Complex Series

Complex Series and Their Partial Sums

A complex series is not defined by adding infinitely many terms all at once. Instead, we examine the sequence of its finite partial sums. The previous tutorial established how convergence works for complex sequences and proved that \(\mathbb{C}\) is complete. Those ideas let us define convergence of a series and translate it into a useful condition on its tails.

Definition (Complex Series and Partial Sums): Given a sequence \((z_n)_{n\geq1}\) in \(\mathbb{C}\), the series \(\sum_{n=1}^{\infty}z_n\) has \(N\)th partial sum \(S_N=\sum_{n=1}^{N}z_n\). The series converges to \(S\in\mathbb{C}\) if \(S_N\to S\) as \(N\to\infty\). In that case we write \(\sum_{n=1}^{\infty}z_n=S\). If the partial sums do not converge, the series diverges.

This definition makes the order of operations explicit: first form each finite sum, then take the limit of those sums. Since a finite sum of complex numbers is well-defined, the question is whether the resulting sequence \((S_N)\) has a limit. In particular, the series converges precisely when its partial sums become arbitrarily close to one fixed complex number.

The Cauchy criterion expresses convergence without requiring us to know the proposed sum. A partial sum far out in the sequence differs from an earlier partial sum by a finite tail of the series.

Theorem (Cauchy Criterion for a Complex Series): The series \(\sum_{n=1}^{\infty}z_n\) converges if and only if, for every \(\varepsilon>0\), there is an integer \(N\) such that \(\left|\sum_{k=q}^{p}z_k\right|<\varepsilon\) whenever \(p\geq q\geq N\).

Proof. Suppose first that the series converges, so its partial sums satisfy \(S_N\to S\). Given \(\varepsilon>0\), convergence gives an \(N\) such that \(|S_j-S|<\varepsilon/2\) for every \(j\geq N-1\), increasing \(N\) if necessary so that \(N\geq2\). For \(p\geq q\geq N\),

\left|\sum_{k=q}^{p}z_k\right| =|S_p-S_{q-1}| \leq |S_p-S|+|S_{q-1}-S| <\varepsilon.

Conversely, suppose the stated tail condition holds. For any \(p\geq q\geq N\), the identity \(S_p-S_{q-1}=\sum_{k=q}^{p}z_k\) shows that the partial sums satisfy the Cauchy condition: for every tolerance, all sufficiently late partial sums differ by less than that tolerance. Thus \((S_N)\) is a Cauchy sequence in \(\mathbb{C}\). By the Completeness of \(\mathbb{C}\), proved earlier in this course, the sequence \((S_N)\) converges. Therefore the series converges. \(\square\)

The criterion is particularly effective when the sum is unknown but the tails can be estimated. It also explains why an estimate for individual terms alone is generally not enough: the criterion concerns sums of consecutive terms, not just one term at a time.

Absolute Convergence

For a complex number \(z_n\), its modulus \(|z_n|\) measures its size without regard to direction in the complex plane. If the real series of these nonnegative sizes converges, then cancellation among the complex terms is not needed to make the original series converge.

Definition (Absolute and Conditional Convergence): The complex series \(\sum_{n=1}^{\infty}z_n\) is absolutely convergent if the real series \(\sum_{n=1}^{\infty}|z_n|\) converges. It is conditionally convergent if it converges but is not absolutely convergent.

Theorem (Absolute Convergence Theorem): If \(\sum_{n=1}^{\infty}|z_n|\) converges, then \(\sum_{n=1}^{\infty}z_n\) converges.

Proof. Since the real series \(\sum |z_n|\) converges, its partial sums converge and therefore satisfy the Cauchy criterion. Fix \(\varepsilon>0\). There is an \(N\) such that \(\sum_{k=q}^{p}|z_k|<\varepsilon\) whenever \(p\geq q\geq N\). The triangle inequality for the complex modulus gives

\left|\sum_{k=q}^{p}z_k\right| \leq \sum_{k=q}^{p}|z_k| <\varepsilon \qquad (p\geq q\geq N).

