The Timpson Lyceum · Real Analysis (2020)

Lecture 5 (Real Analysis 2020)

Overview

This lecture opens Chapter 2 with the central object of real analysis: the limit of a sequence. It builds up four foundational definitions — sequence, series, convergence, and limit — and then turns to the proofs that make limits usable. The heart of the material is the $\varepsilon\text{-}N$ definition of convergence and the technique for proving a given sequence converges to a given value. From there it establishes that limits are unique (so "the limit" is well defined), shows how to prove a sequence diverges, and develops the algebraic limit laws (the comparison lemma, sum law, product laws, scalar law) that let you compute new limits from known ones. The lecture closes by previewing how plugging a convergent sequence into a polynomial converges to the polynomial of the limit — the seed of continuity and the limit of a function.

Key Concepts

Sequence. A sequence is a real-valued function whose domain is the natural numbers, $f : \mathbb{N} \to \mathbb{R}$ . We write the $n$ th term as $a_n$ rather than $f(n)$ . Only the output values and their order matter; despite the brace-like notation a sequence is not a set, because order matters and terms may repeat. Examples: $a_n = n^2$ gives $1,4,9,16,\dots$ ; $a_n = \tfrac1n$ gives $1,\tfrac12,\tfrac13,\dots$ ; $a_n = \tfrac{n+1}{n}$ gives $2,\tfrac32,\tfrac43,\dots$ .

Series (infinite series). Given a sequence $a_n$ , define a new sequence $c_n$ by summing the first $n$ terms: $c_n = \sum_{k=1}^{n} a_k.$ Then $c_n$ is an infinite series, and its $n$ th term $c_n$ is the $n$ th partial sum. For example, from $a_n = \tfrac{1}{2^n}$ the series is $\tfrac12,\ \tfrac34,\ \tfrac78,\ \tfrac{15}{16},\dots$ Like any sequence it has infinitely many terms; in this course only infinite series are considered.

Convergence ( $\varepsilon\text{-}N$ definition). A sequence $a_n$ converges to $a$ if for every $\varepsilon > 0$ there exists a natural number $N$ such that $|a_n - a| < \varepsilon \quad \text{for all } n \ge N.$ Intuition: no matter how tiny a tolerance $\varepsilon$ you choose around $a$ , there is some point $N$ in the sequence beyond which every term lies within the band $(a-\varepsilon,\, a+\varepsilon)$ . Smaller $\varepsilon$ just forces a larger $N$ . The inequality is equivalent to $a - \varepsilon < a_n < a + \varepsilon$ .

Limit. Once convergence is shown to produce one and only one value, that value is named the limit: if $a_n$ converges to $a$ we write $\lim a_n = a$ . Naming it as a single number (like defining the additive inverse only after proving it unique) lets us treat $\lim$ as a well-defined unary operation taking a sequence to a real number.

Key tool — the Archimedean property. For every positive real number there is a natural number $N$ with $\tfrac1N$ smaller than it. This is what converts an arbitrary $\varepsilon$ into the index $N$ needed in convergence proofs.

Proof strategy for $\lim a_n = a$ . Do scratch work first: simplify $|a_n - a|$ and bound it above by something simple (e.g. $\tfrac{C}{n}$ ), enlarging the expression as needed (making a denominator smaller, or multiplying a term by $n$ , only increases it). Then use the Archimedean property to choose $N$ so the simple bound is $< \varepsilon$ , and write the clean chain of inequalities.

Uniqueness of limits. A convergent sequence has exactly one limit. Proof idea (by contradiction): if $a_n \to a$ and $a_n \to a'$ with $a' > a$ , choose $\varepsilon < \tfrac{a'-a}{2}$ (less than half the gap). The two bands $(a-\varepsilon,a+\varepsilon)$ and $(a'-\varepsilon,a'+\varepsilon)$ are disjoint, yet beyond $N = \max(N_1,N_2)$ the terms must lie in both — forcing $a_n < a_n$ , a contradiction.

Divergence. To show a sequence does not converge, negate the definition: exhibit one $\varepsilon > 0$ such that no number $a$ and no index $N$ make $|a_n - a| < \varepsilon$ hold from $N$ onward. For $a_n = (-1)^n$ (which alternates $-1, 1, -1, 1, \dots$ ) take $\varepsilon = \tfrac12$ . Any interval of length $1$ (here $2\varepsilon = 1$ ) contains at most one integer, but from any index onward the sequence visits both $-1$ and $1$ , which cannot both lie in such an interval.

