Real Analysis — 01 · Sequences and Series

A sequence is the simplest infinite process in mathematics, and its limit is the idea on which the rest of analysis is built: derivatives, integrals and continuity are all limits in disguise. This post makes "the terms get closer and closer to $a$" precise with the $\varepsilon$–$N$ definition, works through explicit proofs with concrete numbers, and then proves the big structural theorems that rest on the completeness of $\mathbb{R}$ from Real Analysis 00: the Monotone Convergence Theorem, Bolzano–Weierstrass and the Cauchy criterion. The second half turns to infinite series — sums of infinitely many numbers — with the standard convergence tests and the surprising fact that reordering the terms of some series changes the sum.

Series map: 00 The Real Numbers & Completeness → 01 Sequences & Series (this post) → 02 Topology of ℝ & Continuity → 03 Differentiation & Taylor's Theorem → 04 The Riemann Integral → 05 Uniform Convergence & Power Series.

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Learning Outcome

• State the $\varepsilon$–$N$ definition of convergence and write complete proofs that specific sequences converge, with an explicit $N$.
• Prove that limits are unique and that convergent sequences are bounded.
• Prove the sum and product parts of the Algebraic Limit Theorem, and use the Order Limit Theorem and the squeeze theorem.
• Prove the Monotone Convergence Theorem and apply it to a recursively defined sequence.
• Prove the Bolzano–Weierstrass Theorem by bisection and use subsequences to show divergence.
• Define Cauchy sequences, $\limsup$ and $\liminf$, and state the Cauchy criterion.
• Decide convergence of series with the divergence, comparison, ratio, root and alternating series tests.
• Distinguish absolute from conditional convergence and state the Riemann rearrangement theorem.

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Chapter 1 — The limit of a sequence

Terminology

• Sequence — a function from $\mathbb{N}$ to $\mathbb{R}$, written $(a_n) = (a_1, a_2, a_3, \dots)$.
• $\varepsilon$-neighbourhood of $a$ — the interval $V_\varepsilon(a) = (a - \varepsilon, a + \varepsilon)$, the points within distance $\varepsilon$ of $a$.
• Converges — $(a_n)$ converges to $a$, written $a_n \to a$ or $\lim a_n = a$, if for every $\varepsilon > 0$ there is $N \in \mathbb{N}$ such that $\lvert a_n - a \rvert < \varepsilon$ for all $n \ge N$.
• Diverges — does not converge to any real number.
• Tail — the terms $a_N, a_{N+1}, \dots$ from some index on.
• Bounded — there is $M > 0$ with $\lvert a_n \rvert \le M$ for all $n$.

The definition says: whatever tolerance $\varepsilon$ is demanded, the whole tail from some $N$ on lies inside the band $V_\varepsilon(a)$. The $N$ is allowed to depend on $\varepsilon$ (a smaller band usually needs a larger $N$). Finitely many early terms never matter.

A worked proof with explicit numbers

Claim. $\displaystyle a_n = \frac{2n+1}{n+3} \to 2$.

Scratch work. Compute the distance to the limit:

$$\lvert a_n - 2 \rvert = \left\lvert \frac{2n + 1 - 2(n+3)}{n+3} \right\rvert = \frac{5}{n+3}.$$

This is less than $\varepsilon$ exactly when $n + 3 > 5/\varepsilon$, i.e. $n > 5/\varepsilon - 3$.

Proof. Let $\varepsilon > 0$. By the Archimedean property choose $N \in \mathbb{N}$ with $N > 5/\varepsilon - 3$. For $n \ge N$,

$$\lvert a_n - 2 \rvert = \frac{5}{n+3} \le \frac{5}{N+3} < \varepsilon. \qquad \blacksquare$$

Checking with numbers. The first terms are $a_1 = \tfrac34$, $a_2 = 1$, $a_3 = \tfrac76$, $a_4 = \tfrac97$, $a_5 = \tfrac{11}{8}$, increasing towards $2$.

$\varepsilon$	condition $n > 5/\varepsilon - 3$	smallest $N$	check at $N - 1$	check at $N$
$0.25$	$n > 17$	$18$	$a_{17} = 1.75$, distance $0.25$ (not $< 0.25$)	$a_{18} = \tfrac{37}{21} \approx 1.7619$, distance $\approx 0.2381$
$0.1$	$n > 47$	$48$	$a_{47} = 1.9$, distance $0.1$ (not $< 0.1$)	$a_{48} = \tfrac{97}{51} \approx 1.9020$, distance $\approx 0.0980$
$0.01$	$n > 497$	$498$	$a_{497} = 1.99$, distance $0.01$	$a_{498} = \tfrac{997}{501} \approx 1.9900$, distance $\approx 0.00998$

The term $a_{17}$ lands exactly on the edge of the band for $\varepsilon = 0.25$; since the definition uses a strict inequality, the tail starts at $N = 18$. Any larger $N$ also works — the definition asks for some $N$, not the smallest.

