Complex Series: Comparison, Ratio, and Root Tests

All those who seem stupid, they are, and also so are half of those who do not, Quevedo

Introduction

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).

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).

1. If there exists $N \in \mathbb{N}$ such that for all $|a_n| \leq b_n$ for all $n \geq N$, and if $\sum b_n$ converges, then $\sum |a_n|$ converges. Hence, $\sum a_n$ is absolutely convergent and therefore convergent.
2. If there exists $N \in \mathbb{N}$ such that $b_n \leq |a_n|$ for all $n \geq N$, and if $\sum b_n$ diverges, then $\sum |a_n|$ diverges. Hence $\sum a_n$ does not converge absolutely.

Proof.

Part 1.

Since $\sum_{n=1}^{\infty} b_n$ converges and $b_n \geq 0$, the tail sum satisfies: $\sum_{n=N}^{\infty} b_n < \infty$

By hypothesis, $|a_n| \leq b_n$ for all $n \geq N$.

Therefore, $\sum_{n=N}^{M} |a_n| \leq \sum_{n=N}^{M} b_n \leq \sum_{n=N}^{\infty} b_n < \infty$ for all $M \geq N$.

The partial sums $S_M = \sum_{n=1}^{M} |a_n|$ are:

• Bounded above by $\sum_{n=1}^{N-1}|a_n| + \sum_{n=N}^{\infty} b_n$
• Monotonically increasing (since $|a_n| \geq 0$)

By the monotone convergence theorem, $\sum_{n=1}^{\infty} |a_n|$ converges.

The Monotone Convergence Theorem states that if a sequence of real numbers is both monotone and bounded, then the sequence converges.

Since absolute convergence implies convergence for complex series, $\sum a_n$ converges. $\blacksquare$

Part 2.

Since $b_n \geq 0$ and $\sum b_n$ diverges, we have $\sum_{n=N}^{\infty} b_n = +\infty$.

By hypothesis, $b_n \leq |a_n|$ for all $n \geq N$. Therefore for any $M \geq N$: $\sum_{n=N}^{M} |a_n| \geq \sum_{n=N}^{M} b_n$

Taking the limit as $M \to \infty$: $\sum_{n=N}^{\infty} |a_n| \geq \sum_{n=N}^{\infty} b_n = +\infty$

Thus, $\sum_{n=1}^{\infty} |a_n|$ diverges. $\blacksquare$

The comparison test works by bounding the series of interest by a simpler series whose behavior is known. It is the most elementary convergence test but often the most direct. The key insight is that finitely many terms do not affect convergence — only the “tail” matters, which is why the condition $|a_n| \leq b_n$ need only hold for $n \geq N$.

Futhermore, the reader should notice that Part 2 only states that $\sum a_n$ is not absolutely convergent. The series might still converge conditionally. To establish divergence outright, one typically needs a separate argument (e.g., showing that $a_n \not\to 0$).

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 $b_n = \frac{5}{n^2}$.

1. For all $n \ge 1$, we have $0 \le |a_n| = \dfrac{5}{n^2+n} \le \frac{5}{n^2} = b_n$.
2. The series $\sum \dfrac{5}{n^2}$ converges (it is a constant multiple of the convergent p-series with p = 2 > 1).
3. 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 with $a_n \neq 0$ for all sufficiently large $n$. Consider the limit (if it exists) $L := \lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|}.$

1. Absolute Convergence: If L < 1, the series converges absolutely.
2. Divergence: If L > 1 (including L = +∞), the series diverges because the terms do not tend to zero.
3. Inconclusive: If L = 1 or the limit does not exist, the test is inconclusive. The series may converge or diverge, and another convergence test is needed.

Roughly, the ratio $|a_{n+1}|/|a_n|$ measures how fast the terms are shrinking. If the ratio stabilizes below 1, say at some value $r < 1$, then eventually $|a_n| \lesssim r^n$, the terms decay at least as fast as a convergent geometric series. 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.

Proof

If $L < 1$, pick any real number $r$ such that $L < r < 1$. By the formal definition of a limit, there exists an integer N such that for all $n\geq N, \frac{|a_{n+1}|}{|a_n|} < r$.

From the inequality above we get, for every $n\geq N, |a_{n+1}| < r |a_n|.$

Iterating this inequality yields, for $k\geq 0, |a_{N+k}| < r^k|a_N|$.

Rewriting this by substitution n = N + k gives us a general bound for the “tail” of the series: $|a_n|\leq |a_N|\, r^{\, n-N}\quad \mathrm{for\ all\ }n\geq N.$

To make this bound apply to the entire sequence, we can set $\displaystyle C=\max \{ |a_1|,\dots ,|a_{N-1}|,|a_N|\}$. Then, for every $n\geq 1$ we have $|a_n|\leq Cr^{\, n-N} = \left(\frac{C}{r^N}\right) r^n$

If $1 \leq n < N$. Since r < 1 we have $r^{n-N}=r^{-(N-n)} \geq 1$. Hence $C\leq C\, r^{n-N}$, and because $C\geq |a_n|$ we get $|a_n|\leq C\leq C\, r^{n-N}$.

