CommeUnJeu · L1 MPSI
Real sequences
Real sequences are the bridge between the order on \(\mathbb{R}\) and the analysis of functions. The chapter has four jobs. First, formalise the lycée vocabulary --- bounded, monotone, limit --- into rigorous \(\varepsilon\)-\(N\) definitions. Second, prove the three foundational existence theorems --- monotone-limit theorem, adjacent sequences theorem, Bolzano-Weierstrass. These are exactly the results admitted in Limites et continuité, and they are the workhorses of every later analysis chapter. Third, equip the student with the « particular sequences » toolbox: arithmetic, geometric, arithmético-geometric, linear recurrences of order 2, and \(u_{n+1} = f(u_n)\). Fourth, briefly extend the theory to \(\mathbb{C}\)-valued sequences: limit, operations, and Bolzano-Weierstrass via the componentwise reduction.
\medskip
Convention. \((u_n)_{n \in \mathbb{N}}\) denotes a sequence indexed by \(\mathbb{N}\); statements that ignore the start index are stated « à partir d'un certain rang » (eventually). \(\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}\) is the extended real line. The supremum / infimum properties of \(\mathbb{R}\) are taken as known from Propriétés de \(\mathbb{R}\).
\medskip
Convention. \((u_n)_{n \in \mathbb{N}}\) denotes a sequence indexed by \(\mathbb{N}\); statements that ignore the start index are stated « à partir d'un certain rang » (eventually). \(\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}\) is the extended real line. The supremum / infimum properties of \(\mathbb{R}\) are taken as known from Propriétés de \(\mathbb{R}\).
I
Vocabulary of real sequences
A sequence is the simplest object in analysis: a function from \(\mathbb{N}\) to \(\mathbb{R}\). We re-cast the lycée vocabulary --- bounded, monotone, modes of definition --- in rigorous form, with two operational notions added : « à partir d'un certain rang » (eventually) and the implicit / recurrent / explicit trichotomy.
Definition — Real sequence
A real sequence is a map \(u : \mathbb{N} \to \mathbb{R}\), written \((u_n)_{n \in \mathbb{N}}\) or simply \((u_n)\). The set of real sequences is denoted \(\textcolor{colordef}{\mathbb{R}^{\mathbb{N}}}\). A sequence may also be defined from a rank \(n_0 \in \mathbb{N}\), written \((u_n)_{n \ge n_0}\). Definition — Bounded sequence
A sequence \((u_n)\) is: - majorée (bounded above) if \(\exists M \in \mathbb{R}, \forall n \in \mathbb{N}, u_n \le M\);
- minorée (bounded below) if \(\exists m \in \mathbb{R}, \forall n \in \mathbb{N}, u_n \ge m\);
- bornée if it is both majorée and minorée.
Proposition — Bounded iff \(|u_n|\) bounded above
A sequence \((u_n)\) is bounded if and only if \((|u_n|)_n\) is bounded above, i.e. \(\exists M \ge 0, \forall n \in \mathbb{N}, |u_n| \le M\).
If \(|u_n| \le M\), then \(-M \le u_n \le M\), so \((u_n)\) is bounded by \(m = -M\) below and \(M\) above. Conversely, if \(m \le u_n \le M\), then \(|u_n| \le \max(|m|, |M|)\).
Definition — Monotone sequence
A sequence \((u_n)\) is: - croissante (increasing) if \(\forall n \in \mathbb{N}, u_{n+1} \ge u_n\);
- strictement croissante if \(\forall n \in \mathbb{N}, u_{n+1} > u_n\);
- décroissante / strictement décroissante symmetrically;
- monotone if it is increasing or decreasing.
Definition — Eventually
A property \(P(n)\) holds à partir d'un certain rang (eventually) if \(\exists N \in \mathbb{N}, \forall n \ge N, P(n)\). The vocabulary applies: « \((u_n)\) croissante à partir d'un certain rang » means \(\exists N, \forall n \ge N, u_{n+1} \ge u_n\). Similarly for bornée, monotone, etc. Definition — Modes of definition
A real sequence is typically defined in one of three ways: - Explicit: \(u_n = f(n)\) for some function \(f\) defined on \(\mathbb{N}\).
- Implicit: \(u_n\) is characterised by a property (e.g. \(u_n\) is the unique solution of an equation in a given interval depending on \(n\)).
- Recurrent: \(u_0\) given, \(u_{n+1} = f(u_n)\) for some \(f\). The systematic study of this case is in the section Particular sequences below.
Example
Explicit example. The sequence \((u_n)_{n \ge 1}\) defined by \(u_n = 1/n\) is bounded by \(0\) and \(1\), and is strictly decreasing since \(u_{n+1} - u_n = 1/(n+1) - 1/n = -1/(n(n+1)) < 0\). Example
Recurrent example. Let \(u_0 = 1\) and \(u_{n+1} = u_n + 1/(n+1)\) for \(n \ge 0\). Show that \(u_n > 2\) from a certain rank, and study the monotonicity.
We compute: \(u_1 = 1 + 1 = 2\), \(u_2 = 2 + 1/2 = 5/2 > 2\). Since \(u_{n+1} - u_n = 1/(n+1) > 0\), \((u_n)\) is strictly increasing. Hence for \(n \ge 2\), \(u_n \ge u_2 = 5/2 > 2\), so \(u_n > 2\) from rank \(N = 2\). The sequence is strictly increasing on \(\mathbb{N}\).
Example
Implicit example. For \(n \ge 1\), let \(u_n\) be the unique \(x \in \,]0, +\infty[\) such that \(x + \ln x = n\) (the function \(x \mapsto x + \ln x\) is strictly increasing on \(\,]0, +\infty[\) with limits \(-\infty\) at \(0^+\) and \(+\infty\) at \(+\infty\)). Show that \((u_n)\) is well-defined, increasing, and tends to \(+\infty\).
Existence and uniqueness. The function \(g(x) = x + \ln x\) is continuous and strictly increasing on \(\,]0, +\infty[\), with \(\lim_{x \to 0^+} g = -\infty\) and \(\lim_{x \to +\infty} g = +\infty\). By the théorème de la bijection (Limites et continuité T7.2), \(g\) is a bijection from \(\,]0, +\infty[\) onto \(\mathbb{R}\), so for each \(n \in \mathbb{N}^*\) there is a unique \(u_n \in \,]0, +\infty[\) with \(g(u_n) = n\).
Monotonicity. For \(n \ge 1\), \(g(u_n) = n < n + 1 = g(u_{n+1})\). Since \(g\) is strictly increasing, \(u_n < u_{n+1}\), so \((u_n)\) is strictly increasing.
Limit. If \((u_n)\) were bounded above by some \(M\), then \(g(u_n) \le g(M) < +\infty\) for all \(n\), contradicting \(g(u_n) = n \to +\infty\). Hence \((u_n)\) is unbounded above; combined with strict monotonicity, \(u_n \to +\infty\).
Monotonicity. For \(n \ge 1\), \(g(u_n) = n < n + 1 = g(u_{n+1})\). Since \(g\) is strictly increasing, \(u_n < u_{n+1}\), so \((u_n)\) is strictly increasing.
Limit. If \((u_n)\) were bounded above by some \(M\), then \(g(u_n) \le g(M) < +\infty\) for all \(n\), contradicting \(g(u_n) = n \to +\infty\). Hence \((u_n)\) is unbounded above; combined with strict monotonicity, \(u_n \to +\infty\).
Example
The sequence \(u_n = (-1)^n\) is bounded (by \(-1\) and \(1\)) but not monotone --- not even monotone à partir d'un certain rang, since \(u_{n+1} - u_n = (-1)^{n+1} - (-1)^n = 2 (-1)^{n+1}\) alternates in sign for every \(n\). Example
Two contrasting sequence patterns: monotone bounded vs bounded oscillating. 
Skills to practice
- Establishing boundedness and monotonicity
- Reasoning with eventual properties
II
Limit of a real sequence: \(\varepsilon\)-\(N\) definitions
We give the rigorous \(\varepsilon\)-\(N\) definition of « \(u_n \to \ell\) ». The lycée intuition (« \(u_n\) approaches \(\ell\) as \(n\) grows large ») becomes : for every prescribed closeness to \(\ell\), the sequence eventually stays that close. The four cases (finite or infinite limit) are unified under a single template.
