Topic · Algebra

Sequences and Series

A sequence is an ordered list of numbers with a rule; a series is what you get when you add the list up. From this simple pair of ideas comes recursion, infinite sums with finite values, and the first real taste of what "convergence" means.

What you'll leave with

  • The two ways to define a sequence — recursive (next from previous) and closed-form (term directly from $n$) — and how to convert between them.
  • The two families that cover most of pre-calculus: arithmetic ($a_n = a_1 + (n-1)d$) and geometric ($a_n = a_1 r^{n-1}$).
  • The arithmetic-series sum via Gauss's pairing trick, and the geometric-series sum via the "shift-and-subtract" trick.
  • The infinite geometric series $\dfrac{a}{1-r}$, when it converges ($|r| < 1$), and why $0.\overline{9} = 1$ falls out of it.
  • A first informal picture of convergence, and why the harmonic series breaks the obvious-looking rule.

1. Sequences as functions on the naturals

Sequence

An ordered list of numbers $a_1, a_2, a_3, \ldots$ indexed by the positive integers. Equivalently, a function $a : \mathbb{N} \to \mathbb{R}$ that maps each position $n$ to its term $a_n$.

The word "ordered" carries the whole idea. $\{1, 2, 3\}$ as a set is the same as $\{3, 1, 2\}$, but as sequences they are completely different objects: the first is $a_1=1, a_2=2, a_3=3$; the second is $a_1=3, a_2=1, a_3=2$. The index matters because the index is the input to the function.

A sequence can be finite (ends at some $a_N$) or infinite (continues forever). In this page we mostly mean infinite, but every finite-sum formula we write works just as well for the first $N$ terms of an infinite sequence.

Sequence vs series

A sequence is a list: $1, \tfrac{1}{2}, \tfrac{1}{3}, \ldots$. A series is what you get when you put plus signs between the entries: $1 + \tfrac{1}{2} + \tfrac{1}{3} + \cdots$. They look almost the same on the page and are constantly confused, but they are different objects with different behaviours — the sequence above has limit $0$, while the series (called the harmonic series) actually grows without bound.

2. Two ways to define a sequence

There are two equally legitimate ways to pin down what $a_n$ is.

Recursive definition

Give a starting value (or several) and a rule that produces each new term from the previous one(s).

$$ a_1 = 5, \qquad a_{n+1} = a_n + 4 $$

To find $a_4$ you have to walk: $a_1 = 5$, $a_2 = 9$, $a_3 = 13$, $a_4 = 17$. The rule is easy to state but you can't skip ahead.

Closed-form (explicit) definition

Give a formula for $a_n$ directly in terms of $n$.

$$ a_n = 4n + 1 $$

Now $a_{1000}$ is one line of arithmetic: $4(1000) + 1 = 4001$. The closed-form is harder to discover but much faster to evaluate.

For the two families we're about to meet — arithmetic and geometric — converting between recursive and closed-form is mechanical. For other recurrences (Fibonacci, $F_{n+1} = F_n + F_{n-1}$, comes to mind) it is genuinely hard and a topic for later.

Why bother with both?

Recursive definitions match how processes actually unfold — month-by-month balances, generation-by-generation population counts. Closed-form definitions are how you predict the state at a far-future time without simulating every step in between. Whenever you have a recursive definition and can find its closed-form, you've essentially solved the dynamics.

3. Arithmetic sequences

Arithmetic sequence

A sequence in which the difference between any two consecutive terms is the same constant $d$, called the common difference.

Recursive: $a_{n+1} = a_n + d$, with $a_1$ given.
Closed-form: $a_n = a_1 + (n-1)d$.

The closed-form is almost an identity once you see it: starting at $a_1$ and adding $d$ a total of $n-1$ times lands you at the $n$-th term. The trickiest thing about it is the off-by-one — the exponent on $d$ is $n-1$, not $n$, because the first term has had zero increments applied to it.

