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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 $n \ge N$.
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 $w_m$.
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.
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.
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 $|a| < 1$.
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.
The series \[\sum_{n=1}^\infty z_n\] is said to converges absolutely if the series \[\displaystyle \sum_{n=1}^\infty |z_n|\] converges.
If a series converges absolutely then it converges.
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.
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.
Fix a sequence $z_n$ of non-zero complex numbers.
Verify that \[\displaystyle \sum_{n=1}^\infty \frac{1}{n!}\] converges.
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.
Fix a sequence $z_n$ of complex numbers.
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.