Definition — Finite limit
A sequence \((u_n)\) converges to \(\ell \in \mathbb{R}\) if $$ \forall \varepsilon > 0, \ \exists N \in \mathbb{N}, \ \forall n \ge N, \ |u_n - \ell| \le \varepsilon. $$ We then write \(\textcolor{colordef}{u_n \xrightarrow[n \to +\infty]{} \ell}\) or \(\textcolor{colordef}{\lim_{n \to +\infty} u_n = \ell}\). Definition — Infinite limit
A sequence \((u_n)\) tends to \(+\infty\) if $$ \forall A \in \mathbb{R}, \ \exists N \in \mathbb{N}, \ \forall n \ge N, \ u_n \ge A. $$ Symmetrically, \((u_n)\) tends to \(-\infty\) if \(\forall A \in \mathbb{R}, \exists N \in \mathbb{N}, \forall n \ge N, u_n \le A\). Definition — Convergent / divergent
\((u_n)\) is convergente if it admits a finite limit. Otherwise it is divergente. A sequence tending to \(+\infty\) or \(-\infty\) is divergent in this sense (« diverge vers \(+\infty\) »). Proposition — Uniqueness of the limit
If \((u_n)\) admits a limit \(\ell \in \overline{\mathbb{R}}\), that limit is unique.
Suppose \((u_n)\) admits two limits \(\ell, \ell' \in \overline{\mathbb{R}}\) with \(\ell \ne \ell'\).
- Both finite. Set \(\varepsilon = |\ell - \ell'|/3 > 0\). By the two limit definitions, \(\exists N_1, N_2\) such that for \(n \ge \max(N_1, N_2)\), \(|u_n - \ell| \le \varepsilon\) and \(|u_n - \ell'| \le \varepsilon\). Triangle inequality: \(|\ell - \ell'| \le |u_n - \ell| + |u_n - \ell'| \le 2 \varepsilon = (2/3)|\ell - \ell'|\), contradicting \(\ell \ne \ell'\).
- One finite, one infinite. Suppose \(\ell = +\infty\) and \(\ell' \in \mathbb{R}\). Apply D2.2 with \(A = \ell' + 1\): \(\exists N_1, \forall n \ge N_1, u_n \ge \ell' + 1\). Apply D2.1 with \(\varepsilon = 1/2\): \(\exists N_2, \forall n \ge N_2, |u_n - \ell'| \le 1/2\), hence \(u_n \le \ell' + 1/2\). For \(n \ge \max(N_1, N_2)\): \(\ell' + 1 \le u_n \le \ell' + 1/2\), contradiction. The case \(\ell = -\infty\) is symmetric.
- Both infinite, opposite signs. Suppose \(\ell = +\infty\) and \(\ell' = -\infty\). Apply D2.2 with \(A = 1\) and the symmetric definition with \(A = -1\): for \(n\) large enough, \(u_n \ge 1\) and \(u_n \le -1\), impossible.
Proposition — Convergent \(\Rightarrow\) bounded
Every convergent sequence is bounded.
Let \(u_n \to \ell\) with \(\ell \in \mathbb{R}\). Apply D2.1 with \(\varepsilon = 1\): \(\exists N \in \mathbb{N}, \forall n \ge N, |u_n - \ell| \le 1\), hence \(|u_n| \le |\ell| + 1\) by the triangle inequality. Up to increasing \(N\), assume \(N \ge 1\). The finite tail \(\{u_0, u_1, \dots, u_{N-1}\}\) is then non-empty and bounded by \(M_0 = \max(|u_0|, |u_1|, \dots, |u_{N-1}|)\). Set \(M = \max(M_0, |\ell| + 1)\). Then \(|u_n| \le M\) for all \(n \in \mathbb{N}\), so \((u_n)\) is bounded.
Example
Show by the \(\varepsilon\)-\(N\) definition that \(\lim_{n \to +\infty} 1/n = 0\).
Fix \(\varepsilon > 0\). Choose \(N = \lceil 1/\varepsilon \rceil + 1\). For \(n \ge N\), \(1/n \le 1/N \le \varepsilon\), hence \(|1/n - 0| = 1/n \le \varepsilon\). By D2.1, \(1/n \to 0\).
Example
Show by the definition that \(n^2 \to +\infty\).
Fix \(A \in \mathbb{R}\). If \(A \le 0\), any \(N\) works since \(n^2 \ge 0 \ge A\). If \(A > 0\), choose \(N = \lceil \sqrt{A} \rceil + 1\). For \(n \ge N\), \(n^2 \ge N^2 \ge A\). By D2.2, \(n^2 \to +\infty\).
Example
Show that \(((-1)^n)_n\) is bounded but admits no limit (preview of the two-subsequences method, fully developed in the section Subsequences and Bolzano-Weierstrass below).
Bounded. \(|(-1)^n| = 1\) for all \(n\), so the sequence is bounded by \(M = 1\).
No limit. Since \(((-1)^n)\) is bounded (\(|u_n| = 1\)), it cannot tend to \(+\infty\) or \(-\infty\) (bounded sequences have no infinite limit, by negation of D2.2). It remains to exclude a finite limit. Suppose by contradiction \(u_n \to \ell \in \mathbb{R}\). Apply D2.1 with \(\varepsilon = 1/2\): \(\exists N, \forall n \ge N, |(-1)^n - \ell| \le 1/2\). Pick \(n_1 \ge N\) even and \(n_2 \ge N\) odd: \(u_{n_1} = 1\) and \(u_{n_2} = -1\). Then \(|1 - \ell| \le 1/2\) and \(|-1 - \ell| \le 1/2\), hence by triangle inequality \(|1 - (-1)| = 2 \le 1\), contradiction.
No limit. Since \(((-1)^n)\) is bounded (\(|u_n| = 1\)), it cannot tend to \(+\infty\) or \(-\infty\) (bounded sequences have no infinite limit, by negation of D2.2). It remains to exclude a finite limit. Suppose by contradiction \(u_n \to \ell \in \mathbb{R}\). Apply D2.1 with \(\varepsilon = 1/2\): \(\exists N, \forall n \ge N, |(-1)^n - \ell| \le 1/2\). Pick \(n_1 \ge N\) even and \(n_2 \ge N\) odd: \(u_{n_1} = 1\) and \(u_{n_2} = -1\). Then \(|1 - \ell| \le 1/2\) and \(|-1 - \ell| \le 1/2\), hence by triangle inequality \(|1 - (-1)| = 2 \le 1\), contradiction.
Example
The converse of P2.2 is false: \(((-1)^n)_n\) is bounded (Ex2.3 above) but not convergent. Method — Showing convergence by the \(\varepsilon\)-\(N\) definition
To prove \(u_n \to \ell\) from the definition: - Fix \(\varepsilon > 0\). Treat it as a generic positive number; the choice of \(N\) will depend on it.
- Bound \(|u_n - \ell|\) by an explicit decreasing function of \(n\). Use algebra, the triangle inequality, or known bounds.
- Choose \(N\). Pick the smallest \(N \in \mathbb{N}\) such that the bound is \(\le \varepsilon\) for all \(n \ge N\).
Skills to practice
- Proving convergence by the \(\varepsilon\)-\(N\) definition
- Negating convergence
III
Operations on limits
Sums, products, quotients of sequences with limits behave by rules transferred from the order on \(\mathbb{R}\). The four classical indeterminate forms (\(\infty - \infty\), \(0 \cdot \infty\), \(\infty / \infty\), \(0/0\)) require ad-hoc analysis. Composition with a continuous function is mentioned as a forward reference (sequential characterization of continuity, corollary of Heine's theorem, both proved in the chapter Limites et continuité).
Proposition — Algebraic operations on finite limits
If \(u_n \to \ell\) and \(v_n \to m\) with \(\ell, m \in \mathbb{R}\), then for any \(\lambda, \mu \in \mathbb{R}\): - \(\textcolor{colorprop}{\lambda u_n + \mu v_n \to \lambda \ell + \mu m}\) (linear combination);
- \(\textcolor{colorprop}{u_n v_n \to \ell m}\) (product);
- if \(m \ne 0\), \(v_n \ne 0\) from a certain rank (by the \(\varepsilon = |m|/2\) argument) and \(\textcolor{colorprop}{u_n / v_n \to \ell / m}\) (quotient).
