UNIT 10 · BC ONLY

Infinite Sequences and Series.

What it means to add up infinitely many things, and when that sum is finite.

Unit 10 · BC Only

Infinite Sequences
and Series

The conceptual foundation for the final unit of AP Calculus BC: what sequences are, what it means to sum them, and the tools for determining whether that sum is finite.

CalculusABCs · AP® Calculus BC Review

10.1

Sequences

A sequence is an ordered list of values (or terms).

A sequence is a function whose domain is the positive integers. Each input $n$ produces a single output $a_n$, and we write the entire sequence using curly braces:

$$\{a_n\} = \{a_1,\; a_2,\; a_3,\; \ldots,\; a_n,\; \ldots\}$$

Definition · Convergence of a Sequence

A sequence $\{a_n\}$ converges if $\displaystyle\lim_{n \to \infty} a_n = L$ for some finite number $L$. If no such limit exists, the sequence diverges.

10.2

Series and Partial Sums

A series is the sum of the terms in a sequence. Given a sequence $\{a_n\}$, the corresponding series is the infinite sum:

$$\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots$$

But you cannot simply add infinitely many numbers. Instead, we define convergence through finding the limit of the sequence of partial sums.

Definition · Partial Sum

The $n$th partial sum of a series is the sum of the first $n$ terms:

$$S_n = \sum_{j=1}^{n} a_j = a_1 + a_2 + \cdots + a_n$$

Each partial sum $S_n$ is a single number. As $n$ increases, you get a new number. That means the partial sums themselves form a sequence: $\{S_1,\; S_2,\; S_3,\; \ldots\}$.

Convergence of a Series

$$\sum_{n=1}^{\infty} a_n \;=\; \lim_{n \ o \infty} \sum_{j=1}^{n} a_j \;=\; \lim_{n \ o \infty} S_n \;=\; S$$

The infinite series $\sum a_n$ converges if and only if the sequence of partial sums $\{S_n\}$ converges. If this limit exists and is finite, the series converges to $S$. If the limit does not exist or is infinite, the series diverges.

Read that chain of equalities carefully. It says: the value of an infinite series is defined as the limit of its partial sums. Convergence of a series is convergence of a sequence; the sequence $\{S_n\}$.

S_n

1.527422

L = π²/6

1.644934

L − S_n

0.117512

Partial sums of $\sum 1/n^2$ converging to $L = \pi^2/6 \approx 1.645$. Drag the orange dot, scrub the slider, or click any blue point.

Warning

Be sure not to confuse $a_n$ with $S_n$. The term $a_n$ is the $n$th term of the sequence. The term $S_n$ is the $n$th partial sum of the series; itself a sum of the first $n$ terms. Both generate sequences, but they answer different questions.

Every test and theorem that follows is a tool for answering one question: does $\{S_n\}$ have a finite limit?

On the Exam

The sequence $\{a_n\}$ converging tells you nothing about whether the series $\sum a_n$ converges. These are fundamentally different questions. The AP exam tests whether you understand this distinction.

10.2

Geometric Series

A geometric series converges when $|r| < 1$ and diverges when $|r| \geq 1$. When it converges, the sum is given by:

$$\sum_{n=0}^{\infty} ar^n = \rac{a}{1-r}, \qquad |r| < 1$$

The terms $a_n = (1/2)^n$ of a geometric series with $r = 1/2$, $a = 1$. Each term is half the previous one, decaying toward zero. Their infinite sum equals $a/(1-r) = 2$. Hover any bar to inspect values.

where $a$ is the initial term of the series and $r$ is the common ratio. Convergence depends on the terms approaching zero quickly enough. This is what the interval of convergence tells us: for what values of $r$ do we converge?

10.3 – 10.5

Convergence Tests

Most series cannot be summed exactly. Instead, we determine whether they converge or diverge using a toolkit of tests. Each test has a specific structure it recognizes and conditions under which it is conclusive.

