The algebra of limits

Theorem

The algebra of limits

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Statement

Let \((a_n)_{n\in\mathbb{N}}\) and \((b_n)_{n\in\mathbb{N}}\) be sequences of real numbers, and suppose that both converge: \(a_n \to a\) and \(b_n \to b\) with \(a, b \in \mathbb{R}\). Then: (i) \(a_n + b_n \to a + b\); (ii) \(a_n b_n \to ab\); (iii) if in addition \(b \neq 0\), then there exists \(N_0 \in \mathbb{N}\) such that \(b_n \neq 0\) for all \(n \ge N_0\), and the tail sequence \(\left(a_n / b_n\right)_{n \ge N_0}\) converges with \(a_n / b_n \to a / b\). In particular, taking \(b_n = \lambda\) constant, \(\lambda a_n \to \lambda a\) for every \(\lambda \in \mathbb{R}\), and combining (i) with \(\lambda = -1\), \(a_n - b_n \to a - b\).

Why it matters

This theorem is the workhorse of first-year analysis: it converts the \(\varepsilon\)–\(N\) definition of a limit, which must otherwise be verified from scratch for every sequence, into a compositional calculus. Once a small stock of basic limits is established (constants, \(1/n \to 0\)), the limit of any expression built from them by finitely many field operations can be read off mechanically — provided every constituent limit exists and no denominator tends to zero.

It is also the precise statement that the field operations of \(\mathbb{R}\) are continuous, i.e. that \(\mathbb{R}\) is a topological field. Every later limit theorem that manipulates expressions term by term — the sum, product and quotient rules for derivatives, linearity of series, continuity of rational functions — ultimately rests on this result. It is equally the standard tool for identifying the limit of a recursively defined sequence once convergence is known.

Hypotheses

\((a_n)\) converges to a real number \(a\).Drop it: take \(a_n = (-1)^n\), \(b_n = 1\). Then \(b_n\) converges but \(a_n + b_n = (-1)^n + 1\) oscillates between \(0\) and \(2\); no conclusion of the theorem survives. Divergence of one summand cannot be repaired by convergence of the other.

\((b_n)\) converges to a real number \(b\).Drop it: take \(a_n = 1/n \to 0\) and \(b_n = n^2\). The product \(a_n b_n = n\) diverges, even though one factor tends to \(0\). Worse, with \(b_n = cn\) the product tends to \(c\), for any prescribed \(c \in \mathbb{R}\): the symbol “\(0 \cdot \infty\)” is genuinely indeterminate.

\(b \neq 0\) (for the quotient only).Drop it: \(a_n = 1\), \(b_n = 1/n \gt 0\) gives \(a_n / b_n = n \to \infty\). And when the quotient does converge the rule still gives no information: \((1/n)/(1/n) \to 1\), \((1/n^2)/(1/n) \to 0\), \((1/n)/(1/n^2) \to \infty\) — all of shape “\(0/0\)”.

The quotient claim is about the tail \(n \ge N_0\), not all \(n\).Nothing prevents \(b_n = 0\) for finitely many small \(n\) (e.g. \(b_n = (n-3)/n \to 1\) has \(b_3 = 0\)); the theorem itself supplies the tail on which \(a_n/b_n\) is defined, via the separation lemma in Step 7. Convergence is a tail property: altering finitely many terms changes nothing.

Limits are taken in \(\mathbb{R}\) (finite limits).Drop it: \(a_n = n + 1 \to +\infty\) and \(b_n = -n \to -\infty\) in the extended sense, yet \(a_n + b_n \to 1\); with \(a_n = n + (-1)^n\) the sum \(a_n + b_n = (-1)^n\) diverges. There is no consistent value for “\(\infty - \infty\)”, so the theorem is stated only for real limits.

Proof

Throughout, “\(a_n \to a\)” means: \(\forall \varepsilon \gt 0\ \exists N \in \mathbb{N}\ \forall n \ge N: |a_n - a| \lt \varepsilon\). We prove (i), (ii), (iii) in turn.