Examples and non-examples

Arithmetic
  • $3, 7, 11, 15, \ldots$ — common difference $d = 4$
  • $10, 7, 4, 1, -2, \ldots$ — $d = -3$
  • $\tfrac{1}{2}, 1, \tfrac{3}{2}, 2, \ldots$ — $d = \tfrac{1}{2}$
  • Any constant sequence — $d = 0$
Not arithmetic
  • $2, 4, 8, 16, \ldots$ — ratio is constant, not difference
  • $1, 4, 9, 16, \ldots$ — differences $3, 5, 7, \ldots$ change
  • $1, 1, 2, 3, 5, \ldots$ — Fibonacci; neither difference nor ratio constant

To test whether a sequence is arithmetic, compute differences of consecutive terms. If they're all the same number, you have your $d$ and can write the formula immediately.

4. Summing an arithmetic series: Gauss's trick

A famous (probably apocryphal) story: a teacher trying to keep the seven-year-old Carl Friedrich Gauss busy told the class to add the integers from 1 to 100. Gauss produced the answer in a few seconds. The trick he used solves any arithmetic series.

Write the sum forward and then write it again backward:

$$ \begin{aligned} S_n &= a_1 + a_2 + \cdots + a_{n-1} + a_n \\ S_n &= a_n + a_{n-1} + \cdots + a_2 + a_1 \end{aligned} $$

Now add them column by column. In an arithmetic sequence each column sums to the same thing — $a_1 + a_n$ — because as one row goes up by $d$ per step, the other goes down by $d$ per step, so the pair stays put.

$$ 2 S_n = n (a_1 + a_n) \quad\Longrightarrow\quad \boxed{\, S_n = \frac{n(a_1 + a_n)}{2} \,} $$

"The number of terms, times the average of the first and last, all over two." Substitute $a_n = a_1 + (n-1)d$ if you only know the common difference rather than the last term:

$$ S_n = \frac{n\bigl(2 a_1 + (n-1)d\bigr)}{2} $$

Gauss's original problem: $a_1 = 1$, $a_{100} = 100$, $n = 100$, so $S_{100} = \frac{100 \cdot 101}{2} = 5050$.

forward 1 2 3 98 99 100 backward 100 99 98 3 2 1 column sum 101 101 101 101 101 101 100 × 101 = 10100 ÷ 2 = 5050

5. Geometric sequences

Geometric sequence

A sequence in which the ratio between any two consecutive terms is the same constant $r$, called the common ratio.

Recursive: $a_{n+1} = r \cdot a_n$, with $a_1$ given.
Closed-form: $a_n = a_1 \, r^{n-1}$.

Where arithmetic sequences model linear change (something added each step), geometric sequences model multiplicative change (something scaled each step). Compound interest, population growth, radioactive decay, half-lives — all geometric.

The closed-form has the same off-by-one as the arithmetic one. The first term has had zero multiplications by $r$ applied, so the exponent of $r$ on the $n$-th term is $n-1$, not $n$.

Examples and non-examples

Geometric
  • $2, 6, 18, 54, \ldots$ — common ratio $r = 3$
  • $1, \tfrac{1}{2}, \tfrac{1}{4}, \tfrac{1}{8}, \ldots$ — $r = \tfrac{1}{2}$
  • $3, -6, 12, -24, \ldots$ — $r = -2$ (alternating sign)
  • $\$100, \$105, \$110.25, \ldots$ — 5% compound interest, $r = 1.05$
Not geometric
  • $3, 6, 9, 12, \ldots$ — difference is constant, not ratio
  • $1, 2, 4, 7, 11, \ldots$ — neither
  • $0, 0, 0, \ldots$ from $a_1 = 0$ — the ratio is undefined
Sign of r matters

If $r$ is negative, terms alternate sign. If $|r| > 1$, terms grow in magnitude. If $|r| < 1$, terms shrink toward zero. If $r = 1$, the sequence is constant. These four regimes behave very differently when you start summing, as the next section shows.

6. Summing a geometric series

Adding up a geometric sequence uses a different — but equally clean — trick. Call the partial sum

$$ S_n = a_1 + a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1}. $$

Multiply both sides by $r$:

$$ r S_n = a_1 r + a_1 r^2 + a_1 r^3 + \cdots + a_1 r^{n}. $$

Almost every term on the right of these two equations matches. Subtract the second from the first and watch a long line of cancellations:

$$ S_n - r S_n = a_1 - a_1 r^n. $$

Factor and divide:

$$ \boxed{\, S_n = a_1 \, \frac{1 - r^n}{1 - r} \quad (r \neq 1) \,} $$

(When $r = 1$ the sequence is just $a_1, a_1, a_1, \ldots$ and the partial sum is the boring $S_n = n a_1$.)