Trick: in the scratch work, simplify $\lvert a_n - a \rvert$ and then bound it above by something simple. You do not need the exact solution of $\lvert a_n - a \rvert < \varepsilon$. For $b_n = \frac{n+1}{n^2}$, for example, $\lvert b_n - 0 \rvert = \frac{n+1}{n^2} \le \frac{2n}{n^2} = \frac{2}{n}$, so any $N > 2/\varepsilon$ works.

More examples

$1/\sqrt{n} \to 0$. Given $\varepsilon > 0$ choose $N > 1/\varepsilon^2$. For $n \ge N$, $\sqrt{n} \ge \sqrt{N} > 1/\varepsilon$, so $1/\sqrt{n} < \varepsilon$.

$r^n \to 0$ for $0 \le r < 1$. If $r = 0$ it is trivial. Otherwise write $r = \frac{1}{1 + h}$ with $h > 0$. Bernoulli's inequality (Real Analysis 00) gives $(1+h)^n \ge 1 + nh > nh$, so $0 < r^n < \frac{1}{nh}$, which is less than $\varepsilon$ once $n > \frac{1}{h\varepsilon}$. The same bound on $\lvert r \rvert^n$ covers $-1 < r < 0$.

$(-1)^n$ diverges. Suppose $(-1)^n \to a$. Take $\varepsilon = 1$: there is $N$ with $\lvert (-1)^n - a \rvert < 1$ for $n \ge N$. Using $n = N$ and $n = N + 1$ (one even, one odd),

$$2 = \lvert 1 - (-1) \rvert \le \lvert 1 - a \rvert + \lvert a - (-1) \rvert < 1 + 1 = 2,$$

a contradiction.

Uniqueness and boundedness

Theorem. A sequence has at most one limit.

Proof. Suppose $a_n \to a$ and $a_n \to b$. Let $\varepsilon > 0$. There are $N_1, N_2$ with $\lvert a_n - a \rvert < \varepsilon/2$ for $n \ge N_1$ and $\lvert a_n - b \rvert < \varepsilon/2$ for $n \ge N_2$. For $n = \max(N_1, N_2)$,

$$\lvert a - b \rvert \le \lvert a - a_n \rvert + \lvert a_n - b \rvert < \varepsilon.$$

Since $\varepsilon$ was arbitrary, $a = b$ by the equality lemma of Real Analysis 00. $\blacksquare$

Theorem. Every convergent sequence is bounded.

Proof. Let $a_n \to a$. With $\varepsilon = 1$ there is $N$ such that $\lvert a_n - a \rvert < 1$, hence $\lvert a_n \rvert < \lvert a \rvert + 1$, for $n \ge N$. Put $M = \max\{\lvert a_1 \rvert, \dots, \lvert a_{N-1} \rvert, \lvert a \rvert + 1\}$. Then $\lvert a_n \rvert \le M$ for all $n$. $\blacksquare$

The converse is false: $(-1)^n$ is bounded and divergent.

Common pitfall: choosing $N$ that depends on $n$. In "$\forall \varepsilon\ \exists N\ \forall n \ge N$" the order says $N$ is chosen after $\varepsilon$ but before $n$. "Take $N = n + 1$" is not a proof.

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Chapter 2 — Limit theorems

The Algebraic Limit Theorem

Theorem. Let $a_n \to a$ and $b_n \to b$. Then
(i) $c\,a_n \to c\,a$ for every $c \in \mathbb{R}$;
(ii) $a_n + b_n \to a + b$;
(iii) $a_n b_n \to ab$;
(iv) $a_n / b_n \to a / b$, provided $b \ne 0$ (and $b_n \ne 0$ for all $n$).

