\[ \newcommand{\Arg}{\mathsf{Arg}} \newcommand{\C}{\mathbb{C}} \newcommand{\CC}{\mathbb{C}} \newcommand{\R}{\mathbb{R}} \newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Im}{\mathsf{Im}} \newcommand{\intd}{\,\mathsf{d}} \newcommand{\Re}{\mathsf{Re}} \newcommand{\ball}{\mathsf{B}} \newcommand{\wind}{\mathsf{wind}} \]

1.4 Series

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We have already discussed what it means for a sequence of complex numbers to converge. Recall that if $z_n$ is a sequence in $\C$ then we say $z_n$ converges to $a \in \C$ if for all $\epsilon > 0$ there exists $N \in \N$ such that $|z_n-a| < \epsilon$ for all $n \ge N$.

Definition (Series)

Fix a sequence $z_1,z_2,z_3,\dots$ in $\C$. We can form a new sequence \[ \begin{align*} w_1 &= z_1 \\ w_2 &= z_1 + z_2 \\ w_3 &= z_1 + z_2 + z_3 \\ &\;\;\vdots \\ w_m &= z_1 + z_2 + z_3 + \cdots + z_n \\ &\;\;\vdots \end{align*} \] by adding together terms. We can write \[w_m = \sum_{n=1}^m z_n\] using summation notation. Any sequence $w_1,w_2,w_3,\dots$ defined in this way is caled a series.

We usually write a series as \[\sum_{n=1}^\infty z_n\] without explicitly writing down the summands $w_m$.

Definition (Convergence of a Series)

Let $z_n$ be a sequence in $\C$. We say that the series \[\sum_{n=1}^{\infty} z_n\] converges if the sequence $w_m$ of partial sums \[w_m = \sum_{n=1}^m z_n\] converges. The limit of this sequence of partial sums is called the sum of the series and we say that the series converges to the limit. A series which does not converge is called divergent.

Example

The series \[ \sum_{n=1}^\infty n \] diverges, because the sequence \[w_m = 1 + 2 + \cdots + m = \dfrac{m(m+1)}{2}\] of partial sums diverges.

It is often difficult to determine whether a series converges, and even harder to determine the limit of a convergin series. A lot of what we will be able to calculate about series comes from the geometric series.

Example (The geometric series)

Fix $a \in \C$ with $|a| < 1$. The series \[ \sum_{n=1}^\infty a^n \] is called the geometric series. From \[ \begin{aligned} w_m & = a + a^2 + \cdots + a^m \\ a w_m & = a^2 + \cdots + a^m + a^{m+1} \end{aligned} \] we get $w_m (1-a) = a - a^{m+1}$. Since $|a| < 1$ we have $|a^m| \to 0$ as $m \to \infty$ and \[\lim_{m \to \infty} w_m = \lim_{m \to \infty} \frac{1 - a^{m+1}}{1-a} = \frac{a}{1-a}\] so \[\displaystyle\sum_{n=1}^\infty a^n = \dfrac{a}{1-a}\] whenever $|a| < 1$.

Theorem (Properties of Converging Series)

Suppose that \[\sum_{n=1}^\infty z_n\] and \[\sum_{n=1}^\infty u_n\] converge to $a$ and $b$ respectively. Then the following are all true.

  1. $|z_n| \to 0$ as $n \to \infty$
  2. $\displaystyle \sum_{n=1}^\infty z_n + u_n$ converges to $a+b$
  3. $\displaystyle \sum_{n=1}^\infty c z_n$ converges to $ca$
Definition

The series \[\sum_{n=1}^\infty z_n\] is said to converges absolutely if the series \[\displaystyle \sum_{n=1}^\infty |z_n|\] converges.

Lemma

If a series converges absolutely then it converges.

