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Terms compared with each other rather than with a limit; why the criterion is equivalent to convergence in $\mathbb{R}$ and not in $\mathbb{Q}$.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state the Cauchy condition, prove that a sequence satisfies it by telescoping and bounding a geometric tail, and assemble the proof that in the real numbers Cauchy and convergent are the same thing. You will also be able to say why consecutive terms drawing together proves nothing, and what the same condition means in a field that is not complete.
Bolzano-Weierstrass, which extracts a convergent subsequence from a bounded sequence, and monotone convergence, which needs monotonicity. This lesson removes the last requirement on the sequence's shape: a criterion for convergence that asks nothing but that the terms settle towards each other.
A sequence is Cauchy when its later terms draw close to each other: for every positive amount there is an index beyond which every pair of terms is closer than that. A field is complete when every Cauchy sequence in it converges; the reals are, and the rationals are not.
A sequence is Cauchy when: for every $\varepsilon > 0$ there is $N$ with $|a_m - a_n| < \varepsilon$ for all $m, n \ge N$.
Compare the definition of convergence. That one mentions $L$; this one does not. So the Cauchy condition can be checked on a sequence whose limit nobody knows — which is the situation in almost every application, from a numerical method to the construction of $\mathbb{R}$ itself.
The theorem. In $\mathbb{R}$, a sequence converges if and only if it is Cauchy.
One direction is easy: if $a_n \to L$, then $|a_m - a_n| \le |a_m - L| + |L - a_n| < \varepsilon$ once both are within $\varepsilon/2$.
The other direction is the one that needs completeness. A Cauchy sequence is bounded (take $\varepsilon = 1$); Bolzano-Weierstrass gives a subsequence converging to some $L$; and then the Cauchy condition drags the whole sequence to $L$, because any term past $N$ is close to a subsequence term that is close to $L$.
Completeness is what lets a limit be produced rather than guessed. Boundedness plus monotonicity, or the Cauchy condition, names a number before anyone knows what it is; over the rationals the same argument runs and the number it names is not there.
All later pairs, not just neighbours. The condition quantifies over every $m, n \ge N$. A sequence whose consecutive gaps tend to zero need not be Cauchy, and the harmonic partial sums are the standard proof of that.
Another way: picture
Convergence draws a band about a fixed height and asks the tail to fit inside. The Cauchy condition draws no band at all: it asks that the tail eventually fit inside some band of width $\varepsilon$, wherever that band happens to sit. In a complete space the two requests amount to the same thing.
Another way: steps
To show a sequence is Cauchy:
Step 2 is what converts a hypothesis about neighbours into a statement about distant pairs. Without it, nothing about $m$ and $n$ far apart has been said.
Suppose $|a_{n+1} - a_n| \le K r^n$ with $0 < r < 1$. Then for $m > n$, $$|a_m - a_n| \le \sum_{j=n}^{m-1} |a_{j+1} - a_j| \le K \sum_{j \ge n} r^j = \frac{K r^n}{1 - r},$$ which tends to $0$ as $n$ grows. So the sequence is Cauchy, and therefore converges — and the limit has still not been named.
This is the estimate behind every fixed-point iteration: Newton's method, the proof that a differential equation has a solution, the construction of a fractal. In each case the object being produced is not available in advance, and the Cauchy criterion is what asserts that it exists.
The geometric bound is doing real work. Replace $r^n$ by $1/n$ and the sum becomes a harmonic tail, which does not shrink, and the conclusion fails — as it must, since the harmonic partial sums diverge.
| Hypothesis | In $\mathbb{R}$ | In $\mathbb{Q}$ |
|---|---|---|
| converges | Cauchy, bounded | Cauchy, bounded |
| Cauchy | converges, bounded | bounded; may converge to nothing |
| bounded | has a convergent subsequence | has a Cauchy subsequence |
| consecutive gaps to zero | nothing follows | nothing follows |
The second row is the whole content of completeness, and the right-hand column is why the axiom is an axiom. The truncations $1, 1.4, 1.41, 1.414, \ldots$ are a Cauchy sequence of rationals with no rational limit, and every real number can be reached this way — which is how $\mathbb{R}$ is constructed when it is built rather than assumed.