- Linear combination. Fix \(\varepsilon > 0\) and set \(\varepsilon' = \varepsilon / (|\lambda| + |\mu| + 1) > 0\). By the convergence \(u_n \to \ell\) applied to \(\varepsilon'\), \(\exists N_1, \forall n \ge N_1, |u_n - \ell| \le \varepsilon'\); similarly \(\exists N_2, \forall n \ge N_2, |v_n - m| \le \varepsilon'\). For \(n \ge \max(N_1, N_2)\), the triangle inequality gives $$ |(\lambda u_n + \mu v_n) - (\lambda \ell + \mu m)| \le |\lambda| \cdot |u_n - \ell| + |\mu| \cdot |v_n - m| \le (|\lambda| + |\mu|) \varepsilon' \le \varepsilon. $$
- Product. Write \(u_n v_n - \ell m = u_n (v_n - m) + m (u_n - \ell)\). By P2.2, \((u_n)\) is bounded by some \(M\). Then \(|u_n v_n - \ell m| \le M |v_n - m| + |m| \cdot |u_n - \ell|\), and both terms tend to \(0\) by the linear-combination case applied to constants.
- Quotient. Apply the limit definition to \(v_n \to m\) with \(\varepsilon = |m|/2\): \(\exists N_0, \forall n \ge N_0, |v_n - m| \le |m|/2\). Hence \(|v_n| \ge |m|/2 > 0\), so \(v_n \ne 0\). Then \(1/v_n - 1/m = (m - v_n)/(m v_n)\), with \(|m v_n| \ge |m|^2 / 2\), gives \(|1/v_n - 1/m| \le 2 |v_n - m|/|m|^2 \to 0\). Conclude by the product rule applied to \(u_n \cdot (1/v_n)\).
Method — Operations with infinite limits --- rules table
The operation rules of P3.1 extend to limits in \(\overline{\mathbb{R}}\) whenever the resulting expression has a determined meaning. Determined cases: - Sums. \((+\infty) + (+\infty) = +\infty\) ; \((+\infty) + \ell = +\infty\) for \(\ell \in \mathbb{R}\) ; symmetric for \(-\infty\).
- Products. \(\ell \cdot (+\infty) = +\infty\) if \(\ell > 0\), \(-\infty\) if \(\ell < 0\) ; \((+\infty) \cdot (+\infty) = +\infty\) ; \((+\infty) \cdot (-\infty) = -\infty\) ; sign rules for the rest.
- Quotients. \(1/(+\infty) = 0\) ; \(\ell / (+\infty) = 0\) for \(\ell \in \mathbb{R}\) ; \((+\infty)/\ell = +\infty\) if \(\ell > 0\), \(-\infty\) if \(\ell < 0\).
- Inverses of sequences tending to zero. If \(v_n \to 0\) and \(v_n > 0\) from some rank (notation: \(v_n \to 0^+\)), then \(1/v_n \to +\infty\). Symmetrically, if \(v_n \to 0^-\), then \(1/v_n \to -\infty\). If \(v_n \to 0\) without a fixed sign, \(1/v_n\) has no determined limit.
Remark --- composition with a continuous function (forward reference)
If \(u_n \to a \in \mathbb{R}\) and \(f\) is continuous at \(a\), then \(f(u_n) \to f(a)\). This is the sequential characterization of continuity (treated in the chapter Limites et continuité, as a corollary of Heine's theorem). We use it later in the section Particular sequences for the recurrent case \(u_{n+1} = f(u_n)\).
Example
Compute \(\lim (3 n^2 + 2 n - 1)/(n^2 + 5)\) by factoring (form \(\infty / \infty\)).
Factor \(n^2\) from numerator and denominator: $$ \frac{3 n^2 + 2 n - 1}{n^2 + 5} = \frac{n^2 (3 + 2/n - 1/n^2)}{n^2 (1 + 5/n^2)} = \frac{3 + 2/n - 1/n^2}{1 + 5/n^2}. $$ As \(n \to +\infty\), \(1/n \to 0\) and \(1/n^2 \to 0\). By P3.1, the numerator tends to \(3\) and the denominator to \(1\), hence the quotient to \(3/1 = 3\).
Example
Compute \(\lim (2 + 1/n)(3 - 1/n^2)\) by direct application of P3.1.
\(1/n \to 0\) and \(1/n^2 \to 0\). By P3.1 (linear combination), \(2 + 1/n \to 2\) and \(3 - 1/n^2 \to 3\). By P3.1 (product), \((2 + 1/n)(3 - 1/n^2) \to 2 \cdot 3 = 6\).
Example
Compute \(\lim (\sqrt{n+1} - \sqrt{n})\) by conjugate (form \(\infty - \infty\)).
Multiply by the conjugate \(\sqrt{n+1} + \sqrt{n}\): $$ \sqrt{n+1} - \sqrt{n} = \frac{(\sqrt{n+1})^2 - (\sqrt{n})^2}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}}. $$ The denominator \(\sqrt{n+1} + \sqrt{n} \to +\infty\), so the quotient tends to \(0\) by the rules table above (\(1/(+\infty) = 0\)).
Method — Lifting an indeterminate form for sequences
When the algebra of P3.1 and the rules table for infinite limits yields one of the four indeterminate forms (\(\infty - \infty\), \(0 \cdot \infty\), \(\infty / \infty\), \(0/0\)), proceed in three steps: - Identify the dominant term. The largest power of \(n\) (or the slowest-going-to-zero factor in \(0/0\)).
- Factor it out. Of numerator and denominator, or of each summand for \(\infty - \infty\).
- Take the limit. Of each remaining factor by P3.1 or the infinite-limit rules table; the form is now determinate.
Skills to practice
- Computing limits with the operation rules
- Resolving indeterminate forms
IV
Limits and ordering
Three results connecting limits and inequalities: passage to the limit (a large inequality is preserved), the squeeze theorem (encadrement), and the minoration / majoration trick for infinite limits. Together they form the « comparison toolbox » that justifies most explicit limit calculations.
Proposition — Passage to the limit of a large inequality
If \(u_n \le v_n\) from a certain rank and both sequences admit limits in \(\overline{\mathbb{R}}\), then \(\textcolor{colorprop}{\lim u_n \le \lim v_n}\). The strict inequality is not preserved: \(1/(n+1) < 2/(n+1)\) on \(\mathbb{N}\) but both tend to \(0\).
We prove the case where both limits are finite: \(u_n \to \ell\), \(v_n \to m\). Suppose by contradiction \(\ell > m\). Set \(\varepsilon = (\ell - m)/3 > 0\). By convergence, \(\exists N_1, \forall n \ge N_1, |u_n - \ell| \le \varepsilon\), hence \(u_n \ge \ell - \varepsilon\). Similarly \(\exists N_2, \forall n \ge N_2, v_n \le m + \varepsilon\). For \(n \ge \max(N_1, N_2)\) also above the hypothesis rank, \(u_n \le v_n \le m + \varepsilon < \ell - \varepsilon \le u_n\) (since \(m + \varepsilon = m + (\ell-m)/3\) and \(\ell - \varepsilon = \ell - (\ell-m)/3\), so \(\ell - \varepsilon - (m + \varepsilon) = (\ell - m) - 2 \varepsilon = (\ell-m)/3 > 0\)). Contradiction. The infinite-limit cases follow similarly (e.g. \(\ell = +\infty\) forces \(v_n \to +\infty\) by minoration; \(m = -\infty\) forces \(u_n \to -\infty\) by majoration; mixed cases reduce by inspection).
Theorem — Squeeze theorem (encadrement)
If \(u_n \le v_n \le w_n\) from a certain rank, \(u_n \to \ell\) and \(w_n \to \ell\) with \(\ell \in \mathbb{R}\), then \(\textcolor{colorprop}{v_n \to \ell}\).
Fix \(\varepsilon > 0\). By the limit definitions, \(\exists N_1, N_2\) such that for \(n \ge \max(N_1, N_2)\), \(|u_n - \ell| \le \varepsilon\) and \(|w_n - \ell| \le \varepsilon\), i.e. \(\ell - \varepsilon \le u_n\) and \(w_n \le \ell + \varepsilon\). Let \(N_3\) be the rank from which \(u_n \le v_n \le w_n\). For \(n \ge \max(N_1, N_2, N_3)\): $$ \ell - \varepsilon \le u_n \le v_n \le w_n \le \ell + \varepsilon, $$ so \(|v_n - \ell| \le \varepsilon\). Hence \(v_n \to \ell\).
Proposition — Minoration / majoration with infinite limits
- If \(u_n \le v_n\) from a certain rank and \(u_n \to +\infty\), then \(\textcolor{colorprop}{v_n \to +\infty}\).
- If \(v_n \le u_n\) from a certain rank and \(u_n \to -\infty\), then \(\textcolor{colorprop}{v_n \to -\infty}\).
First case. Fix \(A \in \mathbb{R}\). Since \(u_n \to +\infty\), \(\exists N_1, \forall n \ge N_1, u_n \ge A\). Let \(N_2\) be a rank from which \(u_n \le v_n\). For \(n \ge \max(N_1, N_2)\), \(v_n \ge u_n \ge A\). Hence \(v_n \to +\infty\). The second case follows by replacing \(u, v\) with \(-u, -v\).