This "shift, line up, subtract" idea — multiply $S_n$ by the common ratio so that most terms appear in both copies and collapse on subtraction — is one of the most useful tricks in elementary algebra. The same idea reappears later in telescoping series and generating functions.

Quick sanity check on $2 + 6 + 18 + 54 + 162$: $a_1 = 2$, $r = 3$, $n = 5$:

$$ S_5 = 2 \cdot \frac{1 - 3^5}{1 - 3} = 2 \cdot \frac{1 - 243}{-2} = 2 \cdot 121 = 242. \checkmark $$

7. Sigma notation

The capital-Greek-sigma symbol $\sum$ is shorthand for "add up the following over the given range":

$$ \sum_{k=1}^{n} f(k) \;=\; f(1) + f(2) + f(3) + \cdots + f(n). $$

The letter $k$ underneath is the index variable; it walks from the lower bound to the upper bound, taking integer steps. The expression to the right of $\sum$ is what you plug $k$ into for each step. The two sequence sums we just derived become:

$$ \sum_{k=1}^{n} \bigl( a_1 + (k-1)d \bigr) \;=\; \frac{n(a_1 + a_n)}{2}, \qquad \sum_{k=1}^{n} a_1 r^{k-1} \;=\; a_1 \frac{1 - r^n}{1 - r}. $$

A handful of "classical" closed-form sums are worth remembering:

SumClosed formMnemonic
$\sum_{k=1}^{n} k$ $\dfrac{n(n+1)}{2}$ The triangular numbers — Gauss again
$\sum_{k=1}^{n} k^2$ $\dfrac{n(n+1)(2n+1)}{6}$ "$n$, $n+1$, $2n+1$, over six"
$\sum_{k=1}^{n} k^3$ $\left( \dfrac{n(n+1)}{2} \right)^2$ Sum of cubes is the square of the sum
Watch the bounds

$\sum_{k=1}^{n}$ has $n$ terms. $\sum_{k=0}^{n-1}$ also has $n$ terms but starts at $k = 0$. $\sum_{k=0}^{n}$ has $n + 1$ terms. A surprising number of "obvious" computations go wrong because someone counted the terms one off — always sanity-check the first and last term and the count.

8. Playground: partial sums of a geometric series

Drag the sliders to set the first term $a$ and the common ratio $r$, and the slider for $n$ to choose how many terms to add. Watch the partial sum $S_n$ approach the infinite-sum value $a / (1 - r)$ when $|r| < 1$, and run away when $|r| \geq 1$.

Sn = 2.00
a / (1 − r): 2.00  ·  status: converges
1.0
0.50
10
limit term index k partial sum Sk
Try it

With $a = 1$ and $r = \tfrac{1}{2}$, the partial sums climb $1, 1.5, 1.75, 1.875, \ldots$ — visibly homing in on $2$ but never quite arriving. Now nudge $r$ past $1$ and the bars take off forever; nudge $r$ to $-\tfrac{1}{2}$ and they zig-zag in on $\tfrac{2}{3}$. The dashed green line is where infinity says they're going.

9. Infinite geometric series and convergence

What happens to $S_n = a_1 \dfrac{1 - r^n}{1 - r}$ when $n$ goes to infinity? Everything in that expression is fixed except $r^n$. The behaviour splits cleanly on the size of $|r|$:

  • If $|r| < 1$, then $r^n$ shrinks toward $0$ as $n$ grows, so $S_n$ heads toward a finite number.
  • If $|r| > 1$, then $|r^n|$ blows up and $S_n$ grows without bound.
  • If $r = 1$, every term is $a_1$ and the sum grows linearly: $n a_1$.
  • If $r = -1$, the partial sums oscillate $a_1, 0, a_1, 0, \ldots$ — no single value to land on.

