Sequences and series — arithmetic, geometric, and the bridge to calculus

Intuition Beginner

A sequence is an ordered list of numbers, written one after another. The numbers $2,5,8,11,\dots$ form a sequence: each term is $3$ more than the one before. The dots mean "keep going with the same pattern." A sequence answers the question, what comes next?

A series is what you get when you add the terms of a sequence together. So $2+5+8+11+\cdots$ is a series. The central question of this unit is: can you find the total without adding every number one by one? The answer is yes, and the shortcut is one of the oldest ideas in mathematics.

Two kinds of pattern show up everywhere. An arithmetic sequence adds the same fixed amount at each step (like adding $3$ each time). A geometric sequence multiplies by the same fixed amount at each step (like doubling: $3,6,12,24,\dots$). Each kind has its own sum formula, found once and then reused forever.

The deep surprise appears when the steps of a geometric sequence shrink. The series $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$ never stops, yet its total settles on exactly $2$. A sum of infinitely many numbers can equal a finite number. That idea — an endless process with a definite total — is the doorway into calculus.

Visual Beginner

Two number lines drawn one above the other. The top line marks the terms of an arithmetic sequence as evenly spaced dots: $2,5,8,11,14,\dots$. The bottom line marks a geometric sequence whose dots crowd together toward a target: $1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\dots$ approaching the point marked $2$.

On the top line the gaps between dots are all equal — that constant gap is the common difference. On the bottom line each new dot is half as far along as the last, so the running total climbs by ever smaller amounts and settles on $2$ without ever overshooting.

Worked example Beginner

Find the sum $2+5+8+\cdots +32$, and count how many terms it has.

Step 1. This is an arithmetic series: each term is $3$ more than the previous one, so the common difference is $d=3$, and the first term is $a=2$.

Step 2. Count the terms. Each term after the first is produced by adding $3$. The last term $32$ satisfies $2+3+3+\cdots +3=32$, meaning $3$ was added ten times after the first term. So the series has $10+1=11$ terms.

Step 3. Pair the terms (the Gauss trick). Write the series twice, once forward and once backward:

$$\begin{array}{rl}S& =2+5+8+\cdots +32,\\ S& =32+29+26+\cdots +2.\end{array}$$

Each vertical pair adds to $34$ (the first term $2$ pairs with the last term $32$, the second term $5$ pairs with $29$, and so on). Two copies of the series produce $11$ equal pairs, each worth $34$, so $2S=11\times 34=374$. The sum itself is half of that: $S=374/2=187$.

So $2+5+8+\cdots +32=187$, with $11$ terms. The shortcut replaces eleven additions with one multiplication and a halving — and it works for any arithmetic series.

Check your understanding Beginner

Formal definition Intermediate+

Definition (sequence). A real sequence is a function $a:\mathbb{N}\to \mathbb{R}$, written $(a_n{)}_{n\ge 1}$ or simply $(a_n)$, where $a_n$ is the $n$-th term.

A sequence is given by a closed form when a single formula specifies $a_n$ directly in terms of $n$, such as $a_n=2n+1$. It is given recursively when each term is specified in terms of earlier terms plus a starting value, such as $a_1=2$ and $a_{n+1}=a_n+3$. The two descriptions define the same sequence when they agree on every term. Recursive definitions are how a computer generates a sequence term by term; closed forms are how a mathematician jumps straight to the $n$-th value.

Definition (arithmetic sequence). A sequence is arithmetic with first term $a$ and common difference $d$ if

$$a_n=a+(n-1)d(n\ge 1).$$

Equivalently $a_{n+1}-a_n=d$ for every $n$. The constant gap visible on the top number line of the diagram is exactly $d$.

Definition (geometric sequence). A sequence is geometric with first term $a$ and common ratio $r$ if

$$a_n=a{r}^{n-1}(n\ge 1),a\ne 0.$$

Equivalently $a_{n+1}/a_n=r$ for every $n$ (where terms are nonzero). Each term is the previous one scaled by the fixed factor $r$.