Proposition — Sandwiched at zero
If \(|u_n - \ell| \le \alpha_n\) from a certain rank and \(\alpha_n \to 0\), then \(\textcolor{colorprop}{u_n \to \ell}\).
Fix \(\varepsilon > 0\). Since \(\alpha_n \to 0\), \(\exists N_1, \forall n \ge N_1, |\alpha_n| \le \varepsilon\), hence \(\alpha_n \le \varepsilon\) (as \(\alpha_n \ge |u_n - \ell| \ge 0\)). For \(n\) above this rank and the hypothesis rank, \(|u_n - \ell| \le \alpha_n \le \varepsilon\). So \(u_n \to \ell\).
Example
Compute \(\lim (\sin n)/n\) by squeeze.
For \(n \ge 1\), \(|\sin n| \le 1\), so \(|(\sin n)/n| \le 1/n\), i.e. \(-1/n \le (\sin n)/n \le 1/n\). Both bounds tend to \(0\). By T4.1 (squeeze), \((\sin n)/n \to 0\).
Example
Compute \(\lim (\cos(n^2) + n)\) by minoration.
For all \(n\), \(\cos(n^2) \ge -1\), so \(\cos(n^2) + n \ge n - 1\). The lower bound \(n - 1 \to +\infty\) (P3.1). By P4.2 (minoration), \(\cos(n^2) + n \to +\infty\).
Example
Compute \(\lim n \sin(1/n^2)\) by squeeze, using the elementary bound \(|\sin x| \le |x|\).
For \(n \ge 1\), \(|\sin(1/n^2)| \le 1/n^2\), hence $$ |n \sin(1/n^2)| \le n \cdot \frac{1}{n^2} = \frac{1}{n} \xrightarrow[n \to +\infty]{} 0. $$ By P4.3 (sandwiched at zero), \(n \sin(1/n^2) \to 0\).
Method — Bounding by a known-zero sequence
To show \(u_n \to \ell\), it is often enough to find a sequence \(\alpha_n \to 0\) with \(|u_n - \ell| \le \alpha_n\) from a certain rank. This pattern unifies many seemingly different limit calculations: it is the engine behind « squeeze », « comparison with \(1/n\), \(1/n^2, q^n\) (\(|q| < 1\)), \(1/n!\) », etc. Skills to practice
- Using the squeeze theorem
- Passage to the limit and minoration
V
Existence theorems: monotone-limit and adjacent sequences
Two theorems guaranteeing convergence without computing the limit explicitly: (i) the monotone-limit theorem, proved using the supremum / infimum property of \(\mathbb{R}\) from Propriétés de \(\mathbb{R}\); (ii) the adjacent sequences theorem, proved by reduction to (i). Three debts to the chapter Limites et continuité are paid here and in the section Subsequences and Bolzano-Weierstrass below: the monotone-limit theorem for functions (transferred from T5.1 below), the TVI dichotomy proof (transferred from the adjacent sequences theorem T5.2 below), and the « bornes atteintes » result (transferred from Bolzano-Weierstrass below).
Definition — Adjacent sequences
Two sequences \((u_n), (v_n)\) are adjacent if: - \((u_n)\) is increasing;
- \((v_n)\) is decreasing;
- \(u_n \le v_n\) for all \(n\);
- \(v_n - u_n \to 0\).
Example
Let \(u_n = 1 - 1/(n+1)\) and \(v_n = 1 + 1/(n+1)\) for \(n \ge 0\). Then \(u\) is increasing, \(v\) is decreasing, \(u_n \le 1 \le v_n\) (so \(u_n \le v_n\)), and \(v_n - u_n = 2/(n+1) \to 0\). Hence \((u_n)\) and \((v_n)\) are adjacent; both converge to \(1\). Theorem — Théorème de la limite monotone
Let \((u_n)\) be a real sequence. - If \((u_n)\) is increasing and bounded above, then \(\textcolor{colorprop}{u_n \to \sup_n u_n \in \mathbb{R}}\).
- If \((u_n)\) is increasing and unbounded above, then \(\textcolor{colorprop}{u_n \to +\infty}\).
- Bounded case. Let \(M = \sup_n u_n \in \mathbb{R}\) (exists by the supremum property of \(\mathbb{R}\), Propriétés de \(\mathbb{R}\)). Fix \(\varepsilon > 0\). By the \(\varepsilon\)-characterization of the supremum, \(\exists N \in \mathbb{N}, M - \varepsilon \le u_N \le M\). By monotonicity, for \(n \ge N\), \(u_N \le u_n \le M\), hence \(M - \varepsilon \le u_n \le M\), so \(|u_n - M| \le \varepsilon\). By D2.1, \(u_n \to M\).
- Unbounded case. Fix \(A \in \mathbb{R}\). Since \((u_n)\) is unbounded above, \(\exists N \in \mathbb{N}, u_N \ge A\). By monotonicity, for \(n \ge N\), \(u_n \ge u_N \ge A\). By D2.2, \(u_n \to +\infty\).
Proposition — Corollary --- monotone bounded converges
If \((u_n)\) is monotone and bounded, then \((u_n)\) is convergent. Theorem — Adjacent sequences theorem
If \((u_n), (v_n)\) are adjacent (D5.1), then both converge to a common limit \(\textcolor{colorprop}{\ell \in \mathbb{R}}\) with \(u_n \le \ell \le v_n\) for all \(n\).
\((u_n)\) is increasing and bounded above by \(v_0\) (since \(u_n \le v_n \le v_0\) by monotonicity of \(v\)). By T5.1, \(u_n \to \ell\) for some \(\ell \in \mathbb{R}\). Symmetrically, \((v_n)\) is decreasing and bounded below by \(u_0\), so \(v_n \to \ell'\) for some \(\ell' \in \mathbb{R}\). Pass to the limit in \(v_n - u_n \to 0\): \(\ell' - \ell = 0\), hence \(\ell = \ell'\). The order \(u_n \le \ell \le v_n\) comes from passing to the limit in \(u_n \le u_p\) (\(p \ge n\), so \(u_n \le \lim_p u_p = \ell\)) and similarly \(\ell \le v_n\) from \(v_p \le v_n\) for \(p \ge n\).
Example
Show that \(u_n = \sum_{k=0}^n 1/k!\) converges (its limit is \(\mathrm{e}\), admitted forward-reference).
\((u_n)\) is increasing since \(u_{n+1} - u_n = 1/(n+1)! > 0\). We bound: $$ u_n = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{n!} \le 1 + 1 + \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{n-1}} \le 1 + 2 = 3, $$ using \(k! \ge 2^{k-1}\) for \(k \ge 1\) (induction). Hence \((u_n)\) is bounded above by \(3\). By T5.1 (bounded case), \((u_n)\) converges. (The limit is the Néper constant \(\mathrm{e}\), see Standard functions.)
Example
Let \(u_0 = 0\) and \(u_{n+1} = (u_n + 2)/2\). Show that \((u_n)\) converges and compute its limit.
Bound. Show by induction that \(u_n \in [0, 2]\) for all \(n\). Base: \(u_0 = 0 \in [0, 2]\). Heredity: if \(u_n \in [0, 2]\), then \(u_{n+1} = (u_n + 2)/2 \in [1, 2] \subset [0, 2]\).
Monotonicity. \(u_{n+1} - u_n = (u_n + 2)/2 - u_n = (2 - u_n)/2 \ge 0\) since \(u_n \le 2\). Hence \((u_n)\) is increasing.
Convergence. \((u_n)\) is increasing and bounded above by \(2\). By T5.1 (bounded case), \(u_n \to \ell\) for some \(\ell \in [0, 2]\).
Limit value. Pass to the limit in \(u_{n+1} = (u_n + 2)/2\) using continuity of \(x \mapsto (x + 2)/2\) (by T7.1 in the section Particular sequences below): \(\ell = (\ell + 2)/2\), hence \(\ell = 2\).
Monotonicity. \(u_{n+1} - u_n = (u_n + 2)/2 - u_n = (2 - u_n)/2 \ge 0\) since \(u_n \le 2\). Hence \((u_n)\) is increasing.
Convergence. \((u_n)\) is increasing and bounded above by \(2\). By T5.1 (bounded case), \(u_n \to \ell\) for some \(\ell \in [0, 2]\).
