Sequences and Limits

Tier: U (Undergraduate). Builds on The Real Number System. Everything later - continuity, differentiability, integration, and the whole of measure theory - is built by changing the words in this page’s definitions while keeping the skeleton.

Attempt these from memory before reading. Getting them wrong is the point: the failure is what makes the rest stick.

• Define what it means for $a_n \to L$, using only quantifiers.
• Why is the limit unique?
• Is every convergent sequence bounded? Is every bounded sequence convergent?
• What is a Cauchy sequence, and does every Cauchy sequence in $\mathbb{Q}$ converge?
• State Bolzano–Weierstrass.

Check your answers against the text. Every one you missed is a section to read twice.

” $a_n$ gets close to $L$” is not mathematics, because close is not a number. The first attempt is: for every $\varepsilon > 0$, $|a_n - L| < \varepsilon$. But that is false for every sequence; $|a_n - L|$ is never exactly zero, so it is less than $\varepsilon$ only eventually.

That “eventually” is the whole idea. The definition says the terms get and stay arbitrarily close, and it does so by making you, the prover, supply an $N$ for each $\varepsilon$ your adversary hands you. The adversary picks a tolerance; you must meet it.

Worked example 1 (the adversary, played honestly). Show $a_n = 1/n \to 0$.

Step 1. The adversary names $\varepsilon > 0$.

Step 2. We must find $N$ with $|1/n - 0| < \varepsilon$ for $n \ge N$. Since $|1/n| = 1/n$, we need $1/n < \varepsilon$, i.e. $n > 1/\varepsilon$.

Step 3. Choose $N = \lceil 1/\varepsilon \rceil + 1$. Then $n \ge N$ gives $n > 1/\varepsilon$, so $|1/n| < \varepsilon$. $\blacksquare$

Worked example 2 (the same three steps, harder algebra). Show $a_n = \frac{3n^2 + 2}{n^2 + 1} \to 3$.

Step 1. Adversary gives $\varepsilon > 0$.

Step 2. Bound the error without solving exactly:

Step 3. Choose $N = \lceil 1/\varepsilon \rceil + 1$. $\blacksquare$

Notice the technique: we did not solve $1/(n^2+1) < \varepsilon$ exactly. We overestimated and then solved the easier bound. An overestimate is fine; the definition only asks you to make the error small, not to make it small optimally.

Completion problem. Show $a_n = \frac{\sin n}{n} \to 0$ by filling in the missing step: $|\sin n / n| \le \_\_\_$, so choose $N = \_\_\_$.

Definition (F). A sequence $(a_n)$ converges to $L \in \mathbb{R}$ if for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that

We write $a_n \to L$. A sequence that does not converge diverges.

Each clause earns its place:

• for every $\varepsilon > 0$; arbitrarily close, not just close. Delete it and $a_n = 1/n$ “converges” to $0.1$.
• there exists $N$; the $\varepsilon$ is allowed to depend on nothing; the $N$ may depend on $\varepsilon$. Swapping those quantifiers defines a different (and stronger) statement.
• for all $n \ge N$; stays close. Delete it and $a_n = n + (-1)^n$ “converges” infinitely often.
• $|a_n - L| < \varepsilon$; the metric. Replacing this is how analysis is generalised: to $\mathbb{R}^k$ (Euclidean distance), to functions (sup norm), and to measure theory (convergence in measure).

Proposition 2.1 (Uniqueness of Limits). If $(a_n)$ converges, its limit is unique.

Proof. Suppose $a_n \to L$ and $a_n \to M$ with $L \ne M$. Let $\varepsilon = |L - M|/2 > 0$. There is $N_1$ with $|a_n - L| < \varepsilon$ for $n \ge N_1$, and $N_2$ with $|a_n - M| < \varepsilon$ for $n \ge N_2$. For $n \ge \max(N_1, N_2)$,

a contradiction. $\blacksquare$

The key idea. Two distinct numbers are separated by a fixed gap, and a convergent sequence is eventually inside any gap around its limit. It cannot be inside two disjoint gaps at once. This “two balls around distinct points are disjoint” move is the reason uniqueness holds in every metric space, and it is the reason it fails when the topology is not Hausdorff.

Proposition 2.2. Every convergent sequence is bounded.

Proof. Taking $\varepsilon = 1$ there is $N$ with $|a_n - L| < 1$ for $n \ge N$, so $|a_n| \le |L| + 1$ for $n \ge N$. Let $M = \max\{|a_1|, \dots, |a_{N-1}|, |L| + 1\}$. $\blacksquare$

Counterexample. $a_n = n$ is unbounded and diverges; so boundedness is implied by convergence. The converse is false: $a_n = (-1)^n$ is bounded and diverges, since the subsequences $a_{2k} = 1$ and $a_{2k+1} = -1$ converge to different limits.

Theorem 2.1 (Algebra of Limits). If $a_n \to L$ and $b_n \to M$ then

1. $a_n + b_n \to L + M$
2. $a_n b_n \to LM$
3. $a_n / b_n \to L/M$, provided $M \ne 0$ and $b_n \ne 0$ for all $n$.

Theorem 2.2 (Squeeze Theorem). If $a_n \le b_n \le c_n$ eventually and $a_n \to L$, $c_n \to L$, then $b_n \to L$.

