KindiakMath

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

\displaystyle s_n = \sum_{k=1}^n x_k \equiv x_1 + x_2 + \cdots + x_n,\quad s_\infty = \sum_{k=1}^\infty x_k = x_1+ x_2 + x_3 + \cdots.

We say that the series s_\infty converges to a real number s if s_n \to s, and diverges otherwise.

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 |r^n| = |r|^n \to 0 implies s_\infty = 1/(1-r). If r > 1, then s_n = (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. 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. 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.

What we have shown is that 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.

Theorem 1. 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 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.

Can we can show that (i) \{x_n\} is bounded, then obtain a subsequence \{x_{n_k}\} with limit x, and (ii) use the Cauchy-ness of \{x_n\} to prove that x_n \to x?

For (i), 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. By the Bolzano-Weierstrass theorem, \{x_n\} contains a convergent subsequence x_{n_k} \to x. We proceed to step (ii).

For (ii), 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

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