Limit value. Pass to the limit in \(u_{n+1} = (u_n + 2)/2\) using continuity of \(x \mapsto (x + 2)/2\) (by T7.1 in the section Particular sequences below): \(\ell = (\ell + 2)/2\), hence \(\ell = 2\).
Example
Show that the harmonic sequence \(u_n = \sum_{k=1}^n 1/k\) tends to \(+\infty\).
\((u_n)\) is increasing (terms positive). Show it is unbounded above via the lower bound \(u_{2^p} \ge 1 + p/2\) for \(p \ge 0\). Indeed: $$ u_{2^p} = 1 + \sum_{j=0}^{p-1} \sum_{k = 2^j + 1}^{2^{j+1}} \frac{1}{k} \ge 1 + \sum_{j=0}^{p-1} 2^j \cdot \frac{1}{2^{j+1}} = 1 + \sum_{j=0}^{p-1} \frac{1}{2} = 1 + \frac{p}{2}, $$ using \(1/k \ge 1/2^{j+1}\) for \(k \le 2^{j+1}\) on each block of \(2^j\) terms. As \(p \to +\infty\), the lower bound \(1 + p/2 \to +\infty\), so \((u_n)\) is unbounded above. By T5.1 (unbounded case), \(u_n \to +\infty\).
Method — Convergence by the monotone-limit theorem
To show \((u_n)\) converges (without computing the limit): - Show monotonicity. Sign of \(u_{n+1} - u_n\), or induction, or differential argument for \(u_n = f(n)\).
- Show boundedness (in the appropriate direction). Often by induction, or by an explicit majorisation.
- Conclude by T5.1 (bounded case) — the limit exists and equals \(\sup u_n\) or \(\inf u_n\).
Skills to practice
- Applying the monotone-limit theorem
- Adjacent sequences
VI
Subsequences and Bolzano-Weierstrass
A subsequence (suite extraite) is a sub-selection of indices that preserves order. Two key facts: (i) the limit transfers to subsequences (P6.2), so two subsequences with distinct limits prove non-convergence (P6.3); (ii) every bounded sequence admits a convergent subsequence — this is Bolzano-Weierstrass (T6.1), the chapter's centerpiece. This is the third debt owed to Limites et continuité.
Definition — Suite extraite
A suite extraite of \((u_n)\) is a sequence of the form \((u_{\varphi(n)})_n\) where \(\varphi : \mathbb{N} \to \mathbb{N}\) is strictly increasing (called an extraction). Example
With \(u_n = (-1)^n\) and \(\varphi(n) = 2 n\): the extracted sequence is \(u_{\varphi(n)} = (-1)^{2n} = 1\) (constant). With \(\psi(n) = 2 n + 1\): \(u_{\psi(n)} = (-1)^{2n+1} = -1\). The map \(n \mapsto n^2\) is also a valid extraction; since \(n^2\) has the same parity as \(n\), the extracted sequence \(u_{n^2} = (-1)^{n^2} = (-1)^n\) coincides with the original. Proposition — Extraction lower bound
If \(\varphi : \mathbb{N} \to \mathbb{N}\) is a strictly increasing extraction, then \(\varphi(n) \ge n\) for all \(n\).
By induction on \(n\). Base. \(\varphi(0) \ge 0\) trivially (values in \(\mathbb{N}\)). Heredity. If \(\varphi(n) \ge n\), then \(\varphi(n+1) > \varphi(n) \ge n\) (strict increase), hence \(\varphi(n+1) \ge n+1\) (since both sides are integers and \(\varphi(n+1) > n\)).
Proposition — Limit transfers to subsequences
If \(u_n \to \ell \in \overline{\mathbb{R}}\), then for every extraction \(\varphi\), \(\textcolor{colorprop}{u_{\varphi(n)} \to \ell}\).
Treat the finite case (\(\ell \in \mathbb{R}\)); the infinite cases are analogous. Fix \(\varepsilon > 0\). Since \(u_n \to \ell\), \(\exists N \in \mathbb{N}, \forall n \ge N, |u_n - \ell| \le \varepsilon\). By P6.1, \(\varphi(n) \ge n\), so for \(n \ge N\) we have \(\varphi(n) \ge N\), hence \(|u_{\varphi(n)} - \ell| \le \varepsilon\). The same \(N\) witnesses convergence of \((u_{\varphi(n)})\) to \(\ell\).
Proposition — Two subsequences with distinct limits
If \((u_{\varphi(n)})\) and \((u_{\psi(n)})\) admit distinct limits in \(\overline{\mathbb{R}}\), then \((u_n)\) admits no limit.
By contraposition. If \((u_n) \to \ell \in \overline{\mathbb{R}}\), then by P6.2 (limit transfer), both \((u_{\varphi(n)})\) and \((u_{\psi(n)})\) tend to \(\ell\), contradicting the hypothesis of distinct limits.
Proposition — Even/odd subsequences cover
If \((u_{2n})\) and \((u_{2n+1})\) both tend to the same \(\ell \in \overline{\mathbb{R}}\), then \(\textcolor{colorprop}{u_n \to \ell}\).
Finite case \(\ell \in \mathbb{R}\). Fix \(\varepsilon > 0\). By convergence of \((u_{2n})\), \(\exists N_0, \forall p \ge N_0, |u_{2p} - \ell| \le \varepsilon\). By convergence of \((u_{2n+1})\), \(\exists N_1, \forall p \ge N_1, |u_{2p+1} - \ell| \le \varepsilon\). Set \(N = \max(2 N_0, 2 N_1 + 1)\). For \(n \ge N\): if \(n = 2 p\) is even, then \(p \ge N_0\) so \(|u_n - \ell| \le \varepsilon\); if \(n = 2 p + 1\) is odd, then \(p \ge N_1\) so \(|u_n - \ell| \le \varepsilon\). Hence \(u_n \to \ell\). Infinite cases analogous.
Proposition — Shift corollary
If \(u_n \to \ell\), then for every fixed \(p \in \mathbb{N}\), \(\textcolor{colorprop}{u_{n+p} \to \ell}\).
Apply P6.2 to the extraction \(\varphi(n) = n + p\): \(\varphi\) is strictly increasing \(\mathbb{N} \to \mathbb{N}\), so \(u_{\varphi(n)} = u_{n+p} \to \ell\).
Proposition — Sequential characterization of adherence --- useful complement
Let \(A \subset \mathbb{R}\) and \(x \in \mathbb{R}\). Then \(x\) is adherent to \(A\) (every neighborhood of \(x\) meets \(A\)) if and only if there exists a sequence \(\textcolor{colorprop}{(a_n) \in A^{\mathbb{N}}}\) with \(a_n \to x\).
Forward. Suppose \(x\) adherent to \(A\). For each \(n \in \mathbb{N}\), the neighborhood \([x - 1/(n+1), x + 1/(n+1)]\) meets \(A\); pick \(a_n\) in the intersection. Then \(|a_n - x| \le 1/(n+1) \to 0\), so \(a_n \to x\).
Reverse. Suppose \((a_n) \in A^{\mathbb{N}}\) with \(a_n \to x\). Fix any neighborhood \(V\) of \(x\); \(\exists \varepsilon > 0\) with \([x - \varepsilon, x + \varepsilon] \subset V\). By \(a_n \to x\), \(\exists N\) with \(a_N \in [x - \varepsilon, x + \varepsilon] \subset V\). Since \(a_N \in A\), \(V\) meets \(A\).
Reverse. Suppose \((a_n) \in A^{\mathbb{N}}\) with \(a_n \to x\). Fix any neighborhood \(V\) of \(x\); \(\exists \varepsilon > 0\) with \([x - \varepsilon, x + \varepsilon] \subset V\). By \(a_n \to x\), \(\exists N\) with \(a_N \in [x - \varepsilon, x + \varepsilon] \subset V\). Since \(a_N \in A\), \(V\) meets \(A\).
Theorem — Bolzano-Weierstrass
Every bounded real sequence admits a convergent subsequence.
By dichotomy. Let \((u_n) \in [a_0, b_0]^{\mathbb{N}}\) with \(a_0 \le b_0\) in \(\mathbb{R}\). We build sequences \((a_k), (b_k)\) and an extraction \((\varphi(k))\) maintaining the invariant $$ (\mathcal{I}_k): \quad \text{infinitely many } n > \varphi(k) \text{ satisfy } u_n \in [a_k, b_k]. $$ Set \(\varphi(0) = 0\), \(a_0, b_0\) given; \((\mathcal{I}_0)\) holds since every \(n \in \mathbb{N}\) satisfies \(u_n \in [a_0, b_0]\). At step \(k\), assuming \((\mathcal{I}_k)\):
- Let \(m_k = (a_k + b_k)/2\). Look at \([a_k, m_k]\) and \([m_k, b_k]\).
