Sequences: convergence, monotone, Bolzano–Weierstrass, Cauchy

In one sentence

Four properties — convergent, Cauchy, bounded, has a convergent subsequence — sit in a chain of one-way implications, and the exam tests whether you know which arrows reverse and which do not.

Why the exam asks it

Because it can be asked as a decision. Here is a sequence: is it bounded? Cauchy? convergent? Does it have a convergent subsequence? Four options, four independent judgements, and no partial credit. It is Part C's ideal shape, which is why it recurs.

The idea

Take a real sequence $(a_n)$. The implications that always hold:

Convergent $\Rightarrow$ Cauchy. If the terms approach a limit, they approach each other. Triangle inequality; no completeness needed.

Cauchy $\Rightarrow$ bounded. Beyond some $N$ all terms sit within $1$ of each other; before $N$ there are finitely many. A finite set is bounded.

Bounded $\Rightarrow$ a convergent subsequence. This is Bolzano–Weierstrass, and it is a theorem about $\mathbb{R}$, not a generality. It fails in infinite-dimensional spaces: the sequence of standard basis vectors in $\ell^2$ is bounded and has no convergent subsequence, because any two of them are $\sqrt{2}$ apart.

Cauchy $\Rightarrow$ convergent is the one that needs completeness. In $\mathbb{Q}$ the decimal approximations of $\sqrt{2}$ are Cauchy and converge to nothing rational. In $\mathbb{R}$ it holds, and it is equivalent to the completeness axiom.

Monotone convergence

If $(a_n)$ is increasing and bounded above, it converges — to $\sup_n a_n$. This is the workhorse for recursively defined sequences, and the routine is always the same: prove monotone by induction, prove bounded by induction, then solve $L = f(L)$ for the limit.

For $a_{n+1} = \sqrt{2 + a_n}$ with $a_1 = \sqrt2$: increasing, bounded above by $2$, so it converges; and $L = \sqrt{2 + L}$ gives $L = 2$. Note the order — you may only solve for $L$ after establishing that a limit exists. Solving first and concluding convergence is a genuinely common error, and one the paper is happy to reward with a wrong option.

Sequences worth carrying in

• $(1 + 1/n)^n$ — increasing, bounded by $3$, limit $e$.
• $n^{1/n} \to 1$.
• $(n!)^{1/n}/n \to 1/e$.
• $\sin n$ — bounded, divergent, and its set of subsequential limits is the whole of $[-1,1]$, because $\{n \bmod 2\pi\}$ is dense.

Where intuition breaks

"Bounded, so it converges." $(-1)^n$. This is the first spoiler to reach for, and it disposes of a surprising number of options on its own.

"Every subsequence converges, so the sequence converges." True — but read it carefully, because the exam prefers the near-miss: some subsequence converging says nothing. $(-1)^n$ has two convergent subsequences and does not converge. Bounded plus a unique subsequential limit does give convergence.

"Cauchy is just convergence written differently." In $\mathbb{R}$ they coincide, which is precisely what makes this dangerous: the equivalence is a property of the space, not of the definition. Move to $\mathbb{Q}$, or to $(0,1)$ with $a_n = 1/n$, and the two come apart. Options that shift the ambient space are testing exactly this — a hypothesis dropped, where the hypothesis is completeness of the space.

"$a_{n+1} - a_n \to 0$, so the sequence is Cauchy." False, and the counterexample is the one everyone half-remembers: $a_n = \sqrt n$, or the partial sums of the harmonic series. Consecutive terms get arbitrarily close while the sequence marches off to infinity. Cauchy requires all later pairs to be close, not merely adjacent ones.

"Cesàro means converge, so the sequence converges." The forward direction is a genuine theorem: $a_n \to L$ implies the averages tend to $L$. The converse fails — $(-1)^n$ has averages tending to $0$ and does not converge. A converse assumed, and one of the most reliably productive distractors in the whole paper.

"$a_{n+1}/a_n \to L$, so $a_n^{1/n} \to L$." Also a genuine one-way theorem, also frequently reversed in the options, and also false in reverse: $1, 2, 1, 2, \ldots$ has $n$-th roots tending to $1$ while the ratios oscillate.

The exam's angle

Work in this order and most questions collapse quickly.

1. Is it bounded? If not, nothing else is worth checking — it is not Cauchy and not convergent, though it may still have a monotone subsequence.
2. Is it monotone? Monotone plus bounded finishes it immediately.
3. Can you name two subsequential limits? If so it diverges, whatever else is true.
4. Which space are we in? If the option says $\mathbb{Q}$, or an open interval, or a function space, the Cauchy–convergent arrow is the thing being tested.

Step 4 is the one that costs marks. When two options are identical except for the ambient set, that difference is the question.

The night before

• convergent $\Rightarrow$ Cauchy $\Rightarrow$ bounded $\Rightarrow$ convergent subsequence. None of these reverse without help.
• Cauchy $\Rightarrow$ convergent needs completeness. Fails in $\mathbb{Q}$ and in $(0,1)$.
• Bolzano–Weierstrass is about $\mathbb{R}$; it fails in $\ell^2$.
• Monotone + bounded $\Rightarrow$ convergent. Establish existence before solving $L = f(L)$.
• $(-1)^n$ kills "bounded ⇒ convergent". $\sqrt n$ kills "consecutive differences → 0 ⇒ Cauchy".
• Cesàro and the ratio–root lemma both run one way only.