Checking neighbours only. $|a_{n+1} - a_n| \to 0$ is not the Cauchy condition. The harmonic partial sums satisfy it and diverge, and the gap $H_{2n} - H_n \ge 1/2$ shows exactly which pairs were forgotten.
Thinking a Cauchy sequence has a limit by definition. It has one in $\mathbb{R}$, by a theorem that uses completeness. In $\mathbb{Q}$ the same sequence can have none, and the definition is unchanged.
Treating it as a weaker conclusion. In $\mathbb{R}$ the two are equivalent, so nothing is lost by proving the Cauchy condition instead. What is gained is that no candidate limit is needed.
Let $a_n = \sum_{k=1}^{n} \dfrac{\sin k}{2^k}$. Its limit has no closed form, so the definition is unusable directly.
No candidate limit to aim at.
For $m > n$, $|a_m - a_n| \le \sum_{k=n+1}^{m} 2^{-k} < 2^{-n}$.
Telescope, then bound by a geometric tail.
Given $\varepsilon$, choose $N$ with $2^{-N} < \varepsilon$. So the sequence is Cauchy, hence convergent.
Existence, with the value never mentioned.
Take $H_n = \sum_{k=1}^n 1/k$ and $\varepsilon = 1/2$.
One epsilon is enough to deny the condition.
For any $N$, take $n = N$ and $m = 2N$. Then $H_m - H_n$ is a sum of $N$ terms each at least $1/(2N)$, so it is at least $1/2$.
Terms found beyond every $N$ that stay apart.
So no $N$ works for $\varepsilon = 1/2$, the sequence is not Cauchy, and the harmonic series diverges — proved without computing a single partial sum.
Divergence, from the criterion alone.
Consecutive gaps: $\sqrt{n+1} - \sqrt{n} = \dfrac{1}{\sqrt{n+1} + \sqrt{n}} \to 0$, so neighbours draw together.
The tempting evidence, which is not the condition.
But take $\varepsilon = 1$, $n$ arbitrary and $m = 4n$: then $a_m - a_n = 2\sqrt{n} - \sqrt{n} = \sqrt{n} \ge 1$ for $n \ge 1$.
Test a distant pair, not a neighbouring one.
So it is not Cauchy, and indeed it is unbounded. The lesson is the same as the harmonic one: shrinking steps can still cover unlimited ground, and only a condition on all later pairs rules that out.
Build the proof that every Cauchy sequence of real numbers converges.
This task has no paper form; do it on a device.
$a_n = \dfrac{1}{n}$. For $\varepsilon = \dfrac{1}{8}$, what is the smallest $N$ with $|a_m - a_n| < \varepsilon$ for all $m, n \ge N$?
Answer:
A sequence satisfies $|a_{j+1} - a_j| \le \left(\dfrac{1}{9}\right)^j$ for every $j$. Put the steps that show it is Cauchy into order.
Number the steps in order (write the number in the box):
Match each hypothesis about a sequence to what it entitles you to conclude.
| it converges, with no limit named in advance | it converges, to the supremum of its terms | some subsequence of it converges | nothing at all follows | |
|---|---|---|---|---|
| the sequence is Cauchy | ||||
| the sequence increases and stays below $7$ | ||||
| the sequence is bounded | ||||
| the gap between consecutive terms tends to zero |
Let $H_n = 1 + \tfrac{1}{2} + \cdots + \tfrac{1}{n}$. Here is an argument that it converges. Mark the one step that does not follow.
This task has no paper form; do it on a device.
The decimal truncations of $\sqrt{2}$ form a sequence of rationals that satisfies the Cauchy condition. Regarded as a sequence in $\mathbb{Q}$, what does that show?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Build the proof that every Cauchy sequence of real numbers converges.
This task has no paper form; do it on a device.
You can check the Cauchy condition on a sequence, prove convergence from it without naming a limit, and explain why neighbouring terms are not enough. Say in your own words why the harmonic partial sums fail the condition. Next: the same ideas, moved from a sequence of terms to a function of a real variable.
10. Your turn: is $a_n = \sqrt{n}$ Cauchy?, step 3