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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 $(a_n)_{n \in \mathbb{N}}$ 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.

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 $(a_n)_{n \in \mathbb{N}}$.

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 $|a_n - a_m| < \varepsilon$.

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.

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.

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).

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 $S_N = X_N + i Y_N$.

Case 1. 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 $X + iY$.

Cae 2. 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 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 words, both the real series $\sum x_n$ and the imaginary series $\sum y_n$ must converge.

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 $lim_{n \to \infin} a_n = 0.$

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.

(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 $|a_n| \le M, \forall n \in \mathbb{N}$.

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.

Necessity, not sufficiency. $(a_n)_{n \in \mathbb{N}} = (1)_{n \in \mathbb{N}}$ 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 $\sum_{n=1}^{\infty} (a_n+cb_n) = \sum_{n=1}^{\infty} a_n +c\sum_{n=1}^{\infty} b_n$.

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

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). 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. Consider the alternating harmonic series: $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$. 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: (1) Positivity: $b_n > 0$ for all n; (2) 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); (3) Vanishing terms: $\lim_{n \to \infty} b_n = 0$. In our case $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$, $b_n = \frac{1}{n}$. Clearly: $b_n > 0$ for all $n \ge 1$; since $\frac{1}{n+1} < \frac{1}{n}$, the sequence is decreasing; $\lim_{n \to \infty} \frac{1}{n} = 0$.

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

The Grouping Method (Oresme’s Proof). Let’s write out the series and group the terms (in 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) 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.

To study the convergence of a complex series $\sum_{n=1}^{\infty} a_n$, it is often convenient to apply standard convergence tests (e.g., the comparison test, the ratio test, and Cauchy’s n-th root test) from real analysis to the associated real series $\sum_{n=1}^{\infty} |a_n|$. If $\sum_{n=1}^{\infty} |a_n|$ is found to converge by one of these tests, then $\sum_{n=1}^{\infty} a_n$ is absolutely convergent and hence convergent by the previous proposition.

Comparison test (for absolute convergence)

Let $\sum_{n=1}^{\infty} a_n$ be a complex series and consider the real series $\sum_{n=1}^{\infty} |a_n|$ with non-negative terms. Let $(b_n)_{n \in \mathbb{N}}$ be a sequence of non-negative real numbers ($b_n \ge 0$ for all n).

Suppose there exists $N \in \mathbb{N}$ such that for all $n \ge N$ we have $0 \le |a_n| \le b_n$ and the real series $\sum_{n=1}^{\infty} b_n$ converges. Then, the series $\sum_{n=1}^{\infty} |a_n|$ also converges, so $\sum_{n=1}^{\infty} a_n$ is absolutely convergent and hence convergent.
Suppose there exists $N \in \mathbb{N}$ such that for all $n \ge N$ we have $0 \le b_n \le |a_n|$, and the real series $\sum_{n=1}^{\infty} b_n$ diverges. Then, the series $\sum_{n=1}^{\infty} |a_n|$ also diverges, so $\sum_{n=1}^{\infty} a_n$ cannot be absolutely convergent.

Consider the complex series $\sum_{n=1}^{\infty} \frac{3 + 4i}{n^2+n}$. In this particular case $|a_n| = \dfrac{|3 + 4i|}{n^2+n} = \dfrac{5}{n^2+n}$. Let’s choose $b_n = \frac{5}{n^2}$.

For all $n \ge 1$, we have $0 \le |a_n| = \dfrac{5}{n^2+n} \le \frac{5}{n^2} = b_n$.
The series $\sum \dfrac{5}{n^2}$ converges (it is a constant multiple of the convergent p-series with p = 2 > 1).
So by the comparison test, $\sum_{n=1}^{\infty} |a_n|$ converges, and hence $\sum_{n=1}^{\infty} a_n$ converges absolutely.