Letting $C' = \frac{C}{r^N}$, we have $|a_n| \leq C' r^n$ for some constant $C' > 0$

Convergence by Comparison:

Since $0 < r < 1$, the series $\sum C' r^n$ is a convergent geometric series. Because $|a_n| \leq C' r^n$, the Direct Comparison Test implies that $\sum |a_n|$ converges. Therefore, the original series $\sum a_n$ converges absolutely.

Part 2: Divergence ($L > 1$)

If $L > 1$, choose a real number $s$ such that $1 < s < L$.

By the definition of a limit, for sufficiently large $n$, we have: $\frac{|a_{n+1}|}{|a_n|} > s > 1$

This implies that $|a_{n+1}| > |a_n|$, meaning the sequence of magnitudes $\{ |a_n| \}$ is eventually strictly increasing.

An eventually increasing sequence of positive numbers cannot tend to 0. Since $\lim_{n \to \infty} a_n \neq 0$, the series fails the basic $n$-th term test for divergence. Therefore, the series $\sum a_n$ diverges.

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 $z \in \mathbb{C}$.

This series defines the exponential function $e^z$, and this computation shows that $e^z$ is an entire function — analytic on all of $\mathbb{C}$.

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 ordinary limit $\lim_{n\to\infty} \sqrt[n]{|a_n|}$ exists (for most common series, the regular limit exists), it equals the $\limsup$ and can be used directly.

Proof

Case $L < 1$: Choose $\rho$ with $L < \rho < 1$. By definition of $\limsup$, for all sufficiently large $n$, $\sqrt[n]{|a_n|} \leq \rho$.

Raising both sides to the n-th power gives $|a_n| \leq \rho^n, \forall n \ge N$.

The tail $\sum_{n=N}^{\infty} \rho^n$ is a convergent geometric series because $\rho <1$. By comparison, $\sum_{n=N}^{\infty}|a_n|$ converges, hence $\sum_{n=1}^{\infty}|a_n|$ converges.

Case $L > 1$: There exist infinitely many $n$ with $\sqrt[n]{|a_n|} > 1 + \varepsilon$ for some $\varepsilon > 0$.

For each such n we have $|a_n| > (1+\varepsilon)^n$. Since $(1+\varepsilon )^n\rightarrow \infty$ as $n \rightarrow \infty$, those $|a_n|$ do not tend to 0. A necessary condition for $\sum a_n$ to converge is $a_n\rightarrow 0$, so the series diverges.

Examples of the Root Test

• 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 (Of course, this is a geometric series with ratio $\frac{1}{2}$, so convergence is already known; the root test confirms it.)

• 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 $\sqrt[n]{n} = n^{1/n}$. We need the fundamental limit $\lim_{n \to \infty} n^{1/n} = 1$.

Define $L = \lim_{n\to\infty} n^{1/n}$, and take logarithms $\ln(L) = \lim_{n\to\infty} ln(n^{1/n}) = \lim_{n\to\infty} \frac{\ln(n)}{n} =[\text{We know ln(n) grows slower than n, e.g., by L'Hôpital's rule}] = 0$. 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: $a_n=n!$.

By Stirling’s approximation, $n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$ for large $n$.

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

As $n \to \infty$, $\left(2\pi n\right)^{1/(2n)} \to 1$ (by the same reasoning that $n^{1/n} \to 1$), while $\frac{n}{e} \to \infty$.

Hence, $L = \lim_{n \to \infty}\sqrt[n]{n!} = +\infty.$

Since $L > 1$, the root test confirms divergence. This is unsurprising: the terms $n!$ grow faster than any exponential, so they certainly do not tend to zero.

• Geometric series in $\mathbb{C}$. Consider $\sum_{n=1}^{\infty}\left(\frac{z}{3}\right)^n$ for $z \in \mathbb{C}$.

$L = \lim_{n \to \infty} \sqrt[n]{\left| \left(\frac{z}{3}\right)^n \right|} = \lim_{n \to \infty} \left( \left|\frac{z}{3}\right|^n \right)^{1/n} = \lim_{n \to \infty} \left|\frac{z}{3}\right| = \left|\frac{z}{3}\right|.$

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, $B(0; 3)$.

If |z| > 3, then L > 1 and the series does not converge absolutely. $\left| \left( \frac{z}{3}\right) ^n\right| =\left| \frac{z}{3}\right| ^n\rightarrow \infty$ as $n\rightarrow \infty$. In particular the terms do not tend to 0 ($|a_n|\rightarrow 0$ if and only if $a_n\rightarrow 0$. Therefore, whenever the moduli $|a_n|$ fail to tend to zero, the terms $a_n$ themselves also fail to tend to zero).

If |z| = 3, the test is inconclusive (this is the boundary case of the corresponding geometric series). However, we can resolve this case directly: when $|z| = 3$, the terms are $(z/3)^n$ with $|z/3| = 1$, so $|(z/3)^n| = 1$ for all $n$. Since the terms do not tend to zero, the series diverges by the divergence test.