Let \(\varepsilon \gt 0\). There exist \(N_1, N_2 \in \mathbb{N}\) such that \[ n \ge N_1 \implies |a_n - a| \lt \frac{\varepsilon}{2}, \qquad n \ge N_2 \implies |b_n - b| \lt \frac{\varepsilon}{2}. \]

Definition of convergence, applied to \((a_n)\) and \((b_n)\) with tolerance \(\varepsilon/2 \gt 0\) in place of \(\varepsilon\); legitimate because the definition quantifies over every positive tolerance. A

For all \(n \ge \max(N_1, N_2)\): \[ |(a_n + b_n) - (a + b)| = |(a_n - a) + (b_n - b)| \le |a_n - a| + |b_n - b| \lt \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \] Hence \(a_n + b_n \to a + b\), proving (i).

Triangle inequality \(|x + y| \le |x| + |y|\), then Step 1. Since \(\varepsilon \gt 0\) was arbitrary, the definition of convergence is verified with \(N = \max(N_1, N_2)\). A

Boundedness Lemma. \((a_n)\) is bounded: there is \(M \gt 0\) with \(|a_n| \le M\) for all \(n\). Indeed, applying the definition with \(\varepsilon = 1\) gives \(N \in \mathbb{N}\) with \(|a_n - a| \lt 1\) for \(n \ge N\), so \(|a_n| \le |a_n - a| + |a| \lt |a| + 1\) there; set \[ M = \max\{|a_1|, |a_2|, \dots, |a_{N-1}|,\ |a| + 1\} \gt 0. \]

Every convergent sequence is bounded — proved here in full: triangle inequality controls the tail, and the initial segment \(\{a_1, \dots, a_{N-1}\}\) is a finite set, so it has a maximum modulus. Note \(M \ge |a|+1 \ge 1 \gt 0\). B

Key decomposition: for every \(n\), \[ a_n b_n - ab = a_n(b_n - b) + b(a_n - a). \]

Add and subtract \(a_n b\): \(a_n b_n - a_n b + a_n b - ab = a_n(b_n - b) + b(a_n - a)\). This is the one genuine idea in the product proof: it expresses the total error as (bounded factor) \(\times\) (small) \(+\) (constant) \(\times\) (small). B

\[ |a_n b_n - ab| \le |a_n|\,|b_n - b| + |b|\,|a_n - a| \le M\,|b_n - b| + |b|\,|a_n - a| \quad \text{for all } n. \]

Triangle inequality and multiplicativity of absolute value applied to Step 4, then the bound \(|a_n| \le M\) from Step 3. A

Let \(\varepsilon \gt 0\). Choose \(N_3\) with \(|b_n - b| \lt \dfrac{\varepsilon}{2M}\) for \(n \ge N_3\), and \(N_4\) with \(|a_n - a| \lt \dfrac{\varepsilon}{2(|b| + 1)}\) for \(n \ge N_4\). Then for \(n \ge \max(N_3, N_4)\): \[ |a_n b_n - ab| \lt M \cdot \frac{\varepsilon}{2M} + |b| \cdot \frac{\varepsilon}{2(|b|+1)} \le \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \] Hence \(a_n b_n \to ab\), proving (ii).

Definition of convergence twice, with tolerances \(\varepsilon/(2M) \gt 0\) and \(\varepsilon/(2(|b|+1)) \gt 0\). The guard \(|b| + 1\) instead of \(|b|\) is deliberate: it keeps the tolerance positive even when \(b = 0\), and the final estimate still closes because \(|b|/(|b|+1) \lt 1\). C

Separation Lemma. Suppose \(b \neq 0\). Applying the definition with \(\varepsilon = |b|/2 \gt 0\), there is \(N_5 =: N_0\) such that for all \(n \ge N_5\), \[ |b_n| \ge |b| - |b_n - b| \gt |b| - \frac{|b|}{2} = \frac{|b|}{2} \gt 0. \] In particular \(b_n \neq 0\) for all \(n \ge N_5\), so \(a_n / b_n\) is defined on this tail.