Proof of (ii). Let $\varepsilon > 0$. Choose $N_1$ with $\lvert a_n - a \rvert < \varepsilon/2$ for $n \ge N_1$, and $N_2$ with $\lvert b_n - b \rvert < \varepsilon/2$ for $n \ge N_2$. For $n \ge \max(N_1, N_2)$,

$$\lvert (a_n + b_n) - (a + b) \rvert \le \lvert a_n - a \rvert + \lvert b_n - b \rvert < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon. \qquad \blacksquare$$

Proof of (iii). Add and subtract $a_n b$:

$$\lvert a_n b_n - ab \rvert = \lvert a_n (b_n - b) + (a_n - a) b \rvert \le \lvert a_n \rvert \lvert b_n - b \rvert + \lvert b \rvert \lvert a_n - a \rvert.$$

$(a_n)$ converges, so it is bounded: $\lvert a_n \rvert \le M$ for some $M > 0$. Let $\varepsilon > 0$. Choose $N_1$ with $\lvert b_n - b \rvert < \frac{\varepsilon}{2M}$ for $n \ge N_1$, and $N_2$ with $\lvert a_n - a \rvert < \frac{\varepsilon}{2(\lvert b \rvert + 1)}$ for $n \ge N_2$. (The $+1$ avoids dividing by zero when $b = 0$.) For $n \ge \max(N_1, N_2)$,

$$\lvert a_n b_n - ab \rvert < M \cdot \frac{\varepsilon}{2M} + \lvert b \rvert \cdot \frac{\varepsilon}{2(\lvert b \rvert + 1)} < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \qquad \blacksquare$$

Part (iv) is proved by first showing $1/b_n \to 1/b$: the key step is that $\lvert b_n \rvert > \lvert b \rvert / 2$ for all large $n$ (reverse triangle inequality), which keeps $1/\lvert b_n \rvert$ bounded.

Example. Rewriting $a_n = \frac{2n+1}{n+3} = \frac{2 + 1/n}{1 + 3/n}$, the theorem gives $\lim a_n = \frac{2 + 0}{1 + 0} = 2$ at once, since $1/n \to 0$. The explicit $\varepsilon$–$N$ proof of Chapter 1 is what the theorem packages.

Order and squeezing

Order Limit Theorem. Let $a_n \to a$ and $b_n \to b$.
(i) If $a_n \ge 0$ for all $n$, then $a \ge 0$.
(ii) If $a_n \le b_n$ for all $n$, then $a \le b$.
(iii) If $c \le b_n$ for all $n$, then $c \le b$; if $a_n \le c$ for all $n$, then $a \le c$.

Proof of (i). Suppose $a < 0$. Take $\varepsilon = \lvert a \rvert$. For large $n$, $\lvert a_n - a \rvert < \lvert a \rvert$, which gives $a_n < a + \lvert a \rvert = 0$, a contradiction. Part (ii) is (i) applied to $b_n - a_n$, and (iii) is (ii) with a constant sequence. $\blacksquare$

Watch out: strict inequalities are not preserved. $1/n > 0$ for every $n$, but $\lim 1/n = 0$. Limits turn $<$ into $\le$.

Squeeze theorem. If $x_n \le y_n \le z_n$ for all $n$ and $x_n \to \ell$, $z_n \to \ell$, then $y_n \to \ell$. Example: $0 \le \frac{\lvert\sin n\rvert}{n} \le \frac1n$, so $\frac{\sin n}{n} \to 0$.

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Chapter 3 — Monotone sequences

Terminology

• Increasing — $a_n \le a_{n+1}$ for all $n$. Decreasing — $a_n \ge a_{n+1}$ for all $n$.
• Monotone — increasing or decreasing.

The Monotone Convergence Theorem

Theorem. A monotone sequence converges if and only if it is bounded. An increasing bounded sequence converges to $\sup\{a_n : n \in \mathbb{N}\}$.

Proof. Convergent implies bounded (Chapter 1). Conversely, let $(a_n)$ be increasing and bounded. The set $\{a_n\}$ is nonempty and bounded above, so by the Axiom of Completeness $s = \sup\{a_n\}$ exists. Let $\varepsilon > 0$. Since $s - \varepsilon$ is not an upper bound, some $a_N > s - \varepsilon$. For $n \ge N$, monotonicity and $s$ being an upper bound give

$$s - \varepsilon < a_N \le a_n \le s,$$

so $\lvert a_n - s \rvert < \varepsilon$. The decreasing case is the same with the infimum. $\blacksquare$

The theorem proves that a limit exists without knowing its value in advance — the essential use of completeness. In $\mathbb{Q}$ it fails: the increasing bounded sequence $1, 1.4, 1.41, 1.414, \dots$ of decimal truncations of $\sqrt{2}$ has no rational limit.