So the infinite geometric series only "has a sum" in the first case. When $|r| < 1$, plugging $r^n \to 0$ into the formula gives:

$$ \boxed{\, \sum_{k=0}^{\infty} a r^{k} \;=\; \frac{a}{1 - r} \quad \text{when } |r| < 1 \,} $$

This is the workhorse formula. Two famous applications:

Why $0.\overline{9} = 1$, exactly

The repeating decimal $0.\overline{9}$ is shorthand for an infinite sum:

$$ 0.\overline{9} = 0.9 + 0.09 + 0.009 + \cdots = \sum_{k=0}^{\infty} 9 \cdot (0.1)^{k+1}. $$

That's a geometric series with $a = 0.9$ and $r = 0.1$. Since $|r| = 0.1 < 1$:

$$ 0.\overline{9} = \frac{0.9}{1 - 0.1} = \frac{0.9}{0.9} = 1. $$

Not approximately. Exactly. $0.\overline{9}$ and $1$ are two notations for the same real number.

Zeno's paradox dissolves

Zeno argued you can never cross a room: first you must cross half, then half the rest, then half of that, and so on forever — an infinity of steps. Each step is finite work, so surely an infinite number of finite works is infinite?

No: $\tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \cdots = \frac{1/2}{1 - 1/2} = 1$. The infinite sum of shrinking pieces is finite. Geometric series turn out to be the most ordinary thing in the world; Zeno just didn't have the algebra.

What "converges" means, informally

An infinite series converges to a value $L$ if the partial sums $S_1, S_2, S_3, \ldots$ get and stay arbitrarily close to $L$ as $n$ grows. We've been waving at "approaches" with hand gestures here — the formal definition uses limits, which you'll meet in calculus. But the intuition is solid: pick any tolerance you like, however tiny, and after some $N$ the partial sums all lie inside that tolerance around $L$.

10. A warning: the harmonic series

It's natural to guess: "if the terms of a series shrink to zero, the sum must converge." That guess is wrong, and the smallest example that breaks it is famous enough to have a name.

Harmonic series

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n} \;=\; 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots$

The terms $1/n$ shrink to $0$, yet the sum grows without bound — it diverges.

A clean proof from Nicole Oresme (around 1350): group the terms in blocks of length $1, 2, 4, 8, \ldots$ and bound each block below by replacing every term inside with the smallest one in that block.

$$ \underbrace{\frac{1}{2}}_{\geq\, \frac{1}{2}} + \underbrace{\frac{1}{3} + \frac{1}{4}}_{\geq\, \frac{1}{4} + \frac{1}{4} \,=\, \frac{1}{2}} + \underbrace{\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}}_{\geq\, 4 \cdot \frac{1}{8} \,=\, \frac{1}{2}} + \cdots $$

Every block contributes at least $\tfrac{1}{2}$. There are infinitely many blocks. So the total grows without bound — slowly, but inescapably. To get a partial sum of $20$ takes around 250 million terms, but get there it does.

The harmonic series is the canonical counterexample to the natural-but-wrong rule. The correct rule cuts the other way:

If the terms don't shrink to zero, the series definitely diverges. But shrinking to zero is not enough to guarantee convergence.

Sorting out exactly which "shrinking-to-zero" series converge and which don't is the subject matter of the convergence-tests chapter in calculus. For now, the geometric series is the safest infinite sum you can trust.

11. Common pitfalls

Off-by-one in the n-th term formula

Arithmetic: $a_n = a_1 + (n - 1)d$, not $a_1 + nd$. Geometric: $a_n = a_1 r^{n-1}$, not $a_1 r^n$. The reason is the same in both: the first term has had zero increments or multiplications applied. If you ever can't remember which exponent is right, just check $n = 1$ — the formula should reduce to $a_1$.

Confusing sequence and series

The sequence $1, \tfrac{1}{2}, \tfrac{1}{3}, \ldots$ has limit $0$. The series $1 + \tfrac{1}{2} + \tfrac{1}{3} + \cdots$ diverges. Identical-looking notation, opposite outcomes. The sequence is a list of numbers; the series is one number (or none).

Using a/(1−r) when |r| ≥ 1

The formula is only valid for $|r| < 1$. Apply it to, say, $1 + 2 + 4 + 8 + \cdots$ with $r = 2$ and you'll get $\frac{1}{1 - 2} = -1$ — a confident-looking number for a sum that is plainly infinite. Always check $|r| < 1$ before plugging in.

Starting index trips

$\displaystyle \sum_{k=0}^{\infty} a r^k = \frac{a}{1-r}$ starts at $k = 0$. The cousin $\displaystyle \sum_{k=1}^{\infty} a r^k = \frac{ar}{1-r}$ is missing the $k=0$ term and equals $r$ times as much. Read which index the problem starts from.