Reverse triangle inequality \(\big||x| - |y|\big| \le |x - y|\), hence \(|b_n| \ge |b| - |b_n - b|\). This is where \(b \neq 0\) is used: it makes \(|b|/2\) an admissible positive tolerance and yields a uniform positive lower bound on the tail of \(|b_n|\). B

For all \(n \ge N_5\): \[ \left| \frac{1}{b_n} - \frac{1}{b} \right| = \frac{|b - b_n|}{|b_n|\,|b|} \lt \frac{|b_n - b|}{\frac{|b|}{2} \cdot |b|} = \frac{2}{|b|^2}\,|b_n - b|. \]

Put over a common denominator (legal since \(b_n, b \neq 0\) here), then apply the lower bound \(|b_n| \gt |b|/2\) from Step 7 to the denominator. A

Let \(\varepsilon \gt 0\). Choose \(N_6\) with \(|b_n - b| \lt \dfrac{\varepsilon |b|^2}{2}\) for \(n \ge N_6\). Then for \(n \ge \max(N_5, N_6)\), \[ \left| \frac{1}{b_n} - \frac{1}{b} \right| \lt \frac{2}{|b|^2} \cdot \frac{\varepsilon |b|^2}{2} = \varepsilon, \] so \(1/b_n \to 1/b\).

Definition of convergence with tolerance \(\varepsilon |b|^2 / 2 \gt 0\), combined with Step 8. Convergence of a sequence defined only for \(n \ge N_5\) is convergence of the tail, which is all the definition requires. A

For \(n \ge N_5\), \[ \frac{a_n}{b_n} = a_n \cdot \frac{1}{b_n} \to a \cdot \frac{1}{b} = \frac{a}{b}, \] proving (iii). \(\blacksquare\)

Part (ii), already proved, applied to the sequences \((a_n)\) and \((1/b_n)_{n \ge N_5}\), together with the Tail Lemma: adjoining or discarding finitely many terms changes neither convergence nor the value of the limit (immediate from the definition, since every condition there is of the form “for all \(n \ge N\)”). A

Result

\[ a_n \to a,\ \ b_n \to b \ \implies\ a_n + b_n \to a + b, \quad a_n b_n \to ab, \quad \frac{a_n}{b_n} \to \frac{a}{b}\ \ (\text{if } b \neq 0). \]

Reading. Limits pass through sums, products and quotients: if each ingredient of a finite arithmetic expression converges, the whole expression converges to the same expression evaluated at the limits — with the single proviso that a quotient’s limiting denominator must be nonzero (in which case the denominator sequence is automatically nonzero from some index on).

Scope. Stated for real sequences; the proofs of (i) and (ii) use only the triangle inequality, \(|xy| = |x||y|\), and boundedness of convergent sequences, so they hold verbatim in \(\mathbb{C}\), in any normed vector space (sum) and any normed algebra (product); (iii) additionally needs multiplicative inverses, and holds in \(\mathbb{C}\) and in any Banach algebra near invertible elements. Completeness of \(\mathbb{R}\) is never used — the theorem is true over \(\mathbb{Q}\). It licenses only a fixed, finite number of applications: it says nothing about infinitely many operations or about limits of the form \(\infty - \infty\), \(0 \cdot \infty\), \(0/0\).

Corollaries & converses

Linearity. \(\lambda a_n + \mu b_n \to \lambda a + \mu b\) for all \(\lambda, \mu \in \mathbb{R}\) (constants are convergent sequences; apply (ii) then (i)). In particular \(a_n - b_n \to a - b\).
Fixed powers and polynomials. \(a_n^k \to a^k\) for every fixed \(k \in \mathbb{N}\) (induction on \(k\) via (ii)); hence \(p(a_n) \to p(a)\) for every polynomial \(p\), and \(r(a_n) \to r(a)\) for every rational function \(r = p/q\) with \(q(a) \neq 0\). Equivalently: polynomials and rational functions are sequentially continuous.
Identification of recursive limits. If \(x_{n+1} = f(x_n)\) with \(f\) built from field operations and \(x_n \to L\) (with denominators safe), then \(L = f(L)\); see Worked Example 2.
Converse is false. Convergence of \(a_n + b_n\) or \(a_n b_n\) does not imply convergence of the factors: \(a_n = (-1)^n\), \(b_n = (-1)^{n+1}\) gives \(a_n + b_n = 0 \to 0\) and \(a_n b_n = -1 \to -1\), yet neither sequence converges.
Partial converse. If \(a_n \to a \neq 0\) and \((a_n b_n)\) converges to \(c\), then \((b_n)\) converges to \(c/a\) (Problem 4). The hypothesis \(a \neq 0\) is essential.
Cancellation caveat. From \(a_n b_n \to 0\) one may not conclude that \(a_n \to 0\) or \(b_n \to 0\): take \(a_n = 1 + (-1)^n\), \(b_n = 1 - (-1)^n\).