Comparison lemma. If $\lim a_n = a$ and there exist $N_1 \in \mathbb{N}$ and a constant $C \ge 0$ with $|b_n - b| \le C\,|a_n - a| \quad \text{for all } n \ge N_1,$ then $\lim b_n = b$ . This proves new convergences without returning to the definition: a quantity squeezed (within a constant factor) by something tending to $0$ also tends to $0$ .

Algebraic limit laws. For convergent sequences $a_n \to a$ , $b_n \to b$ :

Constant: $\lim c = c$ .
Sum law: $\lim(a_n + b_n) = a + b$ (split $\varepsilon$ as $\tfrac\varepsilon2 + \tfrac\varepsilon2$ and use the triangle inequality).
Product of null sequences: if $a_n \to 0$ and $b_n \to 0$ then $a_n b_n \to 0$ (bound each factor by $\sqrt\varepsilon$ ).
Scalar law: $\lim(c\,a_n) = c\,a$ (immediate from the comparison lemma, since $|c a_n - c a| = |c|\,|a_n - a|$ ).
Product law: $\lim(a_n b_n) = ab$ .
Useful equivalence (homework): $\lim a_n = a \iff \lim(a_n - a) = 0$ .

Limits of polynomials (preview). For a polynomial $p(x) = \sum_{k=0}^{m} c_k x^k$ , if $a_n \to a$ then $p(a_n) \to p(a)$ . Feeding a convergent sequence through $p$ produces a sequence converging to $p$ evaluated at the limit — the foundation for the limit of a function and continuity.

Worked Example

Prove that $\dfrac{n}{n+1} \to 1$ .

Scratch work bounds the error and the Archimedean property supplies $N$ with $\tfrac1N < \varepsilon$ :

\begin{aligned} \left|\frac{n}{n+1} - 1\right| &= \left|\frac{n - (n+1)}{n+1}\right| = \frac{1}{n+1} \\ &\le \frac{1}{n} \le \frac{1}{N} < \varepsilon \qquad \text{for all } n \ge N. \end{aligned}

Since $\left|\frac{n}{n+1} - 1\right| < \varepsilon$ for all $n \ge N$ , by definition $\dfrac{n}{n+1} \to 1$ .

Product law via null sequences (proof sketch). To show $a_n b_n \to ab$ , use $\lim a_n = a \iff \lim(a_n - a)=0$ , so it suffices to show $\lim(a_n b_n - ab) = 0$ . Set the null sequences $\alpha_n = a_n - a$ and $\beta_n = b_n - b$ (so $\alpha_n \to 0$ , $\beta_n \to 0$ ). Substituting $a_n = \alpha_n + a$ , $b_n = \beta_n + b$ :

\begin{aligned} \lim(a_n b_n - ab) &= \lim\big((\alpha_n + a)(\beta_n + b) - ab\big) \\ &= \lim\big(\alpha_n\beta_n + a\,\beta_n + b\,\alpha_n\big) \\ &= \underbrace{\lim(\alpha_n\beta_n)}_{0} + a\underbrace{\lim \beta_n}_{0} + b\underbrace{\lim \alpha_n}_{0} = 0, \end{aligned}

using the null-product law, the sum law, and the scalar law. Hence $\lim(a_n b_n) = ab$ .

Summary

A sequence is a function $\mathbb{N} \to \mathbb{R}$ (terms $a_n$ ); a series is the sequence of partial sums $c_n = \sum_{k=1}^n a_k$ .
$a_n \to a$ means: for every $\varepsilon > 0$ there is an $N$ with $|a_n - a| < \varepsilon$ for all $n \ge N$ — eventually the sequence stays within any tolerance of $a$ .
Convergence proofs bound $|a_n - a|$ by a simple expression, then use the Archimedean property to pick $N$ making that bound $< \varepsilon$ .
Limits are unique, which lets us speak of the limit and treat $\lim$ as a well-defined operation; to show divergence, negate the definition (e.g. $(-1)^n$ with $\varepsilon = \tfrac12$ ).
The comparison lemma and the limit laws (sum, product, scalar, null-product) compute new limits from known ones without reverting to $\varepsilon\text{-}N$ .
Plugging a convergent sequence into a polynomial gives $p(a_n) \to p(a)$ , previewing the limit of a function and continuity.