- At least one of them, say \(I\), contains \(u_n\) for infinitely many \(n > \varphi(k)\) (otherwise both halves would contain only finitely many such \(n\), contradicting \((\mathcal{I}_k)\)).
- Set \([a_{k+1}, b_{k+1}] = I\) and choose \(\varphi(k+1) > \varphi(k)\) with \(u_{\varphi(k+1)} \in [a_{k+1}, b_{k+1}]\). Since \(I\) contains \(u_n\) for infinitely many \(n > \varphi(k+1)\) (still infinitely many after removing the finite set \(\{n \le \varphi(k+1)\}\)), \((\mathcal{I}_{k+1})\) holds and the induction continues.
Example
Show that \(((-1)^n)_n\) admits two convergent subsequences with distinct limits, hence is divergent.
Take \(\varphi(n) = 2n\) and \(\psi(n) = 2n + 1\). Then \(u_{\varphi(n)} = (-1)^{2n} = 1 \to 1\) and \(u_{\psi(n)} = (-1)^{2n+1} = -1 \to -1\). The two subsequences have distinct limits, so by P6.3, \(((-1)^n)\) admits no limit.
Example
Show that \(u_n = \sin(n \pi / 3)\) admits convergent subsequences and identify their limits.
The values cycle with period \(6\): \(\sin(n \pi / 3)\) takes the values \(0, \sqrt{3}/2, \sqrt{3}/2, 0, -\sqrt{3}/2, -\sqrt{3}/2\) for \(n = 0, 1, 2, 3, 4, 5\). The subsequence \(\varphi(n) = 6n\) gives \(u_{\varphi(n)} = 0\) constant, hence \(\to 0\). The subsequence \(\psi(n) = 6n + 1\) gives \(u_{\psi(n)} = \sqrt{3}/2\) constant, hence \(\to \sqrt{3}/2\). Similarly \(6n + 4\) gives constant \(-\sqrt{3}/2\). Three distinct constant subsequences, so \((u_n)\) is divergent (P6.3).
Example
Let \((u_n)\) be a bounded sequence such that every convergent subsequence has the same limit \(\ell\). Show that \(u_n \to \ell\).
By contradiction, suppose \(u_n \not\to \ell\). Negate D2.1: \(\exists \varepsilon_0 > 0, \forall N \in \mathbb{N}, \exists n \ge N, |u_n - \ell| > \varepsilon_0\). Build inductively a strict extraction \(\varphi\) with \(|u_{\varphi(n)} - \ell| > \varepsilon_0\) for all \(n\): take \(\varphi(0)\) as the first index \(\ge 0\) where the bound fails, then \(\varphi(n+1)\) as the first index \(> \varphi(n)\) where it fails. The subsequence \((u_{\varphi(n)})\) is bounded (subsequence of a bounded), so by Bolzano-Weierstrass T6.1, it admits a convergent sub-subsequence \(u_{\varphi(\psi(n))} \to \ell'\). By the hypothesis, \(\ell' = \ell\). But \(|u_{\varphi(\psi(n))} - \ell| > \varepsilon_0\) for all \(n\), so passing to the limit, \(|\ell' - \ell| \ge \varepsilon_0 > 0\), contradicting \(\ell' = \ell\). Hence \(u_n \to \ell\).
Method — Disproving convergence by two subsequences
To show \((u_n)\) has no limit, exhibit two strict extractions \(\varphi, \psi\) with \(u_{\varphi(n)}\) and \(u_{\psi(n)}\) tending to distinct elements of \(\overline{\mathbb{R}}\). By P6.3, \((u_n)\) admits no limit. Method — Extracting a convergent subsequence via Bolzano-Weierstrass
For a bounded sequence \((u_n)\) whose limit is unclear, T6.1 guarantees that some subsequence converges. Identify the limit by a separate argument (passing to the limit in a recurrence, using the hypothesis of the problem, etc.). Useful pattern: combine BW with a contradiction argument when the sequence is « hardly convergent ». Example
The dichotomy of Bolzano-Weierstrass. 
Skills to practice
- Building subsequences with distinct limits
- Using Bolzano-Weierstrass
VII
Particular sequences
Five recurring families: arithmetic, geometric, arithmético-geometric, linear recurrence of order 2 (with constant coefficients), and the iteration \(u_{n+1} = f(u_n)\). Each comes with a closed-form (when available) or a study technique (monotonicity, fixed point, dichotomy). The iteration \(u_{n+1} = f(u_n)\) with \(f\) continuous is the « bridge to functions » : the limit, if it exists, is a fixed point of \(f\).
Definition — Arithmetic sequence
\((u_n)\) is arithmetic of common difference \(r \in \mathbb{R}\) if \(u_{n+1} = u_n + r\) for all \(n\). Closed form: \(\textcolor{colordef}{u_n = u_0 + n r}\). Sum of the first \(n+1\) terms: \(\sum_{k=0}^n u_k = (n+1)(u_0 + u_n)/2\). Definition — Geometric sequence
\((u_n)\) is geometric of common ratio \(q \in \mathbb{R}\) if \(u_{n+1} = q \, u_n\) for all \(n\). Closed form: \(\textcolor{colordef}{u_n = u_0 q^n}\). Sum (for \(q \ne 1\)): \(\sum_{k=0}^n q^k = (1 - q^{n+1})/(1 - q)\). Proposition — Limit of \(q^n\)
For \(q \in \mathbb{R}\): - \(|q| < 1 \Rightarrow \textcolor{colorprop}{q^n \to 0}\) ;
- \(q = 1 \Rightarrow q^n = 1 \to 1\) ;
- \(q > 1 \Rightarrow \textcolor{colorprop}{q^n \to +\infty}\) ;
- \(q = -1\): \((q^n)\) alternates between \(\pm 1\) ; the subsequences \((q^{2n}) = 1\) and \((q^{2n+1}) = -1\) have distinct limits, so \((q^n)\) admits no limit (P6.3) ;
- \(q < -1\): \(|q^n| = |q|^n \to +\infty\), so \((|q^n|)\) diverges to \(+\infty\). But the signs alternate, so \((q^n)\) admits no limit in \(\overline{\mathbb{R}}\) — neither \(+\infty\) (the even subsequence \(|q|^{2n} \to +\infty\) but the odd \(-|q|^{2n+1} \to -\infty\)) nor \(-\infty\). In particular « \(|q^n| \to +\infty\) » does NOT imply « \(q^n \to +\infty\) » — the sign matters.
We prove the two genuine convergence cases.
- \(q > 1\): write \(q = 1 + h\) with \(h > 0\). Bernoulli's inequality (recap from Propriétés de \(\mathbb{R}\)) gives \(q^n = (1 + h)^n \ge 1 + n h\) for \(n \in \mathbb{N}\). Since \(1 + n h \to +\infty\), by P4.2 (minoration) \(q^n \to +\infty\).
- \(|q| < 1\): if \(q = 0\), then \(q^n = 0\) for every \(n \ge 1\); the first term is irrelevant for the limit, so \(q^n \to 0\). Otherwise apply the previous case to \(1/|q| > 1\): \((1/|q|)^n \to +\infty\), so \(|q|^n = 1/(1/|q|)^n \to 0\) by the inverse rule for a sequence tending to \(+\infty\) (rules table). Hence \(|q^n| = |q|^n \to 0\), equivalently \(q^n \to 0\).
Definition — Arithmético-geometric sequence
\((u_n)\) satisfies \(u_{n+1} = a u_n + b\) with \(a, b \in \mathbb{R}\). Two sub-cases: - If \(a = 1\): arithmetic of common difference \(b\), \(u_n = u_0 + n b\).
- If \(a \ne 1\): the equation \(\ell = a \ell + b\) has a unique solution \(\ell = b/(1 - a)\) (fixed point). Then \((u_n - \ell)\) is geometric of ratio \(a\), hence \(u_n - \ell = a^n (u_0 - \ell)\), i.e. \(\textcolor{colordef}{u_n = \ell + a^n (u_0 - \ell)}\).