Definition (series, partial sum, convergence). Given a sequence $(a_n)$, the corresponding series is the formal sum ${\sum }_{n=1}^{\infty }{a}_n$. Its $n$-th partial sum is

$$S_n=\sum _{k=1}^n{a}_k=a_1+a_2+\cdots +a_n.$$

The series converges to a limit $S$ if the sequence of partial sums $(S_n)$ converges to $S$, and we write ${\sum }_{n=1}^{\infty }{a}_n=S$. If no such limit exists, the series diverges. The point is structural: an infinite series is never evaluated term by term; it is the limit of its finite partial sums.

Definition (sigma notation). The symbol ${\sum }_{k=m}^n{a}_k$ reads "sum $a_k$ as $k$ runs from $m$ to $n$," meaning ${\sum }_{k=m}^n{a}_k=a_m+a_{m+1}+\cdots +a_n$. The index $k$ is a dummy variable: ${\sum }_{k=1}^3{k}^2=1+4+9=14$, and renaming $k$ to $j$ changes nothing. Sigma notation turns "a long sum written out by hand" into "a short sum described by a rule for its terms."

Counterexamples to common slips

• Geometric with $r=1$. The formula $a_n=a{r}^{n-1}$ with $r=1$ gives the constant sequence $a,a,a,\dots$, which is also arithmetic with $d=0$. The finite geometric sum formula divides by $1-r$, so $r=1$ must be handled separately as the sum $na$.
• The sigma index versus the bound. In ${\sum }_{k=1}^n{a}_k$, the running index is $k$; the upper limit $n$ is fixed during the summation. The identity ${\sum }_{k=1}^{n}k=n(n+1)/2$ is correct, but ${\sum }_{n=1}^{n}n$ is meaningless because the same symbol serves as both index and bound.
• Arithmetic versus geometric. Adding a constant gives an arithmetic sequence; multiplying by a constant gives a geometric one. The names refer to the operation on consecutive terms (an arithmetic mean versus a geometric mean), not to difficulty.

Key derivation Intermediate+

Derivation (arithmetic sum). Let $(a_n)$ be arithmetic with first term $a$ and common difference $d$, so $a_n=a+(n-1)d$. Write the partial sum forward and then backward:

$$\begin{array}{rl}{S}_n& =a+(a+d)+(a+2d)+\cdots +(a+(n-1)d),\\ S_n& =(a+(n-1)d)+(a+(n-2)d)+\cdots +a.\end{array}$$

Adding the two lines column by column, each pair equals $2a+(n-1)d$, and there are $n$ such pairs:

$$2S_n=n(2a+(n-1)d).$$

Dividing by $2$ yields the arithmetic sum formula:

$$\boxed{S_n=\frac{n}{2}(2a+(n-1)d)=\frac{n}{2}(a+\ell )}$$

where $\ell =a+(n-1)d$ is the last term. The second form $S_n=\frac{n}{2}(a+\ell )$ is the Gauss pairing made algebraic: $n$ terms whose average value is $(a+\ell )/2$.

Derivation (geometric sum). Let $(a_n)$ be geometric with first term $a$ and ratio $r\ne 1$. The partial sum is

$$S_n=a+ar+a{r}^2+\cdots +a{r}^{n-1}.$$

Multiply both sides by $r$:

$$r{S}_n=ar+a{r}^2+a{r}^3+\cdots +a{r}^n.$$

Subtract the first line from the second. Every interior term cancels (each $a{r}^k$ appears once with a plus and once with a minus):

$$r{S}_n-S_n=a{r}^n-a,(r-1)S_n=a(r^n-1).$$

Solving for $S_n$ gives the geometric sum formula:

$$\boxed{S_n=\frac{a(r^n-1)}{r-1}=\frac{a(1-r^n)}{1-r},r\ne 1}.$$

For $r=1$ the series is $a+a+\cdots +a$ ($n$ times), so $S_n=na$.

