Cauchy’s True Utility

Let’s talk about infinite sums. Consider the following sums:

$\displaystyle \begin{aligned}s &= 1 + 1 + 1 + 1 + \cdots, \\ t &= 1 + 2 + 3 + 4 + \cdots, \\ u &= 1 +2 + 4 + 8 + \cdots, \\ v &= 1 + \frac 12 + \frac 14 + \frac 18 + \cdots. \end{aligned}$

Since we add infinitely many terms, our sums should be infinite, right? That is the case for $s, t, u$ , but surprisingly, not for $v$ . But let’s define what we mean by an infinite sum properly.

We will adopt the convention, as with other topics, that $\mathbb N = \mathbb N^+$ , so that $0 \notin \mathbb N$ . Let $\mathbb K$ be an ordered field.

Definition 1. Let $\{x_n\}$ be a $\mathbb K$ -sequence. Inductively define the partial sums sequence $\{s_n\}$ by

$s_1 := x_1,\quad s_{n+1} := s_n + x_{n+1}.$

Often, we denote

$\begin{aligned} s_n &= \sum_{k=1}^n x_k \equiv x_1 + x_2 + \cdots + x_n,\\ s_\infty &= \sum_{k=1}^\infty x_k = x_1+ x_2 + x_3 + \cdots. \end{aligned}$

We say that the series $s_\infty$ converges to a real number $s$ if $s_n \to s$ , and diverges otherwise. In this case, we write

$\displaystyle s_\infty = \lim_{n \to \infty} s_n = s.$

Example 1. Given $\{x_n\} = \{1\}$ with partial sums $s_n = 1 + \cdots + 1 = n$ , $s_n = n\to \infty$ , thus $s_\infty$ diverges.

Example 2. Given $\{x_n\} = \{r^{n-1}\}$ , the sequence of partial sums is given by

$s_n = \begin{cases}\frac{1-r^n}{1-r}, & \quad r \neq 1, \\ 1, & \quad r = 1. \end{cases}$

If $|r| < 1$ , then

$\displaystyle |r^n| = |r|^n \to 0 \quad \Rightarrow \quad s_\infty = \frac 1{1-r}.$

If $r > 1$ , then

$\displaystyle r^n \to \infty \quad \Rightarrow \quad s_n = \frac{ 1-r^n }{ 1-r } \to -\infty.$

If $r < -1$ , then $-r > 1$ implies

$\displaystyle \begin{aligned} s_{2k} &= \frac{1-r^{2k}}{1-r} = \frac{1-(-r)^{2k}}{1+(-r)} \to -\infty, \\ s_{2k+1} &= \frac{1-r^{2k+1}}{1-r} = \frac{1+(-r)^{2k+1}}{1+(-r)} \to \infty. \end{aligned}$

If $r = 1$ , then $s_n = n \to \infty$ . If $r = -1$ , then

$\displaystyle \begin{aligned} s_{2k} &= \frac{1-(-1)^{2k}}{1-(-1)} = \frac{1-1}{2} = 0 \to 0, \\ s_{2k+1} &= \frac{1-(-1)^{2k+1}}{1-(-1)} = \frac{1-(-1)}{2} = 1 \to 1. \end{aligned}$

However, if $s_\infty = s$ , then $0 = s = 1$ , a contradiction.

This gives us our first crucial convergence result.

Theorem 1. For any $r$ , the partial sums of $\{r^n\}$ , called the standard geometric series, converges to $1/(1-r)$ if and only if $|r| < 1$ . In summation notation,

$\displaystyle \sum_{k=1}^\infty r^k \equiv 1 + r + r^2 + r^3 + \cdots = \frac 1{1-r}$

if and only if $|r| < 1$ .

Example 3. In particular, setting $r = 1/2$ so that $|r| < 1$ ,

$\displaystyle 1 + \frac 12 + \frac 14 + \frac 18 + \cdots = \frac 1{1 - 1/2} = 2.$

But what other series converge? Recall that we write

$x_1 + x_2 + x_3 + \cdots = s$

if and only if the partial sums $s_n \to s$ . Let’s expand this latter idea in epsilontics.

Lemma 1. Suppose $\{s_n\}$ converges. Then for any $\epsilon > 0$ , there exists $N \in \mathbb N$ such that for $m,n > N$ ,

$|s_m - s_n| < \epsilon.$

Proof. Fix $\epsilon > 0$ . Then there exists $N \in \mathbb N$ such that

$\displaystyle m,n > N \quad \Rightarrow \quad |s_m - s|< \frac{\epsilon}{2}, \quad |s_n - s| < \frac{\epsilon}{2}.$

By the triangle inequality, for $m, n > N$ ,

$\displaystyle |s_m - s_n| \leq |s_m - s| + |s - s_n| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$

This is what we mean by a Cauchy sequence.