Worked example: a recursive sequence for the square root of 2

Define $x_1 = 1$ and

$$x_{n+1} = \frac12 \left( x_n + \frac{2}{x_n} \right).$$

(This is Newton's method for $x^2 - 2 = 0$.) The first terms are

$n$	$x_n$	decimal	$x_n^2 - 2$
1	$1$	$1$	$-1$
2	$3/2$	$1.5$	$0.25$
3	$17/12$	$1.416667$	$\approx 6.94 \times 10^{-3}$
4	$577/408$	$1.414216$	$\approx 6.01 \times 10^{-6}$
5	$665857/470832$	$1.414214$	$\approx 4.51 \times 10^{-12}$

Step 1 — bounded below by $\sqrt{2}$ from $n = 2$ on. All $x_n > 0$ by induction. For any $x > 0$,

$$\frac12\left(x + \frac2x\right) - \sqrt{2} = \frac{x^2 - 2\sqrt{2}\,x + 2}{2x} = \frac{(x - \sqrt{2})^2}{2x} \ge 0,$$

so $x_{n+1} \ge \sqrt{2}$ for every $n \ge 1$.

Step 2 — decreasing from $n = 2$ on. For $n \ge 2$, $x_n^2 \ge 2$, so

$$x_n - x_{n+1} = \frac{x_n}{2} - \frac{1}{x_n} = \frac{x_n^2 - 2}{2x_n} \ge 0.$$

Step 3 — find the limit. The tail $(x_n)_{n \ge 2}$ is decreasing and bounded below, so by the Monotone Convergence Theorem $x_n \to L$ for some $L \ge \sqrt{2} > 0$. The shifted sequence $x_{n+1}$ has the same limit, and the Algebraic Limit Theorem applied to the recursion gives $L = \frac12 (L + 2/L)$, i.e. $L^2 = 2$. Since $L > 0$, $L = \sqrt{2}$. $\blacksquare$

Common pitfall: solving $L = \frac12(L + 2/L)$ before proving the limit exists. For $x_{n+1} = 2x_n$ with $x_1 = 1$ the same trick "gives" $L = 2L$, so $L = 0$ — but $x_n = 2^{n-1}$ diverges. Existence first (monotone and bounded), then the equation.

The number $e$ arises the same way: $\left(1 + \frac1n\right)^n$ is increasing and bounded above by $3$, so it converges, and its limit is defined to be $e$ (Bartle and Sherbert, Section 3.3).

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Chapter 4 — Subsequences, Bolzano–Weierstrass and Cauchy sequences

Terminology

• Subsequence of $(a_n)$ — a sequence $(a_{n_k}) = (a_{n_1}, a_{n_2}, \dots)$ with $n_1 < n_2 < n_3 < \cdots$. It keeps infinitely many terms, in order.
• Cauchy sequence — for every $\varepsilon > 0$ there is $N$ such that $\lvert a_n - a_m \rvert < \varepsilon$ for all $m, n \ge N$. The terms get close to each other, with no limit mentioned.
• Limit superior — $\limsup a_n = \lim_{n \to \infty} y_n$ where $y_n = \sup\{a_k : k \ge n\}$. Limit inferior — the same with $\inf$.

Theorem. If $a_n \to a$, every subsequence also converges to $a$. (Given $\varepsilon$ and $N$ for $(a_n)$, use the same $N$: $n_k \ge k$, so $k \ge N$ implies $n_k \ge N$.)

This gives the quickest divergence proofs: find two subsequences with different limits. For $(-1)^n$, the even terms tend to $1$ and the odd terms to $-1$.

Bolzano–Weierstrass

Theorem (Bolzano–Weierstrass). Every bounded sequence has a convergent subsequence.

Proof (bisection). Let $\lvert a_n \rvert \le M$ and $I_1 = [-M, M]$. Choose $n_1 = 1$. Split $I_1$ into two closed halves; at least one contains $a_n$ for infinitely many $n$. Call it $I_2$ and choose $n_2 > n_1$ with $a_{n_2} \in I_2$. Continue: $I_{k+1}$ is a half of $I_k$ containing infinitely many terms, and $n_{k+1} > n_k$ has $a_{n_{k+1}} \in I_{k+1}$. The intervals are closed, bounded and nested, with length $2M / 2^{k-1}$ for $I_k$. By the Nested Interval Property there is $x$ in every $I_k$. Both $x$ and $a_{n_k}$ lie in $I_k$, so

$$\lvert a_{n_k} - x \rvert \le \frac{2M}{2^{k-1}} \to 0.$$

Hence $a_{n_k} \to x$. $\blacksquare$

The Cauchy criterion

Theorem (Cauchy criterion). A sequence of real numbers converges if and only if it is a Cauchy sequence.

Proof. ($\Rightarrow$) If $a_n \to a$, choose $N$ with $\lvert a_n - a \rvert < \varepsilon/2$ for $n \ge N$. For $m, n \ge N$, $\lvert a_n - a_m \rvert \le \lvert a_n - a \rvert + \lvert a - a_m \rvert < \varepsilon$.

