos$ and $\cos '=-\sin$, but it will require some work.

11.46   Warning. Defining sine and cosine in terms of infinite series can be dangerous to the well being of the definer. In 1933 Edmund Landau was forced to resign from his position at the University of Göttingen as a result of a Nazi-organized boycott of his lectures. Among other things, it was claimed that Landau's definitions of sine and cosine in terms of power series was ``un-German'', and that the definitions lacked ``sense and meaning''[33, pp 226-227].

11.47   Exercise. Show that $\displaystyle {\sum\left\{ {{(-1)^nz^{2n+1}}\over
{(2n+1)!}}\right\}}$ converges for all $z\in\mbox{{\bf C}}$.

11.48   Exercise. A
a) Does the series $\displaystyle {\sum\left\{ {{\left({3\over 5}+{{4i}\over 5}\right)^n}\over
{n^2}}\right\}_{n\geq 1}}$ converge?
b) Does the sequence $\displaystyle {\left\{ \sum_{j=1}^{4n}{{i^j}\over j}\right\}_{n\geq 1}}$ converge?

11.49   Exercise. A
a) For what complex numbers $z$ does $\sum\{nz^n\}$ converge?
b) For what complex numbers $z$ does $\sum\{z^{(n^2)}\}$ converge?

11.50   Note. The harmonic series was shown to be unbounded by Nicole
Oresme c. 1360 [31, p437]. However, many 17th and 18th century mathematicians believed that (in our terminology) every null sequence is summable. Jacob Bernoulli rediscovered Oresme's result in 1687, and reported that it contradicted his earlier belief that an infinite series whose last term vanishes must be finite[31, p 437]. As late as 1770, Lagrange said that a series represents a number if its $n$th term approaches $0$ [31, p 464].

The ratio test was stated by Jean D'Alembert in 1768, and by Edward Waring in 1776[31, p 465]. D'Alembert knew that the ratio test guaranteed absolute convergence.

The alternating series test appears in a letter from Leibniz to Jacob Bernoulli written in 1713[31, p461].

The series (11.35) for $\ln(1+t)$ is called Mercator's formula after Nicolaus Mercator who published it in 1668. It was discovered earlier by Newton in 1664 when he was an undergraduate at Cambridge. After Newton read Mercator's book, he quickly wrote down some of his own ideas (which were much more general than Mercator's) and allowed his notes to be circulated, but not published. Newton used the logarithm formula to calculate $\ln (1.1)$ to 68 decimals (of which the 28th and 43rd were wrong), but a few years later, he redid the calculation and corrected the errors.

See [22, chapter 2] for a discussion of Newton's work on series.

The series representation for $\arctan$ (11.37) is called Gregory's formula after John Gregory (1638-1675) or Leibniz's formula after Gottfried Leibniz (1646-1716). However, it was known to sixteenth century Indian mathematicians who credited it to Madhava (c. 1340-1425). The Indian version was

\begin{displaymath}\theta={{\sin\theta}\over {\cos\theta}}-{1\over 3}{{\sin^3\th...
...3\theta}}+{1\over 5}{{\sin^5\theta}\over {\cos^5\theta}}\cdots.\end{displaymath}

(See[30, p292].)

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