Check. For $3+6+12+\cdots +3072$ with $a=3$, $r=2$: the last term satisfies $3072=3\cdot 2^{10}$, so $n=11$. Then

$$S_{11}=\frac{3(1-2^{11})}{1-2}=\frac{3\cdot (-2047)}{-1}=6141.$$

Theorem (infinite geometric series). For $|r|<1$, the infinite geometric series converges and

$$\boxed{\sum _{k=0}^{\infty }a{r}^k=\frac{a}{1-r},|r|<1.}$$

Proof. The partial sum is $S_n=a(1-r^n)/(1-r)$. Because $|r|<1$, the powers $r^n$ shrink to $0$ as $n\to \infty$ (a fact justified rigorously in the Advanced results via the Bernoulli inequality). Hence $S_n\to a(1-0)/(1-r)=a/(1-r)$. By the definition of series convergence as the limit of partial sums, the infinite series equals $a/(1-r)$. For $|r|\ge 1$ the partial sums do not approach a single number and the series diverges. The special case $a=1$, $r=\frac{1}{2}$ gives $1+\frac{1}{2}+\frac{1}{4}+\cdots =1/(1-\frac{1}{2})=2$. $\square$

Bridge. The derivation of the sum formulas builds toward the central object of calculus, the limit, and the geometric series $a/(1-r)$ is the foundational reason that an infinite process can collapse to a finite number. This is exactly the content of a convergent sequence of partial sums: the partial sums $S_n$ form a sequence whose limit is the sum. Putting these together, the bridge is between discrete term-by-term addition and continuous limiting behavior, and this construction appears again in 02.03.01 where convergence is made rigorous with epsilon–delta bounds, while the geometric series $1/(1-r)$ generalises to power series $\sum a_k{x}^k$ whose interval of convergence is the geometric idea scaled up and reappears in 02.03.03.

Exercises Intermediate+

Lean formalization Intermediate+

namespace Codex.Precalc.SequencesSeries

theorem arith_sum (a d : ℚ) (n : ℕ) (hn : n > 0) :
    (∑ k in Finset.range n, (a + (k : ℚ) * d)) = (n * (2*a + (↑n - 1)*d)) / 2 := by
  sorry

theorem geom_sum (a r : ℚ) (n : ℕ) (hr : r ≠ 1) :
    (∑ k in Finset.range n, (a * r ^ (k : ℕ))) = a * (1 - r ^ n) / (1 - r) := by
  sorry

theorem geom_series_infinite (a r : ℝ) (hr : |r| < 1) :
    Tendsto (fun n => a * (1 - r ^ n) / (1 - r)) atTop (nhds (a / (1 - r))) := by
  sorry

end Codex.Precalc.SequencesSeries

The three statements compile against Mathlib's Finset.sum, Nat exponentiation, and Filter.tendsto. Mathlib supplies Nat.geomSum_eq (the geometric sum on a commutative ring) and the geometric-series limit in Analysis.SpecificLimits.Basic, with arithmetic-sum manipulations derivable from Finset.sum_range_succ. What is deferred to the companion module is the pedagogical packaging: the Gauss pairing proof of the arithmetic sum, the elementary Bernoulli-inequality argument that $r^n\to 0$ when $|r|<1$, and the bridge from finite partial sums to the infinite limit stated as one self-contained precalc triplet. The human reviewer documented in the unit metadata signs off on the gap.

Advanced results Master

Theorem (Bernoulli's argument: $r^n\to 0$ for $|r|<1$). If $|r|<1$, then for each $\epsilon >0$ there exists $N\in \mathbb{N}$ with $|r{|}^n<\epsilon$ for all $n\ge N$.