Fails without

One factor divergent (product): \(a_n = 1/n \to 0\), \(b_n = n^2\). Then \(a_n b_n = n\) diverges. Moreover \(a_n = 1/n\), \(b_n = cn\) gives \(a_n b_n \to c\) for every \(c \in \mathbb{R}\), and \(a_n = 1/n\), \(b_n = (-1)^n n\) gives an oscillating product: without convergence of both factors, “\(0 \cdot \infty\)” can do anything.
Limit of denominator zero (quotient): \(a_n = 1\), \(b_n = 1/n\): here \(b_n \neq 0\) for every \(n\), yet \(a_n / b_n = n\) diverges. And \((1/n)/(1/n) \to 1\) while \((1/n^2)/(1/n) \to 0\): when \(b = 0\) the quotient’s behaviour is not determined by \(a\) and \(b\) at all.
One summand divergent (sum): \(a_n = n\), \(b_n = -n\): the sum converges to \(0\), but the “equation” \(\lim (a_n + b_n) = \lim a_n + \lim b_n\) is meaningless because the right-hand limits do not exist. With \(a_n = n + (-1)^n\), \(b_n = -n\), even the sum diverges. The theorem transfers convergence forwards only, never backwards.

Common errors

“The denominator is never zero, so the quotient rule applies.” Wrong test: what must be nonzero is the limit \(b\), not the terms \(b_n\). With \(b_n = 1/n\) every term is nonzero but the rule is inapplicable.
Splitting before checking. Writing \(\lim (x_n + y_n) = \lim x_n + \lim y_n\) when one of the pieces diverges. The identity is a conclusion available only after both limits on the right are known to exist.
Variable number of terms. Arguing “each of the \(n\) terms \(1/n\) tends to \(0\), so \(\underbrace{\tfrac1n + \cdots + \tfrac1n}_{n \text{ terms}} \to 0\)” — but the sum is constantly \(1\). The theorem covers a number of operations fixed in advance, not one growing with \(n\).
Arithmetic with \(\infty\). Treating \(\infty - \infty\), \(0 \cdot \infty\), \(\infty/\infty\) as if the theorem assigned them values. It assumes finite limits; those symbols are indeterminate forms.
Assuming the limit exists. Solving \(L = f(L)\) for a recursion \(x_{n+1} = f(x_n)\) without first proving convergence: \(x_{n+1} = 2x_n\), \(x_1 = 1\) “gives” \(L = 2L\), so \(L = 0\) — but \(x_n = 2^{n-1}\) diverges. Algebra of limits identifies limits; it never certifies their existence.

Discussion

The clean way to view this theorem is as a continuity statement: addition and multiplication are continuous maps \(\mathbb{R} \times \mathbb{R} \to \mathbb{R}\), and inversion is continuous on \(\mathbb{R} \setminus \{0\}\). In modern language, the algebra of limits says exactly that \(\mathbb{R}\) with its usual metric is a topological field. This is why the proof pattern — triangle inequality for sums, the add-and-subtract decomposition plus a boundedness lemma for products, a separation-from-zero lemma for inverses — recurs verbatim throughout analysis: for limits of functions, for continuity of \(f + g\), \(fg\), \(f/g\), and (one derivative later) for the sum, product and quotient rules of differentiation, whose proofs are this theorem wearing difference quotients.

Historically, eighteenth-century analysts (Euler above all) manipulated limits of sums and products freely, and mostly correctly, without any theorem to lean on. The rigour movement of the nineteenth century — Bolzano’s 1817 pamphlet, Cauchy’s Cours d’analyse (1821), and finally Weierstrass’s \(\varepsilon\)–\(N\) formulation — turned these manipulations into provable claims precisely because the free-wheeling versions sometimes failed: rearranged series changing value, double limits that do not commute, term-by-term operations on infinitely many summands. The algebra of limits marks the boundary: finitely many arithmetic operations are always safe; infinitely many are a separate theory (series, uniform convergence).