Convergence of a Series

Theorem

$$\ ext{If } \sum_{n=1}^{\infty} a_n \ ext{ converges, then } \lim_{n \ o \infty} a_n = 0.$$

Think about why this is true: In order for the sum of infinite terms to be a finite number, those terms have to eventually go to zero as $n$ increases without bound. In other words, for our series to converge, our $n$th terms must approach zero. If we are not eventually adding zero, we would be continuously adding terms, and we would diverge.

The converse of this theorem is not generally true. If the limit equals zero, the series could still diverge. The harmonic series $\displaystyle\sum \frac{1}{n}$ is an example: $\displaystyle\lim_{n \to \infty} \frac{1}{n} = 0$, but we know by the integral test the harmonic series diverges.

The harmonic series: terms $a_n = 1/n$ (dashed) approach zero, yet the partial sums $S_n$ (solid) grow without bound. Both curves start at the same point. Their divergence is the key counterintuitive result. Hover any dot to inspect values.

Compute it yourself: $a_n$ vs $S_n$

The same picture made tangible. Pick a series, drag $n$, and watch the term $a_n$ shrink while the partial sum $S_n$ either settles at a limit or marches off to infinity. The table tracks every value; the chart shows them in motion.

aₙ (term) Sₙ (running sum)

n	aₙ	Sₙ
1	1.000000	1.000000
2	0.500000	1.500000
3	0.333333	1.833333
4	0.250000	2.083333
5	0.200000	2.283333
6	0.166667	2.450000
7	0.142857	2.592857
8	0.125000	2.717857
9	0.111111	2.828968
10	0.100000	2.928968
11	0.090909	3.019877
12	0.083333	3.103211
13	0.076923	3.180134
14	0.071429	3.251562
15	0.066667	3.318229
16	0.062500	3.380729
17	0.058824	3.439553
18	0.055556	3.495108
19	0.052632	3.547740
20	0.050000	3.597740
21	0.047619	3.645359
22	0.045455	3.690813
23	0.043478	3.734292
24	0.041667	3.775958
25	0.040000	3.815958
26	0.038462	3.854420
27	0.037037	3.891457
28	0.035714	3.927171
29	0.034483	3.961654
30	0.033333	3.994987

n10

aₙ

0.100000

Sₙ

2.928968

after 30 terms

3.995

harmonic series: diverges, $S_n$ grows without bound, even though $a_n \to 0$

While the limit equaling zero is a requirement for convergence, it is not enough to prove convergence.

However, the contrapositive of our statement is true: if $\displaystyle\lim_{n \to \infty} a_n \neq 0$, then the series diverges. We call this contrapositive the $n$th Term Test.

The $n$th Term Test for Divergence

$n$th Term Test (Divergence Test)

$$\text{If } \lim_{n \to \infty} a_n \neq 0, \text{ then } \sum_{n=1}^{\infty} a_n \text{ diverges.}$$

If $\displaystyle\lim_{n \to \infty} a_n = 0$, the test is inconclusive.

On the Exam

According to the Chief Reader reports, the main misconception about convergence and divergence of infinite series on the AP Calculus BC exam is students thinking that $\lim a_n = 0$ means the series converges. It does not. It is necessary, but we have to do more. We have to use a test to show that the series converges.

The Integral Test

Integral Test

Let $f$ be a function that is positive, continuous, and decreasing on $[1, \infty)$, with $a_n = f(n)$. Then:

$$\sum_{n=1}^{\infty} a_n \quad \ ext{and} \quad \int_1^{\infty} f(x)\,dx$$

either both converge or both diverge.

The Integral Test connects series to improper integrals; something you already know how to evaluate. If the integral converges, so does the series. If the integral diverges, so does the series. But note: the test tells you whether the series converges, not what it converges to. The value of the integral is not the sum of the series.

Conditions Matter

The Integral Test requires all three conditions: positive, continuous, decreasing. On the AP exam, you must state that you have verified these conditions. Skipping this step costs points on free response.

