3.1 Sequences, Monotonicity, and Limits

3.1 Sequences, Monotonicity, and Limits

Sequences form the bedrock of real analysis on the GRE Mathematics Subject Test. The examination tests your ability to evaluate limits, analyze recursive relations, and apply foundational theorems including the Monotone Convergence Theorem, Bolzano-Weierstrass, and Cauchy completeness.

Formal Convergence and Boundedness

A real sequence $(a_n){n=1}^\infty$ converges to $L \in \mathbb{R}$, denoted $\lim{n \to \infty} a_n = L$, if for every $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that: $|a_n - L| < \epsilon \quad \text{for all } n > N$ Otherwise, the sequence diverges (by unbounded growth or oscillation, e.g., $a_n = (-1)^n$).

A sequence $(a_n)$ is bounded above if $a_n \le M$, bounded below if $a_n \ge m$, and bounded if $|a_n| \le K$ for all $n$.

• Fundamental Result: Every convergent sequence is bounded.
• Converse Warning: Boundedness does not imply convergence ($a_n = \sin(n\pi/2)$ is bounded in $[-1, 1]$ but diverges).

The Monotone Convergence Theorem (MCT)

A sequence is monotonically increasing if $a_{n+1} \ge a_n$, and monotonically decreasing if $a_{n+1} \le a_n$.

Theorem (Monotone Convergence Theorem)

Every bounded, monotonic sequence of real numbers converges:

1. If $(a_n)$ is non-decreasing and bounded above, $\lim_{n \to \infty} a_n = \sup {a_n : n \in \mathbb{N}}$.
2. If $(a_n)$ is non-increasing and bounded below, $\lim_{n \to \infty} a_n = \inf {a_n : n \in \mathbb{N}}$.

MCT guarantees existence without computing the value in advance, making it ideal for recursive sequences.

Subsequences, Bolzano-Weierstrass, and Limsup / Liminf

A subsequence $(a_{n_k})$ is defined by indices $n_1 < n_2 < n_3 < \cdots$.

• If $\lim a_n = L$, every subsequence converges to $L$. Distinct subsequential limits imply divergence.
• Bolzano-Weierstrass Theorem: Every bounded sequence in $\mathbb{R}$ contains a convergent subsequence.

Limit Superior and Limit Inferior

Define tail bounds $s_n = \sup_{k \ge n} a_k$ and $i_n = \inf_{k \ge n} a_k$. Then $\limsup a_n = \lim s_n = \inf_{n \ge 1} s_n$, and $\liminf a_n = \lim i_n = \sup_{n \ge 1} i_n$. In $[-\infty, \infty]$, $\limsup a_n$ and $\liminf a_n$ represent the greatest and least subsequential limits. A sequence converges in $\mathbb{R}$ if and only if $\liminf a_n = \limsup a_n = L \in \mathbb{R}$.

Cauchy Sequences and Completeness of $\mathbb{R}$

A sequence $(a_n)$ is Cauchy if for every $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that: $|a_n - a_m| < \epsilon \quad \text{for all } n, m > N$

• A metric space is complete if every Cauchy sequence converges to an element of the space.
• In $\mathbb{R}$ with Euclidean metric, a sequence converges if and only if it is Cauchy. This equivalence fails in incomplete spaces like $\mathbb{Q}$.

Recursive Sequences and Fixed-Point Limits

For recursive sequences $x_{n+1} = f(x_n)$:

1. Prove Convergence: Show $(x_n)$ is monotonic and bounded by induction. MCT guarantees the limit $L = \lim x_n$ exists.
2. Solve Fixed Point: For continuous $f$, take limits across the recurrence: $L = f(L)$. Solve for $L$ and reject extraneous roots.

If $|f'(x)| \le k < 1$ on an invariant interval, the Banach Fixed-Point Theorem guarantees unique convergence to $L$.

Squeeze Theorem and Asymptotics

If $a_n \le b_n \le c_n$ for all $n \ge N$ and $\lim a_n = \lim c_n = L$, then $\lim b_n = L$.

• If $a_n \to 0$ and $(b_n)$ is bounded, $\lim (a_n b_n) = 0$.
• For $a, b > 0$, $\lim_{n \to \infty} (a^n + b^n)^{1/n} = \max(a, b)$.

Summary Comparison Table: Sequence Theorems

Worked Problem: Recursive Sequence with Fixed Point

Problem: Let $x_1 = 1$ and $x_{n+1} = \sqrt{3x_n + 4}$ for $n \ge 1$. Prove convergence and find $\lim_{n \to \infty} x_n$.

Solution:

1. Boundedness: We claim $1 \le x_n < 4$. Base case: $x_1 = 1 < 4$. If $x_k < 4$, then $x_{k+1} = \sqrt{3x_k + 4} < \sqrt{3(4) + 4} = 4$. Thus $(x_n)$ is bounded above by 4.
2. Monotonicity: Base case: $x_2 = \sqrt{7} \approx 2.65 > 1 = x_1$. Inductive step: $x_k \ge x_{k-1} \implies 3x_k + 4 \ge 3x_{k-1} + 4 \implies x_{k+1} \ge x_k$. Thus $(x_n)$ is non-decreasing.
3. Limit: By MCT, $L = \lim x_n$ exists with $1 \le L \le 4$. Continuity yields $L = \sqrt{3L + 4} \implies L^2 - 3L - 4 = 0 \implies (L - 4)(L + 1) = 0$. Since $L \ge 1$, $L = 4$.

GRE Exam Traps & Fast Test-Taking Strategies

• Trap 1: Solving $L = f(L)$ without proving convergence. For $x_1 = 3$ and $x_{n+1} = 2x_n - 1$, setting $L = 2L - 1$ gives $L = 1$, yet $x_n = 2^n + 1 \to \infty$. Always verify boundedness and monotonicity before computing limits.
• Trap 2: Ignoring sign restrictions on roots. Quadratic fixed-point equations often give positive and negative roots; discard roots outside sequence bounds.
• Trap 3: Confusing $a_{n+1} - a_n \to 0$ with convergence. Harmonic numbers $H_n = \sum_{k=1}^n 1/k$ satisfy $H_{n+1} - H_n = 1/(n+1) \to 0$, yet $H_n \to \infty$.