Ratio test (d’Alembert’s test)

Let $\sum_{n=1}^{\infty} a_n$ be a complex series, and suppose that for all sufficiently large n we have $a_n \neq 0$. Consider the limit (if it exists) $L := \lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|}.$ Then, (1) If L < 1, then the series $\sum_{n=1}^{\infty} |a_n|$ converges, so $\sum_{n=1}^{\infty} a_n$. is absolutely convergent and hence convergent. (2) If L > 1 (including $L = +\infty$), then the series $\sum_{n=1}^{\infty} |a_n|$ diverges, so $\sum_{n=1}^{\infty} a_n$ is not absolutely convergent. If L = 1 or the limit does not exist, the test is inconclusive; the series may converge or diverge, and another test is needed.

Roughly, the ratio $|a_{n+1}|/|a_n|$ tells you how fast the terms are shrinking. If the ratio stabilizes below 1, the terms decay like a geometric sequence with ratio c < 1, and geometric series with such ratios always converge. If the ratio exceeds 1, the terms don’t shrink fast enough — they may even grow — and the series diverges. If the ratio is near 1, the behaviour is too delicate for this test alone.

Consider the complex series $\sum_{n=1}^{\infty} \frac{z^n}{n!}, z \in \mathbb{C}.$ In this particular example, $a_n = \dfrac{z^n}{n!}$. Then, $\frac{|a_{n+1}|}{|a_n|} = \frac{|z|^{n+1}}{(n+1)!} \cdot \frac{n!}{|z|^n} = \frac{|z|}{n+1} \xrightarrow[n\to\infty]{} 0.$ Thus, L = 0 < 1, so $\sum_{n=1}^{\infty} |a_n|$ converges and $\sum_{n=1}^{\infty} \frac{z^n}{n!}$ is absolutely convergent for every $z \in \mathbb{C}$.

Cauchy’s n-th root test

Root test for absolute convergence. Let $\sum_{n=1}^{\infty} a_n$ be a complex series. Consider the non-negative numbers $|a_n|$ and define $L := \limsup_{n \to \infty} \sqrt[n]{|a_n|}.$ Then, (1) If L < 1, then the series $\sum_{n=1}^{\infty} |a_n|$ converges, so $\sum_{n=1}^{\infty} a_n$ is absolutely convergent. If L > 1 (including $L = +\infty$), then the series $\sum_{n=1}^{\infty} |a_n|$ diverges, so $\sum_{n=1}^{\infty} a_n$ does not converge absolutely. If L = 1, the test is inconclusive, the series may converge or diverge.

If the limit $\lim_{n\to\infty} \sqrt[n]{|a_n|}$ exists (for most common series, the regular limit exists), you can use it instead of the limsup.

If L < 1, pick any $\rho$ with $L<\rho <1$. By the definition of limsup, for all sufficiently large n, $\sqrt[n]{|a_n|}\leq \rho \Rightarrow |a_n|\leq \rho ^n.$ The geometric series $\sum \rho ^n$ converges, so by comparison $\sum |a_n|$ converges.

Terms do not vanish when L > 1: If L > 1, then for infinitely many n, $\sqrt[n]{|a_n|}>1+\varepsilon$ for some $\varepsilon >0$, hence $|a_n|>(1+\varepsilon )^n$, and in particular $|a_n|$ does not tend to 0. A series whose terms do not tend to 0 cannot converge, so $\sum |a_n|$ diverges.

Examples:

Geometric-like decay: $a_n=2^{-n}$.

$$ \begin{aligned} L: &=\lim_{n \to \infty} \sqrt[n]{|2^{-n}|} \\[2pt] &\text{Since } 2^{-n} \text{is already positive, the absolute value does nothing} \\[2pt] &=\lim_{n \to \infty} {2^{-n}}^{\frac{1}{n}} \\[2pt] &\text{We use the exponent rule }(x^a)^b = x^{ab} \\[2pt] &=\lim_{n \to \infty} {2^{(-n)\cdot \frac{1}{n}}} =\lim_{n \to \infty} 2^{-1} = \frac{1}{2}. \end{aligned} $$

Since 2 < 1, the n-th Root Test confirms that the series converges absolutely

Polynomial decay: $a_n=1/n^p$. This is the p-series.