The Cauchy Criterion for a Complex Series now implies that \(\sum z_n\) converges. The real series of moduli controls every complex tail, which proves the result. \(\square\)

Absolute convergence is a sufficient condition for convergence, but it is not a necessary one. When a series converges but the corresponding series of moduli diverges, its convergence depends on cancellation between terms. Such a series is conditionally convergent.

Worked Example: A Geometric Series with a Complex Ratio

Let \(r=(1+i)/3\), and consider \(\sum_{n=0}^{\infty}r^n\). Since \(|r|=\sqrt{2}/3<1\), the finite geometric-sum identity gives, for \(N\geq0\),

\sum_{n=0}^{N}r^n=\frac{1-r^{N+1}}{1-r}.

Indeed, multiplying the finite sum \(1+r+\cdots+r^N\) by \(1-r\) cancels all intermediate powers and leaves \(1-r^{N+1}\). Also, \(|r^{N+1}|=|r|^{N+1}\to0\). Therefore the partial sums converge to \(1/(1-r)\). In this case, \(1-r=(2-i)/3\), and

\frac{1}{1-r} =\frac{3}{2-i} =\frac{3(2+i)}{(2-i)(2+i)} =\frac{6+3i}{5}.

The denominator calculation is \((2-i)(2+i)=4-i^2=5\). Thus the series sums to \((6+3i)/5\). It is also absolutely convergent: its series of moduli is \(\sum_{n=0}^{\infty}(\sqrt{2}/3)^n\), a convergent real geometric series.

Worked Example: An Absolutely Convergent Telescoping Series

Consider \(\sum_{n=1}^{\infty}\dfrac{1+i}{n(n+1)}\). For each \(n\), \[ \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}, \] because the right-hand side is \(\bigl((n+1)-n\bigr)/(n(n+1))=1/(n(n+1))\). Consequently, the \(N\)th partial sum is

\sum_{n=1}^{N}\frac{1+i}{n(n+1)} =(1+i)\sum_{n=1}^{N}\left(\frac1n-\frac1{n+1}\right) =(1+i)\left(1-\frac{1}{N+1}\right).

As \(N\to\infty\), this tends to \(1+i\). To check absolute convergence, note that \[ \left|\frac{1+i}{n(n+1)}\right| =\frac{\sqrt{2}}{n(n+1)}. \] The partial sums of the series of moduli are therefore \(\sqrt{2}\left(1-\frac{1}{N+1}\right)\), which tend to \(\sqrt{2}\). The series is absolutely convergent, and its complex sum is \(1+i\).

A Necessary Condition and Conditional Convergence

Convergence of a series forces its individual terms to approach zero. This necessary condition is useful for detecting divergence, although it does not by itself guarantee convergence.

Theorem (Term Test for Convergence): If \(\sum_{n=1}^{\infty}z_n\) converges, then \(z_n\to0\).

Proof. Let \(S_N=\sum_{k=1}^{N}z_k\), and suppose \(S_N\to S\). For \(n\geq2\), \(z_n=S_n-S_{n-1}\). Both \(S_n\) and \(S_{n-1}\) tend to \(S\), so the triangle inequality gives

|z_n| =|S_n-S_{n-1}| \leq |S_n-S|+|S_{n-1}-S| \longrightarrow 0.

Hence \(z_n\to0\). \(\square\)

The converse fails: terms can tend to zero while their series diverges. The alternating harmonic series provides a more informative example: it converges, but not absolutely. Its convergence comes from cancellation between positive and negative terms.

Worked Example: The Alternating Harmonic Series

Consider the real-valued, and therefore also complex-valued, series \(\sum_{n=1}^{\infty}(-1)^{n+1}/n\). Write its partial sums as \(s_N\). For \(m\geq1\), the even partial sums satisfy

s_{2m} =\sum_{j=1}^{m}\left(\frac{1}{2j-1}-\frac{1}{2j}\right).