Definition — Linear recurrence of order 2
Let \(a, b \in \mathbb{R}\) with \(b \ne 0\) (genuine order-2 recurrence; the case \(b = 0\) degenerates to \(u_{n+2} = a u_{n+1}\), geometric). Given \(u_0, u_1 \in \mathbb{R}\), the sequence \((u_n)\) is defined by $$ u_{n+2} = a u_{n+1} + b u_n \qquad (n \ge 0). $$ The characteristic equation is \(\textcolor{colordef}{r^2 - a r - b = 0}\), with discriminant \(\Delta = a^2 + 4 b\). Example
For the Fibonacci recurrence \(u_{n+2} = u_{n+1} + u_n\) (so \(a = b = 1\)), the characteristic equation is \(r^2 - r - 1 = 0\) with discriminant \(\Delta = 1 + 4 = 5 > 0\). For \(u_{n+2} = 2 u_{n+1} - u_n\) (\(a = 2\), \(b = -1\)): \(r^2 - 2 r + 1 = 0\), \(\Delta = 0\). For \(u_{n+2} = -u_n\) (\(a = 0\), \(b = -1\)): \(r^2 + 1 = 0\), \(\Delta = -4 < 0\). The three discriminant cases are then resolved in P7.2 below. Proposition — Solutions of linear recurrences of order 2
With \(a, b \in \mathbb{R}\), \(b \ne 0\), and \(\Delta = a^2 + 4 b\): - \(\Delta > 0\): two distinct real roots \(r_1, r_2\). General real term \(\textcolor{colorprop}{u_n = \lambda r_1^n + \mu r_2^n}\) with \(\lambda, \mu \in \mathbb{R}\).
- \(\Delta = 0\): double real root \(r\). General real term \(\textcolor{colorprop}{u_n = (\lambda + \mu n) r^n}\) with \(\lambda, \mu \in \mathbb{R}\).
- \(\Delta < 0\): two complex conjugate roots \(\rho \mathrm{e}^{\pm i \theta}\) with \(\rho > 0\), \(\theta \in \,]0, \pi[\). General real term \(\textcolor{colorprop}{u_n = \rho^n (\lambda \cos(n \theta) + \mu \sin(n \theta))}\) with \(\lambda, \mu \in \mathbb{R}\).
Proof idea. For \(\Delta > 0\): one verifies directly that \(r_1^n\) and \(r_2^n\) satisfy the recurrence (\(r_i^{n+2} - a r_i^{n+1} - b r_i^n = r_i^n (r_i^2 - a r_i - b) = 0\)), and so does any real combination \(\lambda r_1^n + \mu r_2^n\). The \(2 \times 2\) linear system in \((\lambda, \mu)\) for \(u_0, u_1\) has Vandermonde-type determinant \(r_2 - r_1 \ne 0\), hence a unique solution. Uniqueness of a sequence defined by an order-2 recurrence (given \(u_0, u_1\)) finishes the argument. The cases \(\Delta = 0\) and \(\Delta < 0\) are similar, using the elementary sequences \(r^n, n r^n\) and \(\rho^n \cos(n \theta), \rho^n \sin(n \theta)\) respectively.
Theorem — Limit transfer for \(u_{n+1} \equal f(u_n)\)
Let \(f : I \to I\) continuous on an interval \(I \subset \mathbb{R}\) stable by \(f\), \(u_0 \in I\), \(u_{n+1} = f(u_n)\). If \((u_n)\) converges to a limit \(\ell \in I\), then \(\ell\) is a fixed point of \(f\), i.e. \(\textcolor{colorprop}{f(\ell) = \ell}\).
Apply the sequential characterization of continuity (Limites et continuité P5.2, corollary of Heine T3.1): \(f\) continuous at \(\ell\) and \(u_n \to \ell\), hence \(f(u_n) \to f(\ell)\). But \(f(u_n) = u_{n+1}\), and by the shift corollary P6.5 (or P6.2 with \(\varphi(n) = n + 1\)), \(u_{n+1} \to \ell\). By uniqueness of the limit (P2.1), \(f(\ell) = \ell\).
Example
Compute \(\lim_{n \to +\infty} (1 + 1/2 + 1/4 + \dots + 1/2^n)\) (geometric series, ratio \(1/2\)).
Use the geometric sum formula: \(\sum_{k=0}^n (1/2)^k = (1 - (1/2)^{n+1})/(1 - 1/2) = 2 (1 - (1/2)^{n+1})\). As \(n \to +\infty\), \((1/2)^{n+1} \to 0\) (P7.1, \(|q| = 1/2 < 1\)), so the sum tends to \(2 \cdot 1 = 2\).
Example
Arithmético-geometric: \(u_0 = 1\), \(u_{n+1} = u_n / 2 + 1\). Compute \(\lim u_n\).
Apply D7.3 with \(a = 1/2\), \(b = 1\). Fixed point: \(\ell = 1/(1 - 1/2) = 2\). Then \(u_n - 2 = (1/2)^n (u_0 - 2) = (1/2)^n \cdot (-1) = -(1/2)^n\), hence \(u_n = 2 - (1/2)^n\). As \(n \to +\infty\), \((1/2)^n \to 0\), so \(u_n \to 2\).
Example
Fibonacci: \(u_0 = 0\), \(u_1 = 1\), \(u_{n+2} = u_{n+1} + u_n\). Give the closed form.
Characteristic equation: \(r^2 - r - 1 = 0\), \(\Delta = 1 + 4 = 5 > 0\). Two distinct real roots: \(r_1 = \varphi = (1 + \sqrt{5})/2\) and \(r_2 = \psi = (1 - \sqrt{5})/2\). By P7.2, \(u_n = \lambda \varphi^n + \mu \psi^n\). Solving with \(u_0 = 0\) and \(u_1 = 1\): \(\lambda + \mu = 0\) and \(\lambda \varphi + \mu \psi = 1\). From the first, \(\mu = -\lambda\); substituting, \(\lambda(\varphi - \psi) = 1\), i.e. \(\lambda \sqrt{5} = 1\), \(\lambda = 1/\sqrt{5}\). Hence the Binet formula: $$ u_n = \frac{\varphi^n - \psi^n}{\sqrt{5}}. $$
Example
Double root case: \(u_0 = 0\), \(u_1 = 1\), \(u_{n+2} = 2 u_{n+1} - u_n\). Give the closed form.
Characteristic equation: \(r^2 - 2 r + 1 = (r - 1)^2 = 0\), \(\Delta = 0\), double root \(r_0 = 1\). By P7.2, \(u_n = (\lambda + \mu n) \cdot 1^n = \lambda + \mu n\). With \(u_0 = 0\): \(\lambda = 0\). With \(u_1 = 1\): \(\mu = 1\). Hence \(u_n = n\) — an arithmetic sequence of common difference \(1\), as a sanity check.
Example
Complex roots case: \(u_0 = 1\), \(u_1 = 0\), \(u_{n+2} = -u_n\). Give the closed form.
Characteristic equation: \(r^2 + 1 = 0\), \(\Delta = -4 < 0\), complex roots \(r = \pm i = \mathrm{e}^{\pm i \pi / 2}\), so \(\rho = 1\), \(\theta = \pi/2\). By P7.2, \(u_n = \rho^n (\lambda \cos(n \theta) + \mu \sin(n \theta)) = \lambda \cos(n \pi / 2) + \mu \sin(n \pi / 2)\). With \(u_0 = 1\): \(\lambda = 1\). With \(u_1 = 0\): \(\mu = 0\). Hence \(u_n = \cos(n \pi / 2)\) — a period-\(4\) sequence \(1, 0, -1, 0, 1, \dots\), as expected from \(u_{n+2} = -u_n\).
Example
Recurrent: \(u_0 = 0\), \(u_{n+1} = \sqrt{u_n + 2}\). Show \((u_n)\) converges and find the limit.
Set \(f(x) = \sqrt{x + 2}\), defined and continuous on \([-2, +\infty[\).
Stable interval. On \([0, 2]\), \(f(x) = \sqrt{x + 2} \in [\sqrt{2}, 2] \subset [0, 2]\). So \(I = [0, 2]\) is stable by \(f\). By induction, \(u_n \in [0, 2]\) for all \(n\) (base \(u_0 = 0\), heredity by stability).
Monotonicity. For \(x \in [0, 2]\), an algebraic computation: $$ f(x) \ge x \iff \sqrt{x + 2} \ge x \iff x + 2 \ge x^2 \iff (x - 2)(x + 1) \le 0, $$ which holds on \([0, 2]\) (the factors have opposite signs there). Hence \(f(x) \ge x\) on \([0, 2]\). Therefore \(u_{n+1} = f(u_n) \ge u_n\), so \((u_n)\) is increasing.
Convergence. \((u_n)\) is increasing and bounded above by \(2\). By T5.1, \(u_n \to \ell\) for some \(\ell \in [0, 2]\).
