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Complex Series: A Comprehensive Guide

Sequences and limits

Definition. Complex sequence A sequence of complex numbers is a function $a: \mathbb{N} \to \mathbb{C}$. We usually denote it by $\{ a_n \}_{n \in \mathbb{N}}$, $(a_n)_{n \in \mathbb{N}}$ or simply $(a_n)$, where $a_n := a(n)$. The value $a_1$ is called the first term of the sequence, $a_2$ the second term, and in general $a_n$ the n-th term of the sequence.

A complex sequence $(a_n)$ can always be decomposed into its real and imaginary parts by writing $a_n = x_n + iy_n$ where $x_n = \operatorname{Re}(a_n)$ and $y_n = \operatorname{Im}(a_n)$.

Examples: The constant sequence $a_n = 3 + 2i$ for all $n$. The sequence $a_n = \frac{1}{n} + \frac{i}{n^2}$, whose real part $\frac{1}{n}$ and imaginary part $\frac{1}{n^2}$ both tend to zero. The sequence $a_n = i^n$, which cycles through $i, -1, -i, 1, i, -1, \dots$ and does not settle down. The sequence $a_n = \left(\frac{1+i}{2}\right)^n$, which spirals toward the origin since $\left|\frac{1+i}{2}\right| = \frac{\sqrt{2}}{2} < 1$.

Definition. Convergent complex sequence. A complex sequence $\{ a_n \}_{n \in \mathbb{N}}$ is said to converge to a complex number $L \in \mathbb{C}$ if for every $\varepsilon > 0$ there exists an integer $N \in \mathbb{N}$ such that for all $n \geq N$ one has $|a_n - L| < \varepsilon$. In this case we write $\lim_{n \to \infty} a_n = L$ or $a_n \to L$ as $n \to \infty$, and L is called the limit of the sequence $\{ a_n \}_{n \in \mathbb{N}}$.

The condition $|a_n - L| < \varepsilon$ means that $a_n$ lies inside the open disk $B(L; \varepsilon)$ of radius $\varepsilon$ centered at $L$. Convergence says: no matter how small a disk you want to draw around $L$, all but finitely many terms of the sequence lie inside that disk. The terms eventually cluster arbitrarily close to $L$.

Uniqueness of limits. If $a_n \to L$ and $a_n \to L'$, then $L = L'$.

Proof:

Indeed, for any $\varepsilon > 0$, choose $N$ large enough so that $|a_n - L| < \varepsilon/2$ and $|a_n - L'| < \varepsilon/2$ for $n \geq N$. Then $|L - L'| \leq |L - a_N| + |a_N - L'| < \varepsilon/2 + \varepsilon/2 = \varepsilon$. Since $\varepsilon > 0$ was arbitrary, $|L - L'| = 0$, so $L = L'$.

Reduction to real and imaginary parts. A complex sequence converges if and only if its real and imaginary parts both converge: $a_n = x_n + iy_n \to L = X + iY \quad \Longleftrightarrow \quad x_n \to X \text{ and } y_n \to Y.$

Proof sketch. The inequalities $|\operatorname{Re}(z)| \leq |z|$ and $|\operatorname{Im}(z)| \leq |z|$ give the forward direction; the reverse follows from $|z| \leq |\operatorname{Re}(z)| + |\operatorname{Im}(z)|$.

Definition. Cauchy sequence. A complex sequence $\{ a_n \}_{n \in \mathbb{N}}$ is called a Cauchy sequence if for every $\varepsilon > 0$ there exists an integer $N \in \mathbb{N}$ such that for all $n, m \geq N$ one has $|a_n - a_m| < \varepsilon$.

The key distinction from the definition of convergence is that a Cauchy sequence is defined without reference to any limit. The condition says only that terms eventually cluster together and become arbitrarily close to each other. This is an intrinsic property of the sequence, not one that requires knowing the answer in advance.

Proposition. Completeness of $\mathbb{C}$ A complex sequence $\{ a_n \}_{n \in \mathbb{N}}$ is convergent if and only if it is a Cauchy sequence.

Proof sketch.

($\Rightarrow$) If $a_n \to L$, then for large $n, m$: $|a_n - a_m| \leq |a_n - L| + |L - a_m| < \varepsilon$.

By completeness of $\mathbb{R}$, they converge to limits $X$ and $Y$ respectively, so $a_n \to X + iY$.