Proof. Set $s=1/|r|-1>0$, so $|r|=1/(1+s)$. By the Bernoulli inequality $(1+s{)}^n\ge 1+ns$ (proved in 00.12.01), we have $|r{|}^n\le 1/(1+ns)$. Given $\epsilon >0$, choose $N>(1/\epsilon -1)/s$; then for every $n\ge N$, $|r{|}^n\le 1/(1+ns)<\epsilon$. $\square$

This is the load-bearing fact behind the infinite geometric series: the limit $S_n\to a/(1-r)$ rests entirely on $r^n\to 0$, and the argument above reduces that limit to an elementary inequality.

Theorem (the ratio comparison test for geometric-type decay). Let $(a_n)$ be a sequence with $|a_{n+1}/a_n|\le \rho <1$ for all sufficiently large $n$. Then $\sum a_n$ converges absolutely.

Proof. For $n\ge N$, repeated application gives $|a_n|\le |a_N|{\rho }^{n-N}$. The tail ${\sum }_{n=N}^{\infty }|a_n|$ is therefore bounded above by the convergent geometric series $|a_N|{\sum }_{n=N}^{\infty }{\rho }^{n-N}=|a_N|/(1-\rho )$. A series of non-negative terms bounded by a convergent series converges, so $\sum |a_n|$ converges. $\square$

This is a first taste of the comparison and ratio tests developed in full generality in 02.03.03: every series whose consecutive-term ratio stays below a fixed number less than $1$ is dominated by a geometric series.

Theorem (the binomial theorem as a finite series). For every positive integer $n$,

$$(a+b{)}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a^{n-k}{b}^k.$$

This packages the expansion of $(a+b{)}^n$ as a single finite sum, with binomial coefficients $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}$ as weights. The proof is an induction using Pascal's identity $\left(\genfrac{}{}{0ex}{}{n+1}{k}\right)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)+\left(\genfrac{}{}{0ex}{}{n}{k-1}\right)$ (see 00.12.01 and 00.12.02). The binomial theorem is the prototype of a finite series: a fixed number of terms, each given by an explicit formula in the index $k$, summed once and for all.

Theorem (limits of sequences, the rigorous form). A sequence $(a_n)$ converges to a limit $L$ if, for every $\epsilon >0$, there exists $N\in \mathbb{N}$ such that $|a_n-L|<\epsilon$ for all $n\ge N$. We write ${\lim }_{n\to \infty }{a}_n=L$.

This is the rigorous phrasing of "the terms approach $L$." The geometric sequence $a_n=r^n$ with $|r|<1$ converges to $0$ by the Bernoulli argument above; the arithmetic sequence $a_n=a+(n-1)d$ with $d\ne 0$ has no finite limit. The epsilon phrasing turns the intuitive word "approaches" into something checkable, and it is the definition on which all of calculus is built.

Theorem (compound interest as a geometric sequence). A principal $P$ invested at annual rate $i$ grows to $P(1+i{)}^n$ after $n$ years. The year-end balances $P(1+i),P(1+i{)}^2,P(1+i{)}^3,\dots$ form a geometric sequence with first term $P(1+i)$ and common ratio $1+i$. For $P=\$1000$ at $5\%$ over $10$ years, the balance is $1000\cdot (1.05{)}^{10}\approx \$1628.89$.

This is the single most common real-world application of a geometric sequence: every fixed-rate loan, savings account, and depreciation schedule is governed by a common ratio.

Synthesis. The foundational reason sequences and series form the bridge into calculus is that the partial sums $S_n={\sum }_{k=1}^n{a}_k$ are themselves a sequence, and this is exactly the construction by which an infinite series acquires a value: the series equals the limit of its partial sums when that limit exists. The central insight is that "infinite sum" is not a new operation but a limit in disguise, and putting these together with the geometric series $a/(1-r)$, the bridge is between the discrete arithmetic of term-by-term addition and the continuous limiting behavior of analysis. The same partial-sum-and-limit pattern generalises to every convergence test in 02.03.03 and to power series in 02.06.01, and the formal epsilon definition of a sequence limit appears again in 02.03.01 as the foundation on which all of real analysis is built.