$L = \lim_{n \to \infty} \sqrt[n]{\left|\frac{1}{n^p}\right|} = \lim_{n \to \infty} \frac{\sqrt[n]{1}}{(\sqrt[n]{n})^p} = \lim_{n \to \infty} \frac{1}{(\sqrt[n]{n})^p}$

Consider $\sqrt[n]{n} = n^{1/n}$. Define $L = \lim_{n\to\infty} n^{1/n}$, and take logarithms $\ln(L) = \lim_{n\to\infty} ln(n^{1/n}) = \lim_{n\to\infty} \frac{\ln(n)}{n} =[\text{We know ln(n) grows slower than n}] = 0$. So $\ln(L) = 0 \Rightarrow L = e^0 = 1$, therefore $lim_{n\to\infty} \sqrt[n]{n} = 1$.

$L = \lim_{n \to \infty} \sqrt[n]{\left|\frac{1}{n^p}\right|} = \lim_{n \to \infty} \frac{1}{(\sqrt[n]{n})^p} = \frac{1}{1^p} = 1$. The n-th Root Test is inconclusive. We already know from the $p$-series test that $\sum 1/n$ (where p = 1) diverges, but $\sum 1/n^2$ (where p = 2) converges. The root test gives L = 1 for both and cannot tell them apart.

Exponential growth: $a_n=n!$. Then, $\sqrt[n]{|a_n|}\sim n/e\rightarrow \infty$. By the root test, the series does not converge absolutely.

$L = \lim_{n \to \infty} \sqrt[n]{|n!|}$.

We are going to use Stirling’s Approximation, which states that for large n, $n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n$

$\sqrt[n]{n!} \approx \sqrt[n]{\sqrt{2\pi n} \left(\frac{n}{e}\right)^n} = \left( (2\pi n)^{1/2} \right)^{1/n} \cdot \left(\frac{n}{e}\right) \approx (2\pi n)^{1/(2n)} \cdot \left(\frac{n}{e}\right)$

As $n \to \infty$, the first term $(2\pi n)^{1/(2n)}$ approaches 1 (for the same reason $\sqrt[n]{n} \to 1$). The second term, $\frac{n}{e}$, goes to infinity.

$\sqrt[n]{n!} \approx (2\pi n)^{1/(2n)} \cdot \left(\frac{n}{e}\right) \to 1 \cdot (\frac{n}{e}) = \infin$ Since L > 1, the n-th Root Test confirms that the series diverges. This is obvious since the terms $a_n=n!$ do not approach zero as $n \to \infin$

Consider $\sum_{n=1}^{\infty} \left(\frac{z^n}{3^n}\right), z \in \mathbb{C}.$ So $L = \lim_{n \to \infty} \sqrt[n]{\left| \left(\frac{z}{3}\right)^n \right|} = \lim_{n \to \infty} \sqrt[n]{\left( \left|\frac{z}{3}\right| \right)^n} = \lim_{n \to \infty} \left( \left|\frac{z}{3}\right|^n \right)^{1/n} = \lim_{n \to \infty} \left|\frac{z}{3}\right| = \left|\frac{z}{3}\right|.$

If $L = \left|\frac{z}{3}\right| < 1 \implies |z| < 3$ and the series converges absolutely when z is any point inside the open disk of radius 3 centered at the origin. If |z| > 3, then L > 1 and the series does not converge absolutely. If |z| = 3, the test is inconclusive (this is the boundary case of the corresponding geometric series). In this particular case, the terms are $(\frac{z}{3})^n$ with modulus 1, the terms themselves are not approaching 0, $\lim_{n \to \infty} a_n \neq 0$, hence the series must diverge.