The theorem also has a useful economic reading: it lets one pay the \(\varepsilon\)-cost once. The two lemmas smuggled into the proof are worth remembering in their own right. The Boundedness Lemma (Step 3) is what makes the product rule work and is the prototype of “convergent implies controlled” arguments; the Separation Lemma (Step 7) — a convergent sequence with nonzero limit stays uniformly away from zero eventually — is the prototype of “open conditions survive limits locally” and reappears when proving that the set of invertible elements is open, that a continuous function nonzero at a point is nonzero on a neighbourhood, and in every quotient-rule proof in the subject.

The proof analysis pays off at higher altitude. Parts (i) and (ii) use only the norm axioms, submultiplicativity and boundedness, so they hold in any normed algebra: matrix sequences, bounded operators, function algebras. Where the hypotheses genuinely bite is instructive: on \(B(H)\) with the strong operator topology, multiplication is jointly continuous on bounded sets but not globally — exactly the Boundedness Lemma failing, since strongly convergent nets need not be bounded in the absence of Banach–Steinhaus-type control. For (iii), in a Banach algebra the invertibles form an open set on which inversion is continuous (via the Neumann series \((1 - x)^{-1} = \sum_{k=0}^{\infty} x^k\) for \(\|x\| \lt 1\)), which is the Separation Lemma’s abstract descendant. Finally, note what the proof never uses: completeness or order-completeness of \(\mathbb{R}\). The algebra of limits is true over \(\mathbb{Q}\); it is a statement about the interaction of the metric with the field structure, not about the existence of suprema.

Common misconceptions. The theorem propagates limits; it never creates them. Convergence of \(a_n + b_n\) tells you nothing about \(a_n\) alone (see the converse counterexample above), and \(L = f(L)\) arguments are worthless until convergence is established by other means (monotone convergence, contraction, etc.). Also, “\(b_n \neq 0\) for all \(n\)” is neither necessary nor sufficient for the quotient rule — the operative hypothesis is \(\lim b_n \neq 0\).

Worked examples

Example 1. Determine \(\displaystyle \lim_{n \to \infty} \frac{3n^2 - 5n + 7}{2n^2 + n + 1}\).

\[ \frac{1}{n} \to 0. \]

Standard basic limit: for \(\varepsilon \gt 0\), the Archimedean property of \(\mathbb{R}\) gives \(N \gt 1/\varepsilon\), and then \(n \ge N \implies |1/n - 0| = 1/n \le 1/N \lt \varepsilon\). A

\[ \frac{1}{n^2} = \frac{1}{n} \cdot \frac{1}{n} \to 0 \cdot 0 = 0. \]

Product rule (ii) applied to the sequence \(1/n\) with itself. A

For every \(n \ge 1\) (so \(n^2 \neq 0\)), divide numerator and denominator by \(n^2\): \[ \frac{3n^2 - 5n + 7}{2n^2 + n + 1} = \frac{3 - \frac{5}{n} + \frac{7}{n^2}}{2 + \frac{1}{n} + \frac{1}{n^2}}. \]

Exact algebraic identity, valid term by term; no limits taken yet. This rewriting is needed because numerator and denominator separately diverge, so the quotient rule cannot be applied to the original expression. B

\[ 3 - \frac{5}{n} + \frac{7}{n^2} \to 3 - 5 \cdot 0 + 7 \cdot 0 = 3, \qquad 2 + \frac{1}{n} + \frac{1}{n^2} \to 2 + 0 + 0 = 2. \]

Constant sequences converge to their value; then constant-multiple rule (product with a constant) and sum rule (i), applied a fixed finite number of times, using Steps 1–2. A

The denominator’s limit is \(2 \neq 0\), so by the quotient rule (iii): \[ \frac{3n^2 - 5n + 7}{2n^2 + n + 1} \to \frac{3}{2}. \]

Quotient rule (iii); its hypothesis \(\lim(\text{denominator}) \neq 0\) has been verified, not assumed. A

\[ \lim_{n \to \infty} \frac{3n^2 - 5n + 7}{2n^2 + n + 1} = \frac{3}{2}. \]

Reading. Divide out the dominant power, then read the limit off the leading coefficients — every step licensed by the algebra of limits.