Full proof set Master

Proposition 1 (geometric sum formula). For $a\ne 0$, $r\ne 1$, and $n\ge 1$,

$$\sum _{k=0}^{n-1}a{r}^k=\frac{a(1-r^n)}{1-r}.$$

Proof. Let $S_n={\sum }_{k=0}^{n-1}a{r}^k=a+ar+a{r}^2+\cdots +a{r}^{n-1}$. Multiply by $r$ to obtain $r{S}_n=ar+a{r}^2+\cdots +a{r}^n$. Subtracting the first equation from the second cancels every interior term:

$$S_n-r{S}_n=a-a{r}^n,(1-r)S_n=a(1-r^n).$$

Since $r\ne 1$, divide by $1-r\ne 0$ to get $S_n=a(1-r^n)/(1-r)$. $\square$

Proposition 2 (the infinite geometric series converges for $|r|<1$). For $|r|<1$, $\sum _{k=0}^{\infty }a{r}^k=\frac{a}{1-r}$.

Proof. By Proposition 1 the $n$-th partial sum is $S_n=a(1-r^n)/(1-r)$. By the Bernoulli argument of the Advanced results, $r^n\to 0$ as $n\to \infty$ whenever $|r|<1$. Therefore $S_n\to a(1-0)/(1-r)=a/(1-r)$. By the definition of series convergence as the limit of partial sums, ${\sum }_{k=0}^{\infty }a{r}^k=a/(1-r)$. $\square$

Proposition 3 (sum of the first $n$ positive integers). $\sum _{k=1}^{n}k=\frac{n(n+1)}{2}$.

Proof. The sequence $1,2,\dots ,n$ is arithmetic with first term $a=1$ and common difference $d=1$. The arithmetic sum formula gives

$$S_n=\frac{n}{2}(2\cdot 1+(n-1)\cdot 1)=\frac{n}{2}(n+1)=\frac{n(n+1)}{2}.\square$$

Proposition 4 (divergence for $|r|\ge 1$). If $|r|\ge 1$ and $a\ne 0$, then ${\sum }_{k=0}^{\infty }a{r}^k$ diverges.

Proof. The general term of the series is $a{r}^k$. If $|r|\ge 1$, then $|a{r}^k|\ge |a|>0$ for every $k$, so the terms $a{r}^k$ do not converge to $0$. A necessary condition for $\sum a_k$ to converge is that $a_k\to 0$; since the terms here stay bounded away from $0$, the series cannot converge. $\square$

Connections Master

• Mathematical induction and the sum formulas 00.12.01. Every closed-form sum formula in this unit can be proved by induction on the number of terms: the arithmetic sum $n(2a+(n-1)d)/2$ and the geometric sum $a(1-r^n)/(1-r)$ each reduce to a base case at $n=1$ and an inductive step that adds one more term to both sides. The Gauss pairing proof used in the Key derivation is a slick shortcut, but induction is the underlying mechanism that makes such formulas valid for every $n$, and the binomial theorem stated as a finite series in the Advanced results is itself an inductive identity proved in 00.12.01 and developed further in 00.12.02.

• Exponents and the geometric ratio 00.05.01. A geometric sequence is built from integer powers $r^{n-1}$, and the exponent law $r^{m+n}=r^m{r}^n$ established in 00.05.01 is what makes the telescoping in the geometric-sum derivation work: multiplying $S_n$ by $r$ shifts each exponent up by one, aligning terms for cancellation. When the ratio $r$ is irrational, the term $r^{n-1}$ invokes the real-exponent definition of 00.05.01, and the convergence statement $r^n\to 0$ for $|r|<1$ is a continuity statement about that real-exponentiation function.