Each increment is positive, since \(1/(2j-1)>1/(2j)\), so \((s_{2m})\) is increasing. Expanding the terms gives the identity

s_{2m} =1-\sum_{j=1}^{m-1}\left(\frac{1}{2j}-\frac{1}{2j+1}\right)-\frac{1}{2m} \leq 1.

For \(m=1\), the sum from \(j=1\) to \(m-1\) is empty and the identity reads \(s_2=1-\tfrac12=\tfrac12\), as required. For general \(m\), expanding the right-hand side gives \(1-\tfrac12+\tfrac13-\tfrac14+\cdots+\tfrac{1}{2m-1}-\tfrac{1}{2m}\), exactly \(s_{2m}\). Each term subtracted in the identity is positive, which proves the upper bound. The increasing even partial sums are bounded above, so they converge to some real number \(L\).

The odd partial sums satisfy \(s_{2m+1}=s_{2m}+1/(2m+1)\). Since \(1/(2m+1)\to0\), it follows that \(s_{2m+1}\to L\) as well. The even and odd partial sums together are the full sequence of partial sums, so that sequence converges to \(L\). Thus the alternating harmonic series converges.

It does not converge absolutely. Its series of moduli is \(\sum_{n=1}^{\infty}1/n\). For each integer \(j\geq1\), the terms with \(2^{j-1}<n\leq2^j\) number \(2^{j-1}\), and each is at least \(1/2^j\). Their sum is therefore at least \(1/2\). The partial sums of the harmonic series grow without bound as these disjoint blocks are added. The alternating harmonic series is consequently conditionally convergent.

Worked Example: Terms Tending to Zero Do Not Ensure Convergence

For \(z_n=1/n\), the terms satisfy \(z_n\to0\). Nevertheless, \(\sum_{n=1}^{\infty}1/n\) diverges. To see this directly, for every \(j\geq1\), the block with \(2^{j-1}<n\leq2^j\) contains \(2^{j-1}\) terms, each at least \(1/2^j\). Hence

\sum_{n=2^{j-1}+1}^{2^j}\frac1n \geq 2^{j-1}\frac{1}{2^j} =\frac12.

The partial sums at powers of two therefore satisfy \(\sum_{n=1}^{2^J}1/n\geq1+J/2\) for every \(J\geq1\), and so they are unbounded. A convergent sequence of partial sums would be bounded, so the series diverges. This example shows why the Term Test can rule out convergence when terms fail to approach zero, but cannot establish convergence when they do.

Using the Criteria Carefully

The Cauchy Criterion, the Absolute Convergence Theorem, and the Term Test answer different questions. The Cauchy Criterion controls whole tails and is equivalent to convergence. Absolute convergence provides a strong, easy-to-use sufficient condition. The Term Test is only necessary: if the terms do not tend to zero, the series must diverge, but if they do, further analysis is needed.

For a complex series, estimates usually use the modulus and the triangle inequality. One should not treat convergence of the real and imaginary parts of the terms as sufficient by itself; it is the real and imaginary parts of the partial sums that must converge. Equivalently, the sequence of complex partial sums must converge in the modulus metric. Absolute convergence is especially useful because it bounds the modulus of every finite tail by the corresponding tail of a nonnegative real series.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

How is convergence of \(\sum_{n=1}^{\infty}z_n\) defined in terms of its partial sums?
State the Cauchy Criterion for a Complex Series in terms of finite tails.
Why does convergence of \(\sum |z_n|\) imply convergence of \(\sum z_n\)?
What is the sum of \(\sum_{n=0}^{\infty}r^n\) when \(|r|<1\), and which finite-sum identity leads to it?
Why is the alternating harmonic series conditionally convergent rather than absolutely convergent?
What does the Term Test say, and why does \(z_n\to0\) not by itself prove that \(\sum z_n\) converges?

What You'll Learn