Sequences & Series

A2.J.1Arithmetic series01/04

Arithmetic series and sigma notation.

Step01of 04

A series is the SUM of a sequence's terms. An arithmetic series sums an arithmetic sequence (constant difference d).

Sum of first n terms.

S_{n} = \frac{n}{2} (a_{1} + a_{n})

"Average of first and last, times count." Equivalent form using the common difference:

S_{n} = \frac{n}{2} (2 a_{1} + (n - 1) d)

A2.J.1 — Arithmetic series and sigma notation.

Step 01 of 04

A series is the SUM of a sequence's terms. An arithmetic series sums an arithmetic sequence (constant difference d).

Sum of first n terms.

S_{n} = \frac{n}{2} (a_{1} + a_{n})

"Average of first and last, times count." Equivalent form using the common difference:

S_{n} = \frac{n}{2} (2 a_{1} + (n - 1) d)

Step 02 of 04

Worked example. Sum the first 50 odd numbers: $1 + 3 + 5 + \dots$ .

$a_{1} = 1$ , $d = 2$ . The 50th term: $a_{50} = 1 + (49) (2) = 99$ .

S_{50} = \frac{50}{2} (1 + 99) = 25 \cdot 100 = 2500

Step 03 of 04

Sigma notation compactly writes a series.

k = 1 \sum n a_{k}

"Sum of $a_{k}$ as $k$ goes from 1 to $n$ ." The symbol $Σ$ (capital sigma) means SUM.

k = 1 \sum 4 (2 k + 1) = 3 + 5 + 7 + 9 = 24

Step 04 of 04

Useful sigma identities.

$\sum_{k = 1}^{n} c$	→ $c n$
$\sum_{k = 1}^{n} k$	→ $\frac{n ( n + 1 )}{2}$
$\sum_{k = 1}^{n} k^{2}$	→ $\frac{n ( n + 1 ) ( 2 n + 1 )}{6}$
$\sum (a_{k} \pm b_{k})$	→= $\sum a_{k} \pm \sum b_{k}$
$\sum c \cdot a_{k}$	→= $c \sum a_{k}$

Key insight

S_{n} = (n /2) (a_{1} + a_{n})

: average × count. Sigma notation is just shorthand for "add these up." The standard sums (1 to n, k², etc.) save you from tedious computation.

A2.J.2 — Geometric series — finite.

Step 01 of 04

Sum of the first n terms of a geometric sequence with first term $a_{1}$ and common ratio $r$ (with $r \neq = 1$ ):

S_{n} = a_{1} \cdot \frac{1 - r ^{n}}{1 - r}

Step 02 of 04

Worked example. Sum the first 6 terms of $3, 6, 12, 24, \dots$ .

$a_{1} = 3$ , $r = 2$ :

S_{6} = 3 \cdot \frac{1 - 2 ^{6}}{1 - 2} = 3 \cdot \frac{- 63}{- 1} = 189

Check: $3 + 6 + 12 + 24 + 48 + 96 = 189$ $✓$ .

Step 03 of 04

Why the formula works. Multiply $S_{n}$ by $r$ , subtract.

S_{n} r S_{n} S_{n} - r S_{n} S_{n} (1 - r) = a_{1} + a_{1} r + a_{1} r^{2} + \dots + a_{1} r^{n - 1} = a_{1} r + a_{1} r^{2} + \dots + a_{1} r^{n} = a_{1} - a_{1} r^{n} = a_{1} (1 - r^{n})

Almost every term cancels — the formula falls right out.

Step 04 of 04

Worked example with a fraction. Sum the first 5 terms of $8, 4, 2, 1, \dots$ .

$a_{1} = 8$ , $r = 1/2$ :

S_{5} = 8 \cdot \frac{1 - ( 1/2 ) ^{5}}{1 - 1/2} = 8 \cdot \frac{31/32}{1/2} = 8 \cdot \frac{31}{16} = \frac{31}{2}

Key insight

S_{n} = a_{1} (1 - r^{n}) / (1 - r)

. Works for any

r \neq = 1

. The trick "multiply by

r

and subtract" is the canonical telescoping argument — it shows up again for infinite series.

A2.J.3 — Geometric series — infinite, with convergence.

Step 01 of 04

What happens as $n \to \infty$ ? Look at $r^{n}$ in the finite-sum formula.

$∣ r ∣ < 1$	→ $r^{n} \to 0$ — the series CONVERGES
$∣ r ∣ \geq 1$	→ $r^{n}$ blows up or oscillates — the series DIVERGES

Convergent infinite geometric sum.

S_{\infty} = \frac{a _{1}}{1 - r} (∣ r ∣ < 1)

Step 02 of 04

Worked example. Sum $8 + 4 + 2 + 1 + \frac{1}{2} + \dots$ .

$a_{1} = 8$ , $r = 1/2$ , $∣ r ∣ < 1$ → converges.

S_{\infty} = \frac{8}{1 - 1/2} = \frac{8}{1/2} = 16

The infinite sum is FINITE. Each term is half the previous, so the total never quite reaches 16 but gets arbitrarily close.

Step 03 of 04

Repeating decimals as geometric series.

Write $0. \overline{3} = 0.333 \dots$ as $\frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \dots$ with $a_{1} = 3/10$ , $r = 1/10$ :

S_{\infty} = \frac{3/10}{1 - 1/10} = \frac{3/10}{9/10} = \frac{1}{3}

Confirms $0. \overline{3} = 1/3$ exactly.

Step 04 of 04

Divergent example. $1 + 2 + 4 + 8 + \dots$ has $r = 2$ , $∣ r ∣ \geq 1$ → diverges. No finite sum exists.

The convergence test is a HARD condition: $∣ r ∣ < 1$ exactly. Even $r = 1$ (constant series) doesn't converge for any nonzero $a_{1}$ .

Key insight

Infinite geometric series converges iff

∣ r ∣ < 1

. When it does, sum is

a_{1} / (1 - r)

. Repeating decimals are the classic application — every repeating decimal is a rational number, and this proves it.

Free diagnostic

Test your understanding: 10 questions, ~10 min.

Pulled live from our bank for A2.J · scored instantly with worked solutions · free, no sign-up.

Start diagnostic →Print worksheets

A2.J.1 – A2.J.3 · CCSS A.SSE.B.4·Full curriculum →