Completeness means there are “no gaps” in $\mathbb{C}$: every sequence that looks like it should converge (its terms huddle together) actually does converge to some point in $\mathbb{C}$. This property is inherited from the completeness of $\mathbb{R}$ and is essential throughout analysis.

Complex Series

Definition. Series and partial sums.Let $\{ a_n \}_{n \in \mathbb{N}}$ be a complex sequence. For each n $\in \mathbb{N}$, the finite sum $s_n := a_1 + a_2 + \cdots + a_n = \sum_{k=1}^n a_k$ is called the n-th partial sum of the (infinite) series $\sum_{k=1}^\infin a_k$ which we also denote simply by $\sum a_n$ when the index is clear from the context.

The infinite series $\sum_{k=1}^{\infty} a_k$ is not defined as the result of “adding up infinitely many numbers”. Rather, it is defined through the sequence of partial sums: each $s_n$ is a perfectly well-defined finite sum, and we study the behavior of the sequence $\{ a_n \}_{n \in \mathbb{N}}$.

Definition. Convergent series. The series $\sum_{n=1}^{\infty} a_n$ is said to converge to the sum $s \in \mathbb{C}$ if the sequence of partial sums $\{ s_n \}_{n \in \mathbb{N}}$ defined by $s_n = a_1 + a_2 + \cdots + a_n = \sum_{k=1}^n a_k$ converges to s, that is, $\lim_{n \to \infty} s_n = s$. In this case we write $s := \sum_{n=1}^\infin a_n$. If the sequence $\{ s_n \}_{n \in \mathbb{N}}$ does not converge, we say that the series $\sum_{n=1}^{\infty} a_n$ diverges (or does not converge).

Thus, the notation “$\sum_{n=1}^{\infty} a_n = s$” is not asserting that we have literally added infinitely many numbers and obtained s. It is a statement about a limit — it asserts that the partial sums $s_n$ get arbitrarily close to s. The “equals sign” here enodes a limiting process.

Proposition. Reduction to real and imaginary part. Let $\{ a_n \}_{n \in \mathbb{N}}$ be a complex sequence and write $a_n = x_n + i y_n$ with $x_n = \mathbb{Re}(a_n), y_n = \mathbb{Im}(a_n)$. Then, the complex series $\sum_{n=1}^{\infty} a_n$ converges if and only if both real series $\sum Re(a_n)$ and $\sum Im(a_n)$ converges. In that case, $\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \mathbb{Re}(a_n) + i \sum_{n=1}^{\infty} \mathbb{Im}(a_n)$.

Proof.

Let $S_N$ be the N-th partial sum of the complex series, $S_N = \sum_{n=1}^{N} a_n = \sum_{n=1}^{N} (x_n + i y_n) = \left(\sum_{n=1}^{N} x_n\right) + i \left(\sum_{n=1}^{N} y_n\right)$

Let’s call $X_N = \sum_{n=1}^{N} x_n$ (the partial sum of the real part) and $Y_N = \sum_{n=1}^{N} y_n$ (the partial sum of the imaginary part). So, we have $S_N = X_N + i Y_N$.

($\Leftarrow$) If $\sum x_n$ and $\sum y_n$ converge, then $\lim_{N \to \infty} X_N = X$ and $\lim_{N \to \infty} Y_N = Y$ for some real numbers X and Y. By the rules of limits, the limit of the complex partial sum is:

$\lim_{N \to \infty} S_N = \lim_{N \to \infty} (X_N + i Y_N) = (\lim_{N \to \infty} X_N) + i(\lim_{N \to \infty} Y_N) = X + iY$

Since the limit $\lim_{N \to \infty} S_N$ exists, the complex series $\sum a_n$ converges and its sum is $X + iY$.

($\Rightarrow$) If $\sum a_n$ converges: $\lim_{N \to \infty} S_N = S$, where S is some complex number S = X + iY.

A fundamental property of complex limits states that a sequence of complex numbers $\sum z_n$ (where each $z_n = x_n + y_n$) converges to z = x + iy if and only if its real and imaginary parts both converge separately to x and y respectively.

Therefore, $\lim_{N \to \infty} S_N = S$ implies that $\lim_{N \to \infty} \text{Re}(S_N) = \text{Re}(S) = X$ and $\lim_{N \to \infty} \text{Im}(S_N) = \text{Im}(S) = Y$.

Since $\text{Re}(S_N) = X_N$ and $\text{Im}(S_N) = Y_N$, this means: $\lim_{N \to \infty} X_N =X \text{ and } \lim_{N \to \infty} Y_N = Y$. In other words, both the real series $\sum x_n$ and the imaginary series $\sum y_n$ must converge.