Definition 2. A $\mathbb K$ -sequence $\{x_n\}$ is Cauchy if for any $\epsilon > 0$ , there exists $N \in \mathbb N$ such that

$m,n > N \quad \Rightarrow \quad |x_m - x_n| < \epsilon.$

By Lemma 1, for any $\mathbb K$ -seqeunce $\{x_n\}$ , if $\{x_n\}$ converges, then $\{x_n\}$ is Cauchy. Does the reverse hold? Interestingly, when $\mathbb K = \mathbb R$ , the answer is Yes. That is one property that sets $\mathbb R$ apart from $\mathbb Q$ , and it boils down back to the completeness property.

Lemma 2. If $\{x_n\}$ is Cauchy, then $\{x_n\}$ is bounded.

Proof. Since $\{x_n\}$ is Cauchy, for $\epsilon = 1$ , there exists $N \in \mathbb N$ such that

$m,n > N \quad \Rightarrow \quad |x_m - x_n| < 1.$

Then for any $n > N$ ,

$|x_n| \leq |x_n - x_{N+1}| + |x_{N+1}| < 1 + |x_{N+1}|.$

This means that if we define $M := \max \{1,|x_1|, \cdots, |x_N|,|x_{N+1}|\}$ , then for any $n \in \mathbb N$ ,

$|x_n| \leq 1 + M.$

Thus, $\{x_n\}$ is bounded.

Theorem 2. Let $\{x_n\}$ be an $\mathbb R$ -sequence. Then $\{x_n\}$ converges if and only if $\{x_n\}$ is Cauchy.

Proof. We have proven $(\Rightarrow)$ and now turn our attention to prove $(\Leftarrow)$ .

Let $\{x_n\}$ be a Cauchy $\mathbb R$ -sequence. Fix $\epsilon > 0$ . What convergence theorems do we have to prove that $x_n$ converges?

We have the Bolzano-Weierstrass theorem, but that theorem asserts the convergence of a subsequence of $\{x_n\}$ , not the convergence of $\{x_n\}$ itself. Furthermore, this only works if $\{x_n\}$ is bounded. Thankfully, Lemma 2 tells us that this is, in fact, true. Therefore, there exists a subsequence $\{ x_{n_k} \}$ of $\{x_n\}$ such that $x_{n_k} \to x$ .

We somehow need to take advantage of the Cauchy property of $\{x_n\}$ . Fix $\epsilon > 0$ . Since $x_{n_k} \to x$ , there exists $K \in \mathbb N$ such that

$\displaystyle k > K \quad \Rightarrow \quad |x_{n_k} - x| < \frac{\epsilon}{2}.$

Since $\{x_n\}$ is Cauchy, there exists $N \in \mathbb N$ such that

$\displaystyle m,n > N \quad \Rightarrow \quad |x_m - x_n| < \frac{\epsilon}{2}.$

Define $M := \max\{K, N\}$ . Then for $n > n_{M+1}$ , since $n_{M+1} \geq M+1 > M$ ,

$\displaystyle |x_n - x| \leq |x_n - x_{n_{M+1}}| + |x_{n_{M+1}} - x| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$

That’s our proof!

In terms of our main subject of study—infinite series—we get the Cauchy criterion for convergent series.

Corollary 1. Let $\{x_n\}$ be an $\mathbb R$ -sequence. Then $\displaystyle s_\infty \equiv \sum_{k=1}^\infty x_k$ converges if and only if $\{s_n\}$ is Cauchy.

We have established some convergent series, some divergent series, and a comprehensive test to check for convergence. But math people (and worse, engineers) are lazy and don’t have all the time to derive every detail from first principles. Are there some useful rules-of-thumb to help us create more convergent series?

That is the content, which is quite crucial in all honesty, of our next post.

—Joel Kindiak, 28 Dec 24, 1218H

Responses

Acquainting with Convergent Series – KindiakMath

March 20, 2026 at 7:32 pm

[…] of convergent series with several poignant examples and a useful convergence criteria known as the Cauchy criterion. In particular, the geometric series comes into exceptional […]

LikeLike

Cauchy-Goursat Theorem – KindiakMath

March 22, 2026 at 6:04 pm

[…] rectangle for each has a center . Since the sequence is Cauchy, it converges to some . Define the map […]

LikeLike

Cauchy’s True Utility

Share this:

Responses

Leave a comment Cancel reply