($\Leftarrow$) Let $(a_n)$ be Cauchy.
Bounded: with $\varepsilon = 1$ there is $N$ with $\lvert a_n - a_N \rvert < 1$ for $n \ge N$, so $\lvert a_n \rvert < \lvert a_N \rvert + 1$ for $n \ge N$; take the maximum with the first $N - 1$ terms.
A subsequence converges: by Bolzano–Weierstrass, $a_{n_k} \to a$ for some $a$.
The whole sequence follows: given $\varepsilon > 0$, choose $N$ with $\lvert a_n - a_m \rvert < \varepsilon/2$ for $m, n \ge N$, and choose $k$ with $n_k \ge N$ and $\lvert a_{n_k} - a \rvert < \varepsilon/2$. For $n \ge N$,

$$\lvert a_n - a \rvert \le \lvert a_n - a_{n_k} \rvert + \lvert a_{n_k} - a \rvert < \varepsilon. \qquad \blacksquare$$

The Cauchy criterion lets you prove convergence using only the terms themselves. In $\mathbb{Q}$ it fails: the Newton sequence of Chapter 3 is a Cauchy sequence of rationals with no rational limit. Completeness of $\mathbb{R}$ can be defined by "every Cauchy sequence converges" together with the Archimedean property; it is equivalent to the Axiom of Completeness.

Watch out: "consecutive terms get close" is not the Cauchy condition. For $a_n = \sqrt{n}$, $a_{n+1} - a_n = \frac{1}{\sqrt{n+1} + \sqrt{n}} \to 0$, yet $a_n \to \infty$. Cauchy requires $\lvert a_n - a_m \rvert$ small for all pairs $m, n \ge N$.

Limit superior and inferior

For a bounded sequence, $y_n = \sup\{a_k : k \ge n\}$ is decreasing (a supremum over a smaller set) and bounded, so $\limsup a_n = \lim y_n$ always exists by the Monotone Convergence Theorem; likewise $\liminf$. Always $\liminf a_n \le \limsup a_n$.

Example. $a_n = (-1)^n\left(1 + \frac1n\right)$: the terms are $-2, \tfrac32, -\tfrac43, \tfrac54, \dots$. The even terms decrease to $1$ and the odd terms increase to $-1$, so $\limsup a_n = 1$ and $\liminf a_n = -1$, although no term equals $1$ or $-1$.

Theorem. A bounded sequence converges if and only if $\liminf a_n = \limsup a_n$, and then the limit is this common value.

$\limsup$ is the largest limit of any subsequence and $\liminf$ the smallest. These notions give the sharp form of the root test in Chapter 6.

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Chapter 5 — Infinite series

Terminology

• Series — an expression $\sum_{k=1}^\infty a_k$. Its partial sums are $s_n = a_1 + a_2 + \cdots + a_n$.
• Converges — the sequence $(s_n)$ converges; its limit $S$ is the sum, written $\sum_{k=1}^\infty a_k = S$.
• Diverges — $(s_n)$ does not converge.

A series is a sequence in disguise: every question about $\sum a_k$ is a question about $(s_n)$.

Geometric and telescoping series

Geometric series. For $r \ne 1$, $s_n = \sum_{k=0}^{n-1} r^k = \frac{1 - r^n}{1 - r}$ (multiply by $1 - r$ and cancel). If $\lvert r \rvert < 1$, then $r^n \to 0$ and

$$\sum_{k=0}^\infty r^k = \frac{1}{1 - r}.$$

If $\lvert r \rvert \ge 1$ the series diverges (its terms do not tend to $0$). Example: $\sum_{k=1}^\infty \left(\tfrac12\right)^k = \frac{1}{1 - 1/2} - 1 = 1$.

Telescoping series. $\frac{1}{k(k+1)} = \frac1k - \frac{1}{k+1}$, so the partial sums collapse: $s_n = 1 - \frac{1}{n+1} \to 1$. Hence $\sum_{k=1}^\infty \frac{1}{k(k+1)} = 1$.

The divergence test

Theorem. If $\sum a_k$ converges, then $a_k \to 0$.

Proof. $a_k = s_k - s_{k-1}$, and both $s_k$ and $s_{k-1}$ tend to the same limit $S$. $\blacksquare$

So $\sum \frac{k}{k+1}$ diverges, since its terms tend to $1$. The converse is false, and the harmonic series is the standard counterexample.

The harmonic series diverges

Theorem. $\sum_{k=1}^\infty \frac1k$ diverges, although $\frac1k \to 0$.