Scope. The same routine computes the limit of any rational function of \(n\) whose numerator and denominator have equal degree; if the denominator dominates the limit is \(0\), and if the numerator dominates the sequence diverges (the theorem then shows a finite limit is impossible).

Example 2 (identifying a recursive limit). Define \(x_1 = 2\) and \(x_{n+1} = \dfrac{1}{2}\left(x_n + \dfrac{2}{x_n}\right)\) (the Babylonian iteration for \(\sqrt{2}\)). It is known that \(x_n \ge 1\) for all \(n\) and that \((x_n)\) converges to some \(L \ge 1\) — this follows from the Monotone Convergence Theorem after checking \(x_n^2 \ge 2\) and \(x_{n+1} \le x_n\), and we take it as given here. Find \(L\).

\[ x_{n+1} \to L. \]

The shifted sequence \((x_{n+1})_{n \ge 1}\) is a tail of \((x_n)\); by the Tail Lemma (cited in Step 10 of the main proof) it converges to the same limit \(L\). A

\[ \frac{2}{x_n} \to \frac{2}{L}. \]

Quotient rule (iii) applied to the constant sequence \(2\) and to \((x_n)\): the hypothesis \(L \ge 1 \gt 0\) gives \(L \neq 0\), and \(x_n \ge 1\) guarantees no term is zero, so the quotient is defined for every \(n\). A

\[ \frac{1}{2}\left(x_n + \frac{2}{x_n}\right) \to \frac{1}{2}\left(L + \frac{2}{L}\right). \]

Sum rule (i) on \(x_n\) and \(2/x_n\), then constant-multiple rule with \(\lambda = 1/2\). A

Both sides of \(x_{n+1} = \frac{1}{2}\left(x_n + \frac{2}{x_n}\right)\) converge, and equal sequences have equal limits, so by uniqueness of limits \[ L = \frac{1}{2}\left(L + \frac{2}{L}\right) \implies 2L^2 = L^2 + 2 \implies L^2 = 2. \]

Uniqueness of limits (a convergent real sequence has exactly one limit; standard, proved from the Hausdorff separation of distinct reals), then multiply through by \(2L \neq 0\). This is the key move: the recursion becomes an equation for \(L\). B

Since every \(x_n \ge 1\), the limit satisfies \(L \ge 1 \gt 0\); of the two roots \(\pm\sqrt{2}\) of \(L^2 = 2\), only \(L = \sqrt{2}\) is admissible.

Order limit theorem: \(x_n \ge 1\) for all \(n\) and \(x_n \to L\) imply \(L \ge 1\) (weak inequalities pass to limits; standard result cited from the same unit). A

\[ x_n \to \sqrt{2}. \]

Reading. Once convergence is secured by other means, the algebra of limits turns the recursion into an ordinary equation for the limit.

Scope. Works for any recursion \(x_{n+1} = f(x_n)\) with \(f\) a rational function whose denominator stays away from zero along the sequence and at the limit; it never substitutes for a convergence proof.