"Terms go to zero" is necessary, not sufficient

If $\lim_{n\to\infty} a_n \neq 0$, the series $\sum a_n$ definitely diverges. But $\lim a_n = 0$ does not guarantee convergence — the harmonic series is a permanent reminder.

12. Worked examples

Try each one before opening the solution. The goal is to recognise which formula applies and to plug in cleanly without off-by-one slip-ups.

Example 1 · Find $a_{20}$ for $5, 8, 11, \ldots$

The differences are $3$, so this is arithmetic with $a_1 = 5$ and $d = 3$.

$$ a_{20} = a_1 + (20 - 1)d = 5 + 19 \cdot 3 = 5 + 57 = 62 $$
Example 2 · Find $a_6$ for $3, 6, 12, 24, \ldots$

The ratios are $2$, so this is geometric with $a_1 = 3$ and $r = 2$.

$$ a_6 = a_1 \, r^{6-1} = 3 \cdot 2^5 = 3 \cdot 32 = 96 $$
Example 3 · Sum of odd numbers $1 + 3 + 5 + \cdots + 99$

Arithmetic with $a_1 = 1$, $d = 2$. The last term $99$ tells us how many terms: $99 = 1 + (n-1)\cdot 2 \Rightarrow n = 50$.

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

Aside: the sum of the first $n$ odd numbers always equals $n^2$. Here $n = 50$ and $50^2 = 2500$.

Example 4 · Finite geometric sum $2 + 6 + 18 + 54 + 162$

$a_1 = 2$, $r = 3$, $n = 5$.

$$ S_5 = a_1 \cdot \frac{1 - r^n}{1 - r} = 2 \cdot \frac{1 - 3^5}{1 - 3} = 2 \cdot \frac{1 - 243}{-2} = 2 \cdot 121 = 242 $$
Example 5 · Infinite sum $1 + \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \cdots$

Geometric with $a = 1$, $r = \tfrac{1}{2}$, and $|r| < 1$ so the series converges.

$$ S_\infty = \frac{a}{1 - r} = \frac{1}{1 - \tfrac{1}{2}} = \frac{1}{\tfrac{1}{2}} = 2 $$
Example 6 · Recursive to closed-form: $a_1 = 5$, $a_{n+1} = a_n + 4$

The recurrence says "add $4$ each time," so it's arithmetic with $d = 4$. The closed-form is:

$$ a_n = a_1 + (n-1)d = 5 + 4(n - 1) = 4n + 1 $$

Sanity check: $a_1 = 4(1) + 1 = 5$ ✓   $a_2 = 4(2) + 1 = 9$ ✓

Example 7 · A bouncing ball

A ball dropped from $10$ m bounces back to $\tfrac{3}{4}$ of its previous height each time. What is the total distance travelled before it comes to rest?

The initial drop is $10$ m. After that, each bounce contributes an "up" and a matching "down" of the same height. The heights form a geometric sequence with first term $10 \cdot \tfrac{3}{4} = 7.5$ and ratio $\tfrac{3}{4}$.

$$ \text{up + down distance} = 2 \sum_{k=0}^{\infty} 7.5 \cdot \left(\tfrac{3}{4}\right)^k = 2 \cdot \frac{7.5}{1 - 3/4} = 2 \cdot 30 = 60 \text{ m} $$

Total distance: $10 + 60 = 70$ m. An infinite number of bounces, a finite total path.

Example 8 · Express $0.\overline{27}$ as a fraction

$0.\overline{27} = 0.27 + 0.0027 + 0.000027 + \cdots$ is geometric with $a = 0.27$ and $r = 0.01$.

$$ 0.\overline{27} = \frac{0.27}{1 - 0.01} = \frac{0.27}{0.99} = \frac{27}{99} = \frac{3}{11} $$

So $0.\overline{27} = \tfrac{3}{11}$ exactly.

Sources & further reading

The treatment here borrows the layout that's now standard in algebra texts: identify the family, derive the closed-form, derive the sum, then meet the infinite case. The primary sources below are where to go when this page has done its job.

Test your understanding

A quiz that builds from easy to hard. Pick an answer to get instant feedback and a worked explanation. Your progress is saved in this browser — come back anytime to continue.

Question 1 of 24
0 correct