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 $\mathbb{C}$ then we say $z_n$ converges to $a\in \mathbb{C}$ if for all $ϵ>0$ there exists $N\in \mathbb{N}$ such that $|z_n-a|<ϵ$ for all $n\ge N$.

Definition (Series)

Fix a sequence $z_1,z_2,z_3,\dots$ in $\mathbb{C}$. We can form a new sequence $$\begin{array}{rl}{w}_1& =z_1\\ w_2& =z_1+z_2\\ w_3& =z_1+z_2+z_3\\ & ⋮\\ w_m& =z_1+z_2+z_3+\cdots +z_n\\ & ⋮\end{array}$$ 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 $$\sum _{n=1}^{\mathrm{\infty }}{z}_n$$ without explicitly writing down the summands $w_m$.

Definition (Convergence of a Series)

Let $z_n$ be a sequence in $\mathbb{C}$. We say that the series $$\sum _{n=1}^{\mathrm{\infty }}{z}_n$$ 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 $$\sum _{n=1}^{\mathrm{\infty }}n$$ diverges, because the sequence $$w_m=1+2+\cdots +m={\displaystyle \frac{m(m+1)}{2}}$$ 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 $a\in \mathbb{C}$ with $|a|<1$. The series $$\sum _{n=1}^{\mathrm{\infty }}{a}^n$$ is called the geometric series. From $$\begin{array}{rl}{w}_m& =a+a^2+\cdots +a^m\\ a{w}_m& =a^2+\cdots +a^m+a^{m+1}\end{array}$$ we get $w_m(1-a)=a-a^{m+1}$. Since $|a|<1$ we have $|a^m|\to 0$ as $m\to \mathrm{\infty }$ and $$\underset{m\to \mathrm{\infty }}{lim}{w}_m=\underset{m\to \mathrm{\infty }}{lim}\frac{1-a^{m+1}}{1-a}=\frac{a}{1-a}$$ so $${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}^n={\displaystyle \frac{a}{1-a}}}$$ whenever $|a|<1$.

Theorem (Properties of Converging Series)

Suppose that $$\sum _{n=1}^{\mathrm{\infty }}{z}_n$$ and $$\sum _{n=1}^{\mathrm{\infty }}{u}_n$$ converge to $a$ and $b$ respectively. Then the following are all true.

1. $|z_n|\to 0$ as $n\to \mathrm{\infty }$
2. ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{z}_n+u_n}$ converges to $a+b$
3. ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}c{z}_n}$ converges to $ca$

Definition

The series $$\sum _{n=1}^{\mathrm{\infty }}{z}_n$$ is said to converges absolutely if the series $${\displaystyle \sum _{n=1}^{\mathrm{\infty }}|z_n|}$$ 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 $ϵ>0$. Since the sequence $$u_m={\displaystyle \sum _{n=1}^m|z_n|}$$ converges by assumption, it is Cauchy and there is $N\in \mathbb{N}$ such that $$ϵ>|u_r-u_s|=|z_{s+1}|+\cdots +|z_r|$$ 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. $◻$

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 $$\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{n}$$ 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 ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{u}_n}$ converges then so does ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{z}_n}$.

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 ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|\frac{z_{n+1}}{z_n}\right|<1}$ the series ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{z}_n}$ converges absolutely.
2. If ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|\frac{z_{n+1}}{z_n}\right|>1}$ the series ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{z}_n}$ diverges.

Proof:

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

Example

Verify that $${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n!}}$$ converges.

Solution:

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

Theorem (Root test)

Fix a sequence $z_n$ of complex numbers.

1. If ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}|z_n{|}^{1/n}<1}$ the series ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{z}_n}$ converges absolutely.
2. If ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}|z_n{|}^{1/n}>1}$ the series ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{z}_n}$ 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. $◻$

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.