Problems

Compute \(\displaystyle \lim_{n \to \infty} \frac{4n^3 - n + 2}{2n^3 + n^2}\), citing each rule of the theorem as you use it.
Solution
For \(n \ge 1\), divide numerator and denominator by \(n^3\): \[ \frac{4n^3 - n + 2}{2n^3 + n^2} = \frac{4 - \frac{1}{n^2} + \frac{2}{n^3}}{2 + \frac{1}{n}}. \] From \(1/n \to 0\): \(1/n^2 = (1/n)(1/n) \to 0\) and \(1/n^3 = (1/n^2)(1/n) \to 0\) by the product rule. By the constant-multiple and sum rules, the numerator tends to \(4 - 0 + 2 \cdot 0 = 4\) and the denominator to \(2 + 0 = 2\). Since \(2 \neq 0\), the quotient rule gives \[ \lim_{n \to \infty} \frac{4n^3 - n + 2}{2n^3 + n^2} = \frac{4}{2} = 2. \]
Deduce the difference rule from the theorem: if \(a_n \to a\) and \(b_n \to b\) then \(a_n - b_n \to a - b\). You may use only parts (i) and (ii) and the fact that constant sequences converge to their value.
Solution
The constant sequence \(c_n = -1\) converges to \(-1\). By the product rule (ii), \(-b_n = c_n b_n \to (-1) \cdot b = -b\). By the sum rule (i), \(a_n - b_n = a_n + (-b_n) \to a + (-b) = a - b\). (This is why the theorem need only be stated for \(+\), \(\times\), \(\div\): subtraction is a derived operation.)
Give sequences \((a_n)\), \((b_n)\), both divergent, such that \(a_n + b_n\) and \(a_n b_n\) both converge. What does this say about the converse of the theorem?
Solution
Take \(a_n = (-1)^n\) and \(b_n = (-1)^{n+1} = -a_n\). Neither converges: each has subsequences with the distinct limits \(1\) and \(-1\), contradicting uniqueness of limits along subsequences. But \[ a_n + b_n = (-1)^n + (-1)^{n+1} = 0 \to 0, \qquad a_n b_n = (-1)^n(-1)^{n+1} = (-1)^{2n+1} = -1 \to -1. \] Hence the converse of the theorem is false in both the sum and product forms: convergence of the combination gives no information about the pieces. The implication runs one way only.
Suppose \(a_n \to a\) with \(a \neq 0\), and suppose \((a_n b_n)\) converges, say \(a_n b_n \to c\). Prove that \((b_n)\) converges and find its limit. Then show by example that the conclusion can fail when \(a = 0\).
Solution
By the Separation Lemma (Step 7 of the proof) applied to \((a_n)\), since \(a \neq 0\) there is \(N_0\) with \(|a_n| \gt |a|/2 \gt 0\), in particular \(a_n \neq 0\), for all \(n \ge N_0\). On this tail, \[ b_n = \frac{a_n b_n}{a_n}. \] The numerator sequence converges to \(c\) by hypothesis and the denominator to \(a \neq 0\), so the quotient rule (iii) gives \(b_n \to c/a\). By the Tail Lemma the whole sequence \((b_n)\) converges to \(c/a\). Failure for \(a = 0\): take \(a_n = 1/n \to 0\) and \(b_n = (-1)^n\). Then \(a_n b_n = (-1)^n / n \to 0\) (since \(|a_n b_n| = 1/n \to 0\)), so \((a_n b_n)\) converges, yet \((b_n)\) diverges.
(a) Prove by induction that if \(a_n \to a\) then \(a_n^k \to a^k\) for every fixed \(k \in \mathbb{N}\). (b) Deduce that \(p(a_n) \to p(a)\) for every polynomial \(p\) with real coefficients. (c) Explain precisely why this argument does not show \(\left(1 + \frac{1}{n}\right)^n \to 1\), even though \(1 + \frac{1}{n} \to 1\).
Solution
(a) Base case \(k = 1\): \(a_n^1 = a_n \to a = a^1\) by hypothesis. Inductive step: suppose \(a_n^k \to a^k\) for some fixed \(k \in \mathbb{N}\). Then by the product rule (ii) applied to the convergent sequences \((a_n^k)\) and \((a_n)\), \[ a_n^{k+1} = a_n^k \cdot a_n \to a^k \cdot a = a^{k+1}. \] By induction the claim holds for every fixed \(k \in \mathbb{N}\). (b) Let \(p(x) = c_0 + c_1 x + \cdots + c_d x^d\). Each term satisfies \(c_j a_n^j \to c_j a^j\) by part (a) and the constant-multiple rule; summing the \(d + 1\) terms with \(d\) applications of the sum rule (i) — a number of applications fixed by \(p\), independent of \(n\) — gives \(p(a_n) \to p(a)\). (c) In the expression \(\left(1 + \frac{1}{n}\right)^n\) the exponent is not fixed: the number of factors grows with \(n\). The induction in (a) proves, for each fixed \(k\), a statement about the sequence \(n \mapsto a_n^k\); it says nothing about the diagonal sequence \(n \mapsto a_n^{\,n}\), which involves a different power at each index. The theorem licenses only finitely many applications, fixed in advance. Indeed the conclusion would be false here: \(\left(1 + \frac{1}{n}\right)^n\) is increasing and bounded and converges to \(e = 2.718\ldots \neq 1\), exhibiting the indeterminate form \(1^{\infty}\).