Math Core

You've met arithmetic and geometric sequences before. In precalculus the questions get sharper: where is a sequence heading as $n$ grows, how can you add up a hundred (or infinitely many) terms without adding them one by one, and when does an infinite sum even make sense? These ideas are the doorway to limits and to the series you'll study in calculus.

Sequences are functions on the positive integers

Definition

Sequence

A sequence is a function whose domain is the positive integers (sometimes the nonnegative integers). Its outputs $a_1, a_2, a_3, \dots$ are the terms, and $a_n$ is the general term.

A sequence can be given explicitly, by a formula for $a_n$ in terms of $n$ , such as $a_n = \dfrac{2n}{n + 1}$ , or recursively, by a starting value and a rule that builds each term from earlier ones, such as $a_1 = 3$ , $a_{n+1} = 2a_n - 1$ .

Two families have simple formulas you should know cold:

	Arithmetic	Geometric
Rule	add $d$ each step	multiply by $r$ each step
$n$ th term	$a_n = a_1 + (n - 1)d$	$a_n = a_1 r^{\,n-1}$
Sum of first $n$ terms	$S_n = \dfrac{n(a_1 + a_n)}{2}$	$S_n = a_1 \cdot \dfrac{1 - r^n}{1 - r}$ , $r \ne 1$

Since a sequence is a function, you can graph it. The graph is a set of isolated dots, one for each $n$ , not a connected curve.

The terms of a_n = 2n/(n + 1) creep up toward the dashed line y = 2 but never reach it.Open in grapher →

Where is a sequence heading?

Look at $a_n = \dfrac{2n}{n+1}$ : the terms are $1, \tfrac43, \tfrac32, \tfrac85, \dots$ and $a_{1000} = \tfrac{2000}{1001} \approx 1.998$ . The terms get as close to $2$ as you like once $n$ is large enough. We say the sequence converges to $2$ and write

\lim_{n \to \infty} a_n = 2.

A sequence that doesn't settle on a single finite number diverges. It might grow without bound ( $a_n = n^2$ ), or bounce between values forever ( $a_n = (-1)^n$ , which alternates $-1, 1, -1, 1, \dots$ ).

A few facts handle most sequences you'll see now:

$\dfrac{1}{n} \to 0$ , and more generally $\dfrac{c}{n^p} \to 0$ for any constant $c$ and $p > 0$ .
$r^n \to 0$ when $|r| < 1$ ; $r^n$ diverges when $|r| > 1$ or $r = -1$ .
For a ratio of polynomials in $n$ , divide the top and bottom by the highest power of $n$ in the denominator. This is the same reasoning you used for horizontal asymptotes of rational functions.

Worked example: Finding a limit by dividing by the highest power

Find $\displaystyle\lim_{n \to \infty} \frac{3n^2 - 5n}{6n^2 + 1}$ .

Divide every term by $n^2$ :

\frac{3n^2 - 5n}{6n^2 + 1} = \frac{3 - \dfrac{5}{n}}{6 + \dfrac{1}{n^2}}.

As $n \to \infty$ , $\dfrac{5}{n} \to 0$ and $\dfrac{1}{n^2} \to 0$ , so the limit is $\dfrac{3}{6} = \dfrac{1}{2}$ .

Sigma notation and its rules

A series is a sum of terms of a sequence. Sigma notation packs a long sum into one line:

\sum_{k=1}^{n} a_k = a_1 + a_2 + \cdots + a_n.

Because addition is commutative and multiplication distributes over it, sigma behaves linearly:

\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\,a_k = c\sum_{k=1}^{n} a_k, \qquad \sum_{k=1}^{n} c = cn.

Combine these with three power-sum formulas and you can add any polynomial sequence quickly.

Power sums

\sum_{k=1}^{n} k = \frac{n(n+1)}{2}, \qquad \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}, \qquad \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2.

The first formula comes from pairing the first and last terms. You'll prove the other two in the next lesson with mathematical induction.

Worked example: Using the power sums

Evaluate $\displaystyle\sum_{k=1}^{20} (3k^2 - 2k)$ .

Split the sum and pull out the constants:

\sum_{k=1}^{20} (3k^2 - 2k) = 3\sum_{k=1}^{20} k^2 - 2\sum_{k=1}^{20} k.