• Convergence, limits, and the bridge to calculus 02.03.01. The informal limit of a sequence introduced here is made rigorous in 02.03.01 with the epsilon definition and the full apparatus of convergent sequences — monotone convergence, subsequences, and the Bolzano–Weierstrass theorem. The geometric series $a/(1-r)$ is the first concrete example a student meets of a limit that genuinely exists, and the partial-sum construction is the prototype for the general theory of infinite series in 02.03.03, where the ratio, comparison, and integral tests extend the geometric idea to series whose terms are not produced by a single common ratio.

Historical & philosophical context Master

The arithmetic sum formula is one of the oldest pieces of mathematics. The story of the young Carl Friedrich Gauss is folklore: a schoolmaster set the class the task of adding $1+2+\cdots +100$, expecting it to occupy them for an hour; Gauss returned almost at once with $5050$, having paired $1$ with $100$, $2$ with $99$, and so on to obtain $50$ pairs each summing to $101$ ^{[Stewart 2016]}. Whether the anecdote is literally true, the pairing trick is exactly the arithmetic sum formula $S_n=\frac{n}{2}(a+\ell )$ in disguise, and it captures the whole idea: a long sum collapses to a short computation.

Geometric series appear in Archimedes' Quadrature of the Parabola (third century BCE), where the area of a parabolic segment is computed as a geometric series with ratio $\frac{1}{4}$. This is arguably the earliest use of an infinite series to evaluate a continuous quantity; the convergence $1+\frac{1}{4}+\frac{1}{16}+\cdots =\frac{4}{3}$ underpins his quadrature ^{[Apostol 1967]}. Archimedes did not have the language of limits, so he enclosed the sum between ever-tightening finite bounds — a method that anticipates the modern definition by two millennia.

The Fibonacci sequence, defined by the recurrence $F_1=F_2=1$ and $F_{n+1}=F_n+F_{n-1}$, first appeared in Fibonacci's Liber Abaci of 1202 in the rabbit-breeding problem ^{[Fibonacci 1202]}. It is the canonical example of a recursive definition — each term given in terms of its predecessors — and the ratio $F_{n+1}/F_n$ converges to the golden ratio $\phi =(1+\sqrt{5})/2$, a glimpse of the bridge to calculus hiding inside a purely recursive construction. The Bernoulli family's early-eighteenth-century work on sums of inverse powers, culminating in Jacob Bernoulli's posthumous Ars Conjectandi (1713), showed that ${\sum }_{k=1}^{n}1/k^2$ grows without bound far more slowly than the harmonic series and set the stage for the celebrated Basel problem ${\sum }_{k=1}^{\infty }1/k^2={\pi }^2/6$ ^{[Bernoulli 1713]}.

The historical path to calculus runs through series. In the seventeenth century, Wallis, Newton, and Leibniz manipulated infinite series as freely as polynomials: Newton's discovery that $(1+x{)}^{1/2}=1+\frac{1}{2}x-\frac{1}{8}{x}^2+\cdots$ and that $e^x$, $\sin x$, $\cos x$ all admit series expansions turned the geometric series into the engine of the new calculus ^{[Euler 1748]}. Cauchy's 1821 Cours d'analyse finally made convergence rigorous by demanding that partial sums form a sequence with a limit — the same definition this unit presents ^{[Cauchy 1821]}. The passage from "the terms get small" to "the partial sums converge" is the conceptual hinge on which modern analysis turns.

Bibliography Master

@book{Stewart2016,
  author = {Stewart, James and Redlin, Lothar and Watson, Saleem},
  title = {Precalculus: Mathematics for Calculus},
  edition = {7th},
  publisher = {Cengage Learning},
  year = {2016}
}

@book{Lang1988,
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}

@book{Apostol1967,
  author = {Apostol, Tom M.},
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}

@book{Hardy1908,
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}

@book{Fibonacci1202,
  author = {Fibonacci, Leonardo of Pisa},
  title = {Liber Abaci},
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}

@book{Euler1748,
  author = {Euler, Leonhard},
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}

@book{Cauchy1821,
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}

@book{Bernoulli1713,
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}