Proof:

We prove that the sequence \[w_m = \displaystyle \sum_{n=1}^m z_n\] is Cauchy. Fix $\epsilon > 0$. Since the sequence \[u_m = \displaystyle \sum_{n=1}^m |z_n|\] converges by assumption, it is Cauchy and there is $N \in \N$ such that \[\epsilon > |u_r - u_s| = |z_{s+1}| + \cdots + |z_r|\] for all $r > s \ge N$. From the triangle inequality \[|w_r - w_s| = |z_{s+1} + \cdots + z_r| \le |z_{s+1}| + \cdots + |z_r|\] so the sequence $w_m$ is also Cauchy. $\square$

Convergence is much more delicate than absolute convergence. It is easy to give an example of a series that is convergent but not absolutely convergent: the alternating series \[\sum_{n=1}^\infty \frac{(-1)^n}{n}\] is such an example.

In this course we will usually be working with absolutely convergent series. This will turn out to be because - very much unlike the situation in real analysis - holomorphic functions will have Taylor expansions on open balls that converge absolutely.

We conclude this section with some techniques for proving that a series converges absolutely. We will apply these tools to power series in the next section.

Theorem (Comparison Test)

If $|z_n| \le |u_n|$ for all $n$ large enough, and $\displaystyle\sum_{n=1}^\infty u_n$ converges then so does $\displaystyle\sum_{n=1}^\infty z_n$.

As is the case with real series, we have ratio and root tests that can sometimes tell us whether a series converges or diverges. Since they boil down to comparisons with geometric series, they can only tell us if a series converges absolutely: they are therefor unsuitable for series with complicated oscillations.

Theorem (Ratio Test)

Fix a sequence $z_n$ of non-zero complex numbers.

  1. If $\displaystyle \lim_{n \to \infty} \left| \frac{z_{n+1}}{z_n} \right| < 1$ the series $\displaystyle \sum_{n=1}^\infty z_n$ converges absolutely.
  2. If $\displaystyle \lim_{n \to \infty} \left| \frac{z_{n+1}}{z_n} \right| > 1$ the series $\displaystyle \sum_{n=1}^\infty z_n$ diverges.
Proof:

  1. There is $\delta > 0$ and $N \in \N$ such that, for all $n \ge N$, one has \[\left| \dfrac{z_{n+1}}{z_n} \right| < 1-\delta\] giving $|z_{N+k}| \le |z_N| (1-\delta)^k$ for all $k \in \N$ by induction. Since $1-\delta < 1$ the series converges by comparison with the geometric series.
  2. There is $\delta > 0$ and $N \in \N$ such that, for all $n \ge N$, one has \[\left| \dfrac{z_{n+1}}{z_n} \right| > 1+\delta\] giving $|z_{N+k}| \ge |z_N| (1+\delta)^k$ for all $k \in \N$ by induction. It follows that $z_{N+k}$ does not converge to 0 as $k \to \infty$. Thus the series diverges.
Example

Verify that \[\displaystyle \sum_{n=1}^\infty \frac{1}{n!}\] converges.

Solution:

We have \[\frac{z_{n+1}}{z_n} = \frac{\dfrac{1}{(n+1)!}}{\dfrac{1}{n!}} = \frac{1}{n+1}\] so \[\displaystyle \lim_{n \to \infty} \left| \dfrac{z_{n+1}}{z_n} \right| = 0\] and the series converges absolutely by the ratio test.

Theorem (Root test)

Fix a sequence $z_n$ of complex numbers.

  1. If $\displaystyle \lim_{n \to \infty} |z_n|^{1/n} < 1$ the series $\displaystyle \sum_{n=1}^\infty z_n$ converges absolutely.
  2. If $\displaystyle \lim_{n \to \infty} |z_n|^{1/n} > 1$ the series $\displaystyle \sum_{n=1}^\infty z_n$ diverges.
Proof:

The proof is similar to that of the ratio test, except that one instead has $|z_n| < (1-\delta)^n$ for all $n$ large enough in the first case, and $|z_n| > (1+\delta)^n$ for all $n$ large enough in the second. $\square$

Both tests are inconclusive when their respective limits equal 1.

The root test does not require all the terms in the series to be non-zero. Often the ratio test is easier to apply, but the root test is more powerful: there are series for which the ratio test is inconcluse and for which the root test is not.