With $n = 20$ : $\displaystyle\sum k^2 = \frac{20 \cdot 21 \cdot 41}{6} = 2870$ and $\displaystyle\sum k = \frac{20 \cdot 21}{2} = 210$ . So the sum is

3(2870) - 2(210) = 8610 - 420 = 8190.

Common mistake

$\displaystyle\sum (a_k b_k)$ is not $\left(\sum a_k\right)\left(\sum b_k\right)$ . For example, $\displaystyle\sum_{k=1}^{2} k \cdot k = 1 + 4 = 5$ , but $(1 + 2)(1 + 2) = 9$ . Sigma splits over sums and pulls out constants, and that's all.

Telescoping sums

Sometimes each term is a difference, and consecutive terms cancel like the sections of a collapsing telescope. The classic example uses the partial-fraction identity

\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}.

Worked example: A telescoping sum

Find $\displaystyle\sum_{k=1}^{n} \frac{1}{k(k+1)}$ and its value when $n = 49$ .

Rewrite each term and write out the sum:

\left(1 - \tfrac12\right) + \left(\tfrac12 - \tfrac13\right) + \left(\tfrac13 - \tfrac14\right) + \cdots + \left(\tfrac1n - \tfrac{1}{n+1}\right).

Every inner fraction appears once with a plus sign and once with a minus sign, so only the first and last pieces survive:

\sum_{k=1}^{n} \frac{1}{k(k+1)} = 1 - \frac{1}{n+1} = \frac{n}{n+1}.

When $n = 49$ the sum is $\dfrac{49}{50}$ .

Infinite series

What could $\dfrac12 + \dfrac14 + \dfrac18 + \cdots$ mean when there's no last term? Look at the partial sums $S_n$ , the sum of the first $n$ terms: $S_1 = \tfrac12$ , $S_2 = \tfrac34$ , $S_3 = \tfrac78$ , and in general $S_n = 1 - \tfrac{1}{2^n}$ . These partial sums form a sequence of their own, and it converges to $1$ . We define the infinite sum to be that limit.

Sum of an infinite series

$\displaystyle\sum_{k=1}^{\infty} a_k = \lim_{n \to \infty} S_n$ when that limit exists; the series then converges. Otherwise the series diverges. For a geometric series,

\sum_{k=1}^{\infty} a_1 r^{\,k-1} = \frac{a_1}{1 - r} \quad \text{if } |r| < 1,

and the series diverges if $|r| \ge 1$ .

The formula follows from the finite sum: $S_n = a_1\dfrac{1 - r^n}{1 - r}$ , and $r^n \to 0$ exactly when $|r| < 1$ . Telescoping series can also have infinite sums: from the example above, $\dfrac{n}{n+1} \to 1$ , so $\displaystyle\sum_{k=1}^{\infty} \frac{1}{k(k+1)} = 1$ .

Worked example: When does a series converge?

For which $x$ does $\displaystyle\sum_{k=0}^{\infty} \left(\frac{x - 1}{3}\right)^{k}$ converge, and what is its sum?

This is geometric with first term $1$ and ratio $r = \dfrac{x-1}{3}$ . It converges when $|r| < 1$ :

\left|\frac{x - 1}{3}\right| < 1 \iff -3 < x - 1 < 3 \iff -2 < x < 4.

On that interval the sum is $\dfrac{1}{1 - \frac{x-1}{3}} = \dfrac{3}{3 - (x - 1)} = \dfrac{3}{4 - x}$ . For example, at $x = 2$ the series is $1 + \tfrac13 + \tfrac19 + \cdots = \tfrac32$ , and $\dfrac{3}{4 - 2} = \dfrac32$ . ✓

Tip

Terms shrinking to $0$ is necessary for an infinite series to converge, but not enough. The harmonic series $1 + \tfrac12 + \tfrac13 + \tfrac14 + \cdots$ has terms going to $0$ , yet it diverges: $\tfrac13 + \tfrac14 > \tfrac12$ , $\tfrac15 + \cdots + \tfrac18 > \tfrac12$ , and you can keep finding groups worth more than $\tfrac12$ forever.