Proof. Group the terms in blocks that end at powers of $2$:

$$s_{2^m} = 1 + \tfrac12 + \left(\tfrac13 + \tfrac14\right) + \left(\tfrac15 + \cdots + \tfrac18\right) + \cdots + \left(\tfrac{1}{2^{m-1}+1} + \cdots + \tfrac{1}{2^m}\right).$$

Each bracket has $2^{j-1}$ terms, each at least $\frac{1}{2^j}$, so each bracket is at least $\frac12$. Hence $s_{2^m} \ge 1 + \frac{m}{2}$. The partial sums are increasing and unbounded, so the series diverges. $\blacksquare$

The actual values confirm the bound: $s_2 = 1.5$, $s_4 \approx 2.083$, $s_8 \approx 2.718$, $s_{16} \approx 3.381$, $s_{32} \approx 4.058$, each at least $1 + m/2$. The growth is very slow — $s_n$ grows like $\ln n$.

The Cauchy criterion for series

Applying the Cauchy criterion to $(s_n)$: $\sum a_k$ converges if and only if for every $\varepsilon > 0$ there is $N$ such that for all $n > m \ge N$,

$$\lvert a_{m+1} + a_{m+2} + \cdots + a_n \rvert < \varepsilon.$$

"The blocks far out add up to almost nothing." The harmonic series fails this: the block from $m+1$ to $2m$ is always at least $m \cdot \frac{1}{2m} = \frac12$.

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Chapter 6 — Convergence tests

Comparison

Comparison test. Suppose $0 \le a_k \le b_k$ for all $k$.
(i) If $\sum b_k$ converges, so does $\sum a_k$.
(ii) If $\sum a_k$ diverges, so does $\sum b_k$.

Proof. The partial sums of $\sum a_k$ are increasing and bounded by $\sum b_k$; apply the Monotone Convergence Theorem. Part (ii) is the contrapositive of (i). $\blacksquare$

Example: $\sum 1/k^2$ converges. For $k \ge 2$, $\frac{1}{k^2} < \frac{1}{k(k-1)} = \frac{1}{k-1} - \frac1k$, a telescoping series with sum $1$. So $\sum_{k=1}^\infty \frac{1}{k^2} < 1 + 1 = 2$ converges. (Its exact value, $\pi^2/6 \approx 1.645$, is Euler's solution of the Basel problem.)

$p$-series. $\sum \frac{1}{k^p}$ converges if and only if $p > 1$. (Cauchy condensation or the integral test, Abbott Section 2.4.)

Ratio and root tests

Ratio test. Suppose $a_k \ne 0$ and $\lim \left\lvert \frac{a_{k+1}}{a_k} \right\rvert = r$. If $r < 1$, $\sum a_k$ converges absolutely; if $r > 1$, it diverges; if $r = 1$, the test says nothing.

Root test. Let $r = \limsup \lvert a_k \rvert^{1/k}$. If $r < 1$, $\sum a_k$ converges absolutely; if $r > 1$, it diverges; if $r = 1$, no conclusion.

Both tests compare with a geometric series: if $r < 1$, pick $q$ with $r < q < 1$; eventually $\lvert a_k \rvert \le C q^k$.

Example. $\sum_{k=1}^\infty \frac{k}{2^k}$: $\frac{a_{k+1}}{a_k} = \frac{k+1}{2k} \to \frac12 < 1$, so it converges. (Its sum is $2$.)

Watch out: $r = 1$ is genuinely inconclusive. Both $\sum 1/k$ (divergent) and $\sum 1/k^2$ (convergent) have ratio limit $1$.

Alternating series

Alternating Series Test. If $(a_k)$ is decreasing and $a_k \to 0$, then $\sum_{k=1}^\infty (-1)^{k+1} a_k = a_1 - a_2 + a_3 - \cdots$ converges.

Proof sketch. The even partial sums $s_{2n} = (a_1 - a_2) + \cdots + (a_{2n-1} - a_{2n})$ increase; the odd ones $s_{2n+1} = a_1 - (a_2 - a_3) - \cdots$ decrease; and $s_{2n} \le s_{2n+1}$ with $s_{2n+1} - s_{2n} = a_{2n+1} \to 0$. So the intervals $[s_{2n}, s_{2n+1}]$ are nested with lengths going to $0$, and all partial sums converge to the single common point. $\blacksquare$

The proof also gives the error bound: the sum lies between any two consecutive partial sums, so $\lvert S - s_n \rvert \le a_{n+1}$.

Example. The alternating harmonic series converges:

$$1 - \frac12 + \frac13 - \frac14 + \cdots = \ln 2 \approx 0.6931.$$

(The value $\ln 2$ comes from the Taylor series of $\ln(1 + x)$ evaluated at $x = 1$; the test itself gives only convergence.)

Absolute and conditional convergence

• $\sum a_k$ converges absolutely if $\sum \lvert a_k \rvert$ converges.
• It converges conditionally if it converges but $\sum \lvert a_k \rvert$ diverges.