This proposition is conceptually important: it tells us that convergence of complex series is “nothing new” beyond pairs of real series.

Basic Properties of Convergent Complex Series

Let $\sum_{n=1}^{\infty} a_n$ be a complex series.

(1) Necessary condition for convergence of a series. If $\sum_{n=1}^{\infty} a_n$ converges, then the sequence of terms $\{ a_n \}_{n \in \mathbb{N}}$ satisfies $lim_{n \to \infin} a_n = 0.$

Proof.

If $s_n = \sum_{k=1}^{n} a_k \to s$, then $a_n = s_n - s_{n-1} \to s - s = 0.$ Hence $lim_{n \to \infin} a_n = 0$ is necessary for convergence of $\sum_{n=1}^{\infty} a_n$, but not sufficient.

The converse is false. The condition $a_n \to 0$ is necessary but not sufficient for convergence. The harmonic series $\sum \frac{1}{n}$ has terms tending to zero, yet diverges (see below). The divergence test is therefore primarily useful as a quick way to establish divergence: if $a_n \not\to 0$, then $\sum a_n$ definitely diverges.

(2) Boundedness of the terms. If $\sum_{n=1}^{\infty} a_n$ converges, then the sequence of terms $\{ a_n \}_{n \in \mathbb{N}}$ is bounded, meaning there exists a real non-negative number M such that $|a_n| \le M, \forall n \in \mathbb{N}$.

Proof.

If $\sum_{n=1}^{\infty} a_n$ converges, its partial sums $s_n = \sum_{k=1}^{n} a_k$ form a convergent sequence. Every convergent sequence is bounded, so there exists $K\geq 0$ such that $|s_n|\leq K$ for all n.

$∣a_n∣=∣s_n−s_{n−1}∣ \le[\text{Triangle inequality}] ∣s_n∣+∣s_{n−1}∣ \le K + K = 2K$, therefore the sequence $\{ a_n \}_{n \in \mathbb{N}}$ is bounded by M = 2K. $\blacksquare$

Necessity, not sufficiency. The constant sequence $(a_n)_{n \in \mathbb{N}} = (1)_{n \in \mathbb{N}}$ is bounded but $\sum_{n=1}^{\infty} 1$ diverges.

(3) Linearity. If $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ are convergent complex series and $c\in \mathbb{C}$, then the series $\sum_{n=1}^{\infty} (a_n+cb_n)$ also converges, and $\sum_{n=1}^{\infty} (a_n+cb_n) = \sum_{n=1}^{\infty} a_n +c\sum_{n=1}^{\infty} b_n$.

Proof.

This follows immediately from the corresponding properties of limits of sequences: $\sum_{k=1}^n (a_k + cb_k) = \sum_{k=1}^n a_k + c\sum_{k=1}^n b_k \to \sum_{n=1}^{\infty} a_n + c\sum_{n=1}^{\infty} b_n. \quad \blacksquare$

Remark. Linearity says that the set of convergent series forms a vector space (over $\mathbb{C}$), and the map “take the sum” $\sum a_n \mapsto s$ is a linear functional on this space.

Definition. A complex series $\sum_{n=1}^{\infty} a_n$ is said to be absolutely convergent if the real series of moduli $\sum_{n=1}^{\infty} |a_n|$ converges.

(4) Absolute convergence. If the series $\sum_{n=1}^{\infty} |a_n|$ converges, then the complex series $\sum_{n=1}^{\infty} a_n$ converges (in $\mathbb{C}$). In this case we say that $\sum_{n=1}^{\infty} a_n$ converges absolutely.

This is powerful because absolute convergence guarantees ordinary convergence, but not vice versa. If $\sum_{n=1}^{\infty} |a_n|$ converges, then the terms $a_n$ must shrink (get small) quickly enough. The oscillations (positive/negative or complex direction changes) in $a_n$ cannot prevent convergence, because the absolute values already ensure the series is bounded and well-behaved.

Proof.

