Sequences & Series

Sequences are functions

A sequence is a function $a: \mathbb{N} \to \mathbb{R}$, usually written $a_1, a_2, a_3, \ldots$ or $(a_n)_{n=1}^{\infty}$. The index $n$ is the input; $a_n$ is the output.

Two ways to define a sequence:

• Explicit (closed-form): $a_n = 3n + 1$ gives the value directly from $n$.
• Recursive: $a_1 = 4$, $a_{n+1} = a_n + 3$ — gives the next term from the previous.

Same sequence, two descriptions. The explicit form is almost always more useful for finding distant terms; the recursive form is usually easier to discover. Pick your weapon based on the problem. Don't be the person who grinds through a recursive formula 47 times when the explicit form was right there.

Arithmetic sequences: linear in disguise

An arithmetic sequence has a constant difference $d$ between consecutive terms: $a_{n+1} - a_n = d \quad \text{for all } n.$

The general term: $a_n = a_1 + (n-1)d.$

This is a linear function of $n$ — slope $d$, y-intercept (in terms of $n=0$) would be $a_1 - d$. Everything you know about lines applies: constant rate of change, predictable behavior.

Common difference $d > 0$: increasing. $d < 0$: decreasing. $d = 0$: constant sequence.

Geometric sequences: exponential in disguise

A geometric sequence has a constant ratio $r$ between consecutive terms: $\frac{a_{n+1}}{a_n} = r \quad \text{for all } n.$

The general term: $a_n = a_1 \cdot r^{n-1}.$

This is an exponential function of $n$ — the same constant-ratio structure from the Exponential Functions lesson, now counting discrete steps $n = 1, 2, 3, \ldots$ instead of continuous $x$.

$r > 1$: exponential growth. $0 < r < 1$: exponential decay. $r < 0$: oscillating (alternating signs).

Sigma notation

The sum $a_1 + a_2 + \cdots + a_n$ is written: $\sum_{k=1}^{n} a_k.$

Read: "sum from $k = 1$ to $n$ of $a_k$." The variable $k$ is a dummy index — it runs through $1, 2, \ldots, n$ and doesn't exist outside the sum. Properties:

$\sum_{k=1}^{n}(a_k + b_k) = \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} b_k, \qquad \sum_{k=1}^{n}(c \cdot a_k) = c \sum_{k=1}^{n} a_k.$

These are linearity properties, and they follow from the commutativity and associativity of addition.

Sum of an arithmetic series: the pairing trick

Theorem. $\displaystyle\sum_{k=1}^{n} a_k = \frac{n(a_1 + a_n)}{2}$ for an arithmetic sequence.

Proof. Let $S = a_1 + a_2 + \cdots + a_n$. Write $S$ forward AND backward: $S = a_1 + a_2 + \cdots + a_n$ $S = a_n + a_{n-1} + \cdots + a_1.$

Add the two expressions, pairing corresponding terms. Each pair sums to $a_1 + a_n$ (the first plus the last — and since the sequence is arithmetic, every pair has the same sum): $2S = n(a_1 + a_n).$ $S = \frac{n(a_1 + a_n)}{2}. \quad \square$

This is the Gauss pairing argument — allegedly the trick Gauss used as a child to compute $1 + 2 + \cdots + 100 = 5050$ in seconds while his teacher was expecting him to suffer through it. The proof is one line once you see it, and once you see it you can't unsee it. That's how all great mathematics works.

Induction cameo. We could also prove $\sum_{k=1}^{n} k = n(n+1)/2$ by induction (base case $n=1$: $1 = 1(2)/2$; inductive step: assume it holds for $k$, show for $k+1$). The pairing proof is cleaner, but induction works — see the Induction node.

Sum of a geometric series: the multiply-and-subtract trick

Finite sum. $\displaystyle\sum_{k=0}^{n-1} a r^k = a \cdot \frac{r^n - 1}{r - 1}$ for $r \ne 1$.

Proof. Let $S = a + ar + ar^2 + \cdots + ar^{n-1}$. Multiply by $r$: $rS = ar + ar^2 + \cdots + ar^{n-1} + ar^n.$ Subtract $S$ from $rS$: almost everything cancels: $rS - S = ar^n - a.$ $S(r-1) = a(r^n - 1).$ $S = a \cdot \frac{r^n - 1}{r - 1}. \quad \square$

This "multiply by $r$ and subtract" trick is the standard proof — elegant, surgical, and worth understanding deeply. Memorize the trick, not the formula. The formula follows from the trick in 30 seconds, and if you forget the formula on an exam you can re-derive the whole damn thing on a napkin. That's the power of actually understanding the proof.

Infinite geometric series

When $|r| < 1$, the terms $ar^{n-1}$ shrink toward 0 as $n \to \infty$. The finite sum becomes: $\sum_{k=0}^{\infty} ar^k = \lim_{n \to \infty} a \cdot \frac{r^n - 1}{r - 1} = a \cdot \frac{0 - 1}{r - 1} = \frac{a}{1 - r}.$

(Since $|r| < 1$: $r^n \to 0$, so the numerator $r^n - 1 \to -1$.)

When $|r| \ge 1$, the terms don't shrink; the series diverges (the sum grows without bound or oscillates). No finite answer.

The $0.\overline{9} = 1$ fight. $0.999\ldots = 9/10 + 9/100 + 9/1000 + \cdots = \dfrac{9/10}{1 - 1/10} = \dfrac{9/10}{9/10} = 1$.

This is not a philosophical opinion, not a limit "approaching" something, not a rounding error. It is an exact equality, derived directly from the infinite geometric series formula. The decimal $0.\overline{9}$ and the integer $1$ are the same real number, full stop. Anyone who argues otherwise in the comments section is wrong. Provably, mathematically, demonstrably wrong.