Theorem. Absolute convergence implies convergence.

Proof. By the triangle inequality $\lvert a_{m+1} + \cdots + a_n \rvert \le \lvert a_{m+1} \rvert + \cdots + \lvert a_n \rvert$. If $\sum \lvert a_k \rvert$ satisfies the Cauchy criterion, so does $\sum a_k$. $\blacksquare$

The alternating harmonic series converges conditionally: it converges, but $\sum \frac1k$ diverges.

Rearrangements

A rearrangement of $\sum a_k$ is $\sum a_{f(k)}$ for a bijection $f : \mathbb{N} \to \mathbb{N}$ — the same terms in a different order.

Theorem. If $\sum a_k$ converges absolutely, every rearrangement converges to the same sum.

Riemann rearrangement theorem. If $\sum a_k$ converges conditionally, then for every $L \in \mathbb{R}$ there is a rearrangement that converges to $L$ (and there are rearrangements that diverge).

The idea: the positive terms alone and the negative terms alone both diverge. Add positive terms until you pass $L$, then negative terms until you drop below $L$, and repeat; since $a_k \to 0$, the overshoots shrink to $0$.

Example. Rearrange the alternating harmonic series as "one positive term, then two negative terms":

$$1 - \frac12 - \frac14 + \frac13 - \frac16 - \frac18 + \frac15 - \frac{1}{10} - \frac{1}{12} + \cdots$$

Group each block of three. Since $\frac{1}{2k-1} - \frac{1}{4k-2} = \frac{1}{4k-2}$, the $k$-th block is

$$\frac{1}{4k-2} - \frac{1}{4k} = \frac12 \left( \frac{1}{2k-1} - \frac{1}{2k} \right).$$

So the block sums add up to half of the alternating harmonic series: the rearranged series converges to $\tfrac12 \ln 2$, not $\ln 2$. (The partial sums that stop inside a block differ from the block sums by terms tending to $0$, so they have the same limit.)

Common pitfall: treating an infinite sum like a finite one. Regrouping, reordering and term-by-term manipulations are safe for absolutely convergent series, and can fail for conditionally convergent ones.

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Practice Exercises

1. Prove from the definition that $\frac{3n - 1}{n + 2} \to 3$. Find the smallest valid $N$ for $\varepsilon = 0.1$.

Solution

$\left\lvert \frac{3n-1}{n+2} - 3 \right\rvert = \frac{7}{n+2}$. Given $\varepsilon$, choose $N > 7/\varepsilon - 2$; then for $n \ge N$, $\frac{7}{n+2} \le \frac{7}{N+2} < \varepsilon$. For $\varepsilon = 0.1$ the condition is $n > 68$, so the smallest $N$ is $69$ (at $n = 68$ the distance is $\frac{7}{70} = 0.1$ exactly, which is not $< 0.1$).

2. Prove that if $a_n \to a$ then $\lvert a_n \rvert \to \lvert a \rvert$. Is the converse true?

Solution

By the reverse triangle inequality, $\big\lvert \lvert a_n \rvert - \lvert a \rvert \big\rvert \le \lvert a_n - a \rvert$, so the same $N$ works. The converse is false: $a_n = (-1)^n$ has $\lvert a_n \rvert \to 1$ but diverges. (It is true when $a = 0$.)

3. Let $x_1 = 3$ and $x_{n+1} = \frac12 x_n + 1$. Prove that $(x_n)$ converges and find its limit.

Solution

Claim: $x_n > 2$ and decreasing. If $x_n > 2$ then $x_{n+1} = \frac12 x_n + 1 > 2$. And $x_n - x_{n+1} = \frac12 x_n - 1 = \frac12(x_n - 2) > 0$. So $(x_n)$ is decreasing and bounded below by $2$, hence converges to some $L$. Taking limits in the recursion: $L = \frac12 L + 1$, so $L = 2$. (Terms: $3, 2.5, 2.25, 2.125, \dots$; in fact $x_n = 2 + 2^{1-n}$.)

4. Use subsequences to show that $a_n = \cos\left(\frac{n\pi}{2}\right)$ diverges, and find its $\limsup$ and $\liminf$.

Solution

The terms cycle $0, -1, 0, 1, \dots$ (for $n = 1, 2, 3, 4, \dots$). The subsequence $a_{4k} = 1 \to 1$ and $a_{4k+2} = -1 \to -1$ have different limits, so $(a_n)$ diverges. Every tail contains both $1$ and $-1$, so $\limsup a_n = 1$ and $\liminf a_n = -1$.