For m > n,

$$ \begin{aligned} \left|\sum_{k=n+1}^{m} a_k\right| \leq[\text{Triangle inequality for complex numbers}] &\sum_{k=n+1}^{m} |a_k| \\[2pt] &[\text{Since } \sum_{n=1}^{\infty} |a_n| \text{ converges, its partial sums form a Cauchy sequence} ] \\[2pt] &[\forall \varepsilon \gt 0, \exists N: \forall m \gt n \ge N]\\[2pt] &\lt \varepsilon. \end{aligned} $$

The right-hand side can be made arbitrarily small (since $\sum |a_k|$ is Cauchy), so the partial sums of $\sum a_k$ also form a Cauchy sequence in $\mathbb{C}$. By completeness of $\mathbb{C}$, the series $\sum a_n$ converges. $\blacksquare$

Example: $\sum_{n=1}^{\infty} \frac{i^n}{n^2}$

$\sum_{n=1}^{\infty} |\frac{i^n}{n^2}| = \sum_{n=1}^{\infty} \frac{|i^n|}{n^2} = \sum_{n=1}^{\infty} \frac{1}{n^2}$ which converges (it is a p-series with p = 2 > 1). Since the complex series converges absolutely, the original series $\sum_{n=1}^{\infty} \frac{i^n}{n^2}$ is guaranteed to converge.

Recall. A p-series is any infinite series of the form: $\sum_{k=1}^\infin \frac{1}{k^p}$ where p is a positive real number. The p-Series Test. If p > 1, the series converges (approaches a finite sum). If p ≤ 1, the series diverges (grows without bound).

Counterexample. Absolute convergence guarantees ordinary convergence, but not vice versa.

The canonical example is the alternating harmonic series: $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \ln 2$. This series converges to $\ln 2$ (Alternating Series Test), but not absolutely, because $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges.

Alternating Series Test (Leibniz Criterion). A series of the form $\sum (-1)^{n+1} b_n$ or $\sum (-1)^n b_n$ converges (conditionally, meaning a series converges, but not absolutely) if the following hold:

Positivity: $b_n > 0$ for all n.
Monotonicity: $b_n$ is eventually non-increasing, i.e. there exists N such that for all $n\geq N, b_{n+1} \le b_n$ for all n (this refinement allows for small fluctuations at the beginning, which don’t affect convergence).
Vanishing terms: $\lim_{n \to \infty} b_n = 0$.

For the alternating harmonic series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$, $b_n = \frac{1}{n}$ satisfies all three conditions. Clearly: $b_n > 0$ for all $n \ge 1$; since $\frac{1}{n+1} < \frac{1}{n}$, the sequence is decreasing; $\lim_{n \to \infty} \frac{1}{n} = 0$.

Divergence of the Harmonic Series

$\sum_{n=1}^{\infty} \frac{1}{n}$ diverges.

The Grouping Method (Oresme’s Proof). Let’s write out the series and group terms in blocks whose sizes are powers of 2: $1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \left(\frac{1}{9} + \dots + \frac{1}{16}\right) + \dots$. In the group $(\frac{1}{3} + \frac{1}{4})$, each term is at least $\frac{1}{4}$, so the sum is $(\frac{1}{4} + \frac{1}{4}) = \frac{2}{4} = \frac{1}{2}$. In the group $(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8})$ each term is at least $\frac{1}{8}$ and there are 4 terms, so the sum is $(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}) = \frac{4}{8} = \frac{1}{2}$. $(\frac{1}{9} + \dots + \frac{1}{16}) > (\frac{1}{16} + \dots + \frac{1}{16}) =[ \text{there are 8 terms }] \frac{8}{16} = \frac{1}{2}$ …and so on. So each block contributes at least $\frac{1}{2}$. Since there are infinitely many blocks, the partial sums grow beyond any bound (it diverges to infinity).
Integral test states that if a function f(x), $f: [1, \infty) \to \mathbb{R}$, is positive, continuous, and decreasing, then the series $\sum_{n=1}^{\infty} f(n)$ and the improper integral $\int_1^\infty f(x) dx$ either both converge or both diverge.

$f(x) = \frac{1}{x}$ is positive, continuous, and decreasing for all $x \ge 1$. We compare our series $\sum_{n=1}^{\infty} \frac{1}{n}$ to the integral $\int_1^\infty \frac{1}{x} dx$

$\int_1^\infty \frac{1}{x} dx = \lim_{b \to \infty} \int_1^b \frac{1}{x} dx = \lim_{b \to \infty} [\ln(x)]_1^b = \lim_{b \to \infty} (\ln(b) - \ln(1)) = \lim_{b \to \infty} (\ln(b) - 0) = \infty$. Since the improper integral $\int_1^\infty \frac{1}{x} dx$ diverges (it goes to infinity), the Integral Test guarantees that the harmonic series $\sum_{n=1}^{\infty} \frac{1}{n}$ must also diverge.