Limit. By T7.1, \(\ell\) is a fixed point of \(f\) in \([0, 2]\): \(\sqrt{\ell + 2} = \ell\), i.e. \(\ell + 2 = \ell^2\), i.e. \(\ell^2 - \ell - 2 = 0\), with roots \(\ell = 2\) and \(\ell = -1\). The root in \([0, 2]\) is \(\ell = 2\).
Stable interval. On \([0, 2]\), \(f(x) = \sqrt{x + 2} \in [\sqrt{2}, 2] \subset [0, 2]\). So \(I = [0, 2]\) is stable by \(f\). By induction, \(u_n \in [0, 2]\) for all \(n\) (base \(u_0 = 0\), heredity by stability).
Monotonicity. For \(x \in [0, 2]\), an algebraic computation: $$ f(x) \ge x \iff \sqrt{x + 2} \ge x \iff x + 2 \ge x^2 \iff (x - 2)(x + 1) \le 0, $$ which holds on \([0, 2]\) (the factors have opposite signs there). Hence \(f(x) \ge x\) on \([0, 2]\). Therefore \(u_{n+1} = f(u_n) \ge u_n\), so \((u_n)\) is increasing.
Convergence. \((u_n)\) is increasing and bounded above by \(2\). By T5.1, \(u_n \to \ell\) for some \(\ell \in [0, 2]\).
Limit. By T7.1, \(\ell\) is a fixed point of \(f\) in \([0, 2]\): \(\sqrt{\ell + 2} = \ell\), i.e. \(\ell + 2 = \ell^2\), i.e. \(\ell^2 - \ell - 2 = 0\), with roots \(\ell = 2\) and \(\ell = -1\). The root in \([0, 2]\) is \(\ell = 2\).
Method — Studying \(u_{n+1} \equal f(u_n)\)
For a recurrent sequence \(u_{n+1} = f(u_n)\) with \(u_0\) given: - Find a stable interval \(I\). An interval \(I\) such that \(f(I) \subset I\), with \(u_0 \in I\). Then \(u_n \in I\) for all \(n\) (induction).
- Find candidate fixed points. Solve \(f(\ell) = \ell\) in \(I\).
- Study monotonicity of \((u_n)\). Sign of \(f(x) - x\) on \(I\) : \(> 0 \Rightarrow\) increasing ; \(< 0 \Rightarrow\) decreasing. Combined with \(f\) increasing or decreasing, conclude on the sign of \(u_{n+1} - u_n\).
- Conclude convergence by T5.1. Monotone bounded \(\Rightarrow\) convergent.
- Identify the limit by T7.1. The limit is a fixed point of \(f\) in \(I\). If multiple, pick the one consistent with monotonicity and the start value.
- If \(f\) is decreasing on \(I\): \((u_n)\) is generally not monotone. However \(f \circ f\) is increasing on \(I\), and the even subsequence \((u_{2n})\) and the odd subsequence \((u_{2n+1})\) are both governed by the recurrence \(w_{n+1} = (f \circ f)(w_n)\). Study each subsequence separately by the previous bullets: find a stable interval, then study the sign of \((f \circ f)(x) - x\) on it to obtain monotonicity, then apply T5.1. If the two subsequences share the same limit \(\ell\), conclude \(u_n \to \ell\) by P6.4 (even/odd cover).
Example
Cobweb diagram for \(u_{n+1} = \sqrt{u_n + 2}\), \(u_0 = 0\). 
Skills to practice
- Arithmetic\(\virgule\) geometric\(\virgule\) arithmético-geometric
- Linear recurrences of order 2
- Recurrent sequences \(u_{n+1} \equal f(u_n)\)
VIII
Brief extension to complex sequences
When \((u_n) \in \mathbb{C}^{\mathbb{N}}\), the limit notion transfers via the modulus \(|u_n - \ell| \to 0\), equivalently via componentwise convergence of \(\mathrm{Re}(u_n)\) and \(\mathrm{Im}(u_n)\). Operations and Bolzano-Weierstrass extend without difficulty. The chapter's order-based theorems (monotone-limit, adjacent) do not extend, since \(\mathbb{C}\) has no natural order.
Definition — Complex sequence
A complex sequence is a map \(u : \mathbb{N} \to \mathbb{C}\), written \((u_n)\). It is bounded if \((|u_n|)_n\) is bounded above (equivalently: \(\mathrm{Re}(u_n)\) and \(\mathrm{Im}(u_n)\) both bounded). Definition — Limit of a complex sequence
For \(\ell \in \mathbb{C}\): \(u_n \to \ell\) if any of the following equivalent conditions holds: - \(|u_n - \ell| \to 0\) in \(\mathbb{R}\);
- \(\mathrm{Re}(u_n) \to \mathrm{Re}(\ell)\) and \(\mathrm{Im}(u_n) \to \mathrm{Im}(\ell)\).
Proposition — Operations on complex limits
If \(u_n \to \ell\) and \(v_n \to m\) with \(\ell, m \in \mathbb{C}\), then for any \(\lambda, \mu \in \mathbb{C}\): \(\textcolor{colorprop}{\lambda u_n + \mu v_n \to \lambda \ell + \mu m}\), \(\textcolor{colorprop}{u_n v_n \to \ell m}\), and \(\textcolor{colorprop}{u_n / v_n \to \ell / m}\) if \(m \ne 0\). Same statements as P3.1 with \(\mathbb{C}\) in place of \(\mathbb{R}\).
Linear combination: decompose into real and imaginary parts. Writing \(\lambda = \alpha + i \beta\), \(u_n = a_n + i b_n\) etc., \(\lambda u_n + \mu v_n\) has real and imaginary parts that are real linear combinations of \(a_n, b_n, c_n, d_n\), each converging by P3.1 applied componentwise. The componentwise characterization D8.2 concludes. The product case: write \(u_n v_n = (a_n + i b_n)(c_n + i d_n) = (a_n c_n - b_n d_n) + i (a_n d_n + b_n c_n)\), and apply P3.1 to each component. Quotient: same with the additional argument that \(|v_n| \to |m| > 0\) (since \(m \ne 0\)), so \(|v_n|\) is bounded away from \(0\) from a certain rank; then \(1/v_n = \overline{v_n}/|v_n|^2\) has bounded denominator and converges componentwise.
Theorem — Bolzano-Weierstrass for complex sequences
Every bounded complex sequence admits a convergent subsequence.
Let \((u_n) \in \mathbb{C}^{\mathbb{N}}\) bounded. Then \((\mathrm{Re}(u_n))\) and \((\mathrm{Im}(u_n))\) are bounded real sequences (by D8.1).
- Apply T6.1 (BW for \(\mathbb{R}\)) to \((\mathrm{Re}(u_n))\): extract a convergent subsequence \(\mathrm{Re}(u_{\varphi(n)}) \to \alpha\).
- The sequence \((\mathrm{Im}(u_{\varphi(n)}))_n\) is still bounded; apply T6.1 again to extract \(\mathrm{Im}(u_{\varphi(\psi(n))}) \to \beta\).
- \(\mathrm{Re}(u_{\varphi(\psi(n))})\) is a subsequence of \(\mathrm{Re}(u_{\varphi(n)})\), so by P6.2 it still tends to \(\alpha\).
Example
Show that \((\mathrm{e}^{i n \pi / 3})_n\) is bounded, and exhibit two subsequences with distinct limits.
\(|\mathrm{e}^{i n \pi / 3}| = 1\) for all \(n\), so the sequence is bounded by \(1\). The values cycle with period \(6\). Take \(\varphi(n) = 6 n\): \(\mathrm{e}^{i \cdot 6n \pi / 3} = \mathrm{e}^{i \cdot 2n\pi} = 1\), hence \(u_{\varphi(n)} \to 1\). Take \(\psi(n) = 6n + 3\): \(\mathrm{e}^{i \cdot (6n+3) \pi / 3} = \mathrm{e}^{i \pi} \cdot \mathrm{e}^{i \cdot 2n\pi} = -1\), hence \(u_{\psi(n)} \to -1\). Two subsequences with distinct limits in \(\mathbb{C}\), so \((u_n)\) is divergent.
Example
Compute \(\lim ((1 + i)^n / 2^n)\).
\(|1 + i| = \sqrt{2}\), so \(|(1+i)^n / 2^n| = (\sqrt{2}/2)^n = (1/\sqrt{2})^n\). Since \(1/\sqrt{2} < 1\), by P7.1 (real case), \((1/\sqrt{2})^n \to 0\). Hence \(|u_n| \to 0\), and by D8.2 (the modulus characterization), \(u_n \to 0\) in \(\mathbb{C}\).
Skills to practice
- Complex limits and Bolzano-Weierstrass for complex sequences
Jump to section