5. Decide whether each series converges: (a) $\sum \frac{1}{2^k + k}$; (b) $\sum \frac{k^2}{3^k}$; (c) $\sum \frac{1}{\sqrt{k}}$; (d) $\sum (-1)^k \frac{1}{\sqrt{k}}$.

Solution

(a) Converges: $0 < \frac{1}{2^k + k} < \frac{1}{2^k}$, a convergent geometric series. (b) Converges by the ratio test: $\frac{(k+1)^2}{3k^2} \to \frac13 < 1$. (c) Diverges: $\frac{1}{\sqrt{k}} \ge \frac1k$ and the harmonic series diverges. (d) Converges by the Alternating Series Test ($1/\sqrt{k}$ decreases to $0$), but only conditionally, by (c).

6. Show that $\sum_{k=1}^\infty \frac{1}{k(k+2)} = \frac34$.

Solution

$\frac{1}{k(k+2)} = \frac12\left(\frac1k - \frac{1}{k+2}\right)$. The partial sum telescopes, leaving the first two positive and last two negative terms: $s_n = \frac12\left(1 + \frac12 - \frac{1}{n+1} - \frac{1}{n+2}\right) \to \frac12 \cdot \frac32 = \frac34$.

7. Give an example of a bounded sequence with no convergent subsequence, or explain why none exists. Then give an unbounded sequence with a convergent subsequence.

Solution

None exists: Bolzano–Weierstrass says every bounded sequence has a convergent subsequence. An unbounded example: $a_n = n$ for odd $n$ and $a_n = 0$ for even $n$; the even terms form a constant, convergent subsequence.

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Cheat-sheet

Result	Statement	Key idea
Limit	$\forall \varepsilon\ \exists N\ \forall n \ge N:\ \lvert a_n - a\rvert < \varepsilon$	simplify, bound, solve for $n$
Example	$\frac{2n+1}{n+3} \to 2$; $\varepsilon = 0.1$ needs $N = 48$	distance $= \frac{5}{n+3}$
Unique, bounded	limits unique; convergent implies bounded	$\varepsilon/2$ trick; $\varepsilon = 1$
Algebraic limits	sums, products, quotients ($b \ne 0$)	add and subtract a middle term
Order limits	$a_n \le b_n$ implies $a \le b$	$<$ becomes $\le$
Monotone Convergence	monotone + bounded implies convergent	the limit is the sup
Bolzano–Weierstrass	bounded implies a convergent subsequence	bisection + nested intervals
Cauchy criterion	convergent iff Cauchy	BW + one subsequence
Divergence test	$a_k \not\to 0$ implies divergence	converse false
Harmonic	$\sum 1/k = \infty$	$s_{2^m} \ge 1 + m/2$
Geometric	$\sum_{k \ge 0} r^k = \frac{1}{1-r}$ for $\lvert r\rvert < 1$	$s_n = \frac{1 - r^n}{1 - r}$
Ratio / root	limit $< 1$: converges; $> 1$: diverges	compare with geometric
Alternating	decreasing to $0$: converges; error $\le a_{n+1}$	nested partial sums
Rearrangement	absolute: same sum; conditional: any sum	$\tfrac12 \ln 2$ example

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What's next

Real Analysis 02 moves from sequences to sets and functions: open and closed sets, compactness (where Bolzano–Weierstrass becomes Heine–Borel), limits of functions via the $\varepsilon$–$\delta$ definition and its sequential form, continuity, the Extreme and Intermediate Value Theorems, and uniform continuity.

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References

• Stephen Abbott, Understanding Analysis, 2nd edition, Springer, 2015 — Chapter 2 (sequences and series), Section 2.7 (rearrangements).
• Walter Rudin, Principles of Mathematical Analysis, 3rd edition, McGraw-Hill, 1976 — Chapter 3 (numerical sequences and series; Theorem 3.54 is Riemann's theorem).
• Robert G. Bartle and Donald R. Sherbert, Introduction to Real Analysis, 4th edition, Wiley, 2011 — Chapter 3 (sequences, including the number $e$) and Chapter 9 (infinite series).
• Terence Tao, Analysis I, 4th edition, Springer, 2022 — Chapters 6–7.
• Konrad Knopp, Theory and Application of Infinite Series, Dover reprint, 1990 — the classic reference for convergence tests.
• MIT OpenCourseWare, 18.100A Real Analysis, Fall 2020 (Casey Rodriguez), lecture notes: https://ocw.mit.edu/courses/18-100a-real-analysis-fall-2020/

Note: this post is in progress. Planned additions: full proofs of the quotient rule for limits and of the ratio and root tests; Cauchy condensation; and the proof of the Riemann rearrangement theorem.