The squeeze theorem is the working analyst’s favourite tool: it converts a hard estimate into an easy one by sandwiching the unknown between two known limits. Worked example 2 above used exactly this idea.

Theorem 2.3 (Monotone Convergence Theorem). Every bounded monotone sequence in $\mathbb{R}$ converges; to its supremum if increasing, its infimum if decreasing.

Proof. Let $(a_n)$ be bounded and increasing. By completeness, $s = \sup\{a_n\}$ exists. Given $\varepsilon > 0$, the approximation property gives $N$ with $s - \varepsilon < a_N \le s$. Since $(a_n)$ increases, $a_n \ge a_N > s - \varepsilon$ for $n \ge N$, while $a_n \le s$ throughout. So $|a_n - s| < \varepsilon$. $\blacksquare$

Why this theorem is the most important on the page. Its proof uses completeness; the existence of suprema; and nothing else. That is the property separating $\mathbb{R}$ from $\mathbb{Q}$, and it is the engine behind every existence theorem in analysis: the intermediate value theorem, the extreme value theorem, Riemann integrability, and the convergence of Fourier series all trace back to it.

Counterexample. In $\mathbb{Q}$, the sequence $a_n = (1 + 1/n)^n$ is increasing and bounded above (by 3, say), but its limit is $e \notin \mathbb{Q}$. Bounded and monotone, yet not convergent. Completeness is exactly what was missing.

Definition. $(a_n)$ is Cauchy if for every $\varepsilon > 0$ there exists $N$ such that $|a_n - a_m| < \varepsilon$ for all $m, n \ge N$.

The definition replaces “close to a limit” with “close to each other”; a statement you can verify without knowing the limit. That is what makes it useful for proving things exist.

Theorem 2.4. Every convergent sequence is Cauchy.

Proof. Given $\varepsilon > 0$, choose $N$ with $|a_n - L| < \varepsilon/2$ for $n \ge N$. Then for $m, n \ge N$, $|a_n - a_m| \le |a_n - L| + |a_m - L| < \varepsilon$. $\blacksquare$

Theorem 2.5 (Cauchy Completeness of $\mathbb{R}$). Every Cauchy sequence in $\mathbb{R}$ converges.

Proof. A Cauchy sequence is bounded, so by Bolzano–Weierstrass it has a convergent subsequence $a_{n_k} \to L$. Given $\varepsilon > 0$, choose $N_1$ with $|a_n - a_m| < \varepsilon/2$ for $m,n \ge N_1$, and $K$ with $|a_{n_k} - L| < \varepsilon/2$ for $k \ge K$. For $n \ge N_1$ pick $k \ge K$ with $n_k \ge N_1$. Then

Why this matters. A space in which Cauchy sequences converge is called complete. Completeness of $\mathbb{R}$ is what makes Newton’s method, fixed point iteration, power series and the fundamental theorem of calculus work. In functional analysis the same property defines a Banach space, and in that setting Picard iteration gives existence and uniqueness of solutions to ODEs.

A subsequence of $(a_n)$ is $(a_{n_k})$ with $n_1 < n_2 < \cdots$.

Proposition 2.3. If $a_n \to L$ then every subsequence $a_{n_k} \to L$.

Proposition 2.4. If $(a_n)$ has two subsequences converging to different limits, then $(a_n)$ diverges.

Theorem 2.6 (Bolzano–Weierstrass). Every bounded sequence in $\mathbb{R}$ has a convergent subsequence.

The proof idea is a bisection argument: split a bounded interval in half, at least one half contains infinitely many terms, keep that half, and repeat. The midpoints converge; choosing one term from each nested interval gives a convergent subsequence. Bolzano–Weierstrass is used to prove Theorem 2.5, and it is the reason every continuous function on a closed bounded interval is bounded and attains its bounds.

The fourth row is the one to remember. It is a sequence of rationals, Cauchy in $\mathbb{Q}$, with no rational limit; and it is the reason $\mathbb{R}$ has to be constructed as a completion rather than assumed.

Mix these with problems from The Real Number System and Series: deciding which tool to use is the part blocked practice never trains.

1. Show $a_n = \sqrt{n^2 + n} - n \to \tfrac12$, by rationalising and then bounding.
2. Show that if $a_n > 0$ and $a_{n+1}/a_n \to L < 1$ then $a_n \to 0$. (Hint: compare with a geometric sequence; this is the ratio test in disguise.)
3. Show a sequence with $|a_{n+1} - a_n| \le 2^{-n}$ is Cauchy, and deduce it converges. (This is how most existence proofs in analysis actually run.)
4. Give a sequence $(a_n)$ with no convergent subsequence. What property of $\mathbb{R}$ did you use?
5. Show that if every subsequence of $(a_n)$ has a further subsequence converging to $L$, then $a_n \to L$. (This “subsequence of subsequence” argument is used constantly in probability.)

• The Real Number System - completeness, the engine behind this page
• Series; sequences of partial sums
• Continuity; the same definition, one abstraction up
• Measure Theory: where convergence gets harder and more useful
• How to Use These Notes; the method, and why recall prompts come first

• Continuity
• Differentiability
• Problem Set
• Riemann Integration