$p$-Series

$p$-Series Test

$$\sum_{n=1}^{\infty} \rac{1}{n^p} \ ext{ converges if and only if } p > 1.$$

Two cases worth memorizing: when $p = 1$, you get the harmonic series $\sum \frac{1}{n}$, which diverges. When $p = 2$, you get $\sum \frac{1}{n^2}$, which converges. The boundary is $p = 1$, and the boundary itself diverges.

10.4

Comparison Tests

The big idea: if you can relate a complicated series to a simpler one whose behavior you already know, you can draw conclusions about the complicated series without evaluating it directly. The two simpler series we lean on are the $p$-series and the geometric series; nearly every comparison problem on the AP exam reduces to one of those.

Direct Comparison Test

Let $0 < a_n \leq b_n$ for all $n$ (both sequences positive).

Case 1 · the bigger one converges

If $\displaystyle\sum_{n=1}^{\infty} b_n$ converges, then $\displaystyle\sum_{n=1}^{\infty} a_n$ converges. The bigger series is a finite ceiling; the smaller one cannot exceed it, so it must also be finite.

Case 2 · the smaller one diverges

If $\displaystyle\sum_{n=1}^{\infty} a_n$ diverges, then $\displaystyle\sum_{n=1}^{\infty} b_n$ diverges. The smaller series already adds to infinity; whatever sits above it has no choice but to do the same.

The two cases that are not listed are the dangerous ones. If the bigger one diverges, the smaller one could go either way. If the smaller one converges, the bigger one could go either way. Direct comparison only resolves these two configurations; if your inequality runs the wrong direction, you are not done.

Choosing a comparison series

Look at the dominant terms in the numerator and denominator for large $n$. The simplified expression almost always points to a $p$-series or a geometric series. Those are the two pitfalls to recognize, and they are the two parents you compare against.

Limit Comparison Test

Suppose $\displaystyle\sum a_n$ is a transformed series and $\displaystyle\sum b_n$ is a parent series, both with positive terms. Compute:

$$\lim_{n \ o \infty} \rac{a_n}{b_n} = L.$$

If $0 < L < \infty$ (that is, $L$ is a positive, finite number), then either both series converge or both series diverge.

Read the conclusion carefully. The existence of a finite, positive limit justifies the comparison; it is the reason the test applies. It does not tell you whether the series converges. To answer that question, you go look at the parent.

Common Trap

Concluding that because $L$ exists, the series converges. The limit existing only tells you the two series share a fate. You still have to identify $\sum b_n$, recognize it as a $p$-series or geometric series, and report its verdict; whatever happens to the parent happens to your series.

Workflow: form the ratio $a_n / b_n$ → take the limit → verify $0 < L < \infty$ → conclude from the parent's behavior.

What if the limit does not exist, or is $0$ or $\infty$?

The limit comparison test only applies when $L$ is a positive, finite number. If $L = 0$, $L = \infty$, or the limit fails to exist, the test is silent; you have learned nothing about $\sum a_n$, and you cannot conclude that it diverges from a failed limit comparison alone.

When that happens, fall back: pick a different parent series whose ratio does give a positive finite limit, or switch tools. The $n$th term test is often the next move (if $a_n$ does not approach zero, the series diverges immediately), followed by the integral test or direct comparison with a tighter parent.

Why the limit comparison test exists

Direct comparison demands an inequality $a_n \leq b_n$ that holds for every $n$. That can be ugly to prove. The limit comparison test trades the inequality for a ratio: if the two sequences grow at the same rate, you skip the inequality bookkeeping and inherit the parent's verdict. It is direct comparison's more forgiving cousin.

10.7

Alternating Series

An alternating series is one whose terms alternate in sign:

$$\sum_{n=1}^{\infty} (-1)^n a_n \quad\ ext{or}\quad \sum_{n=1}^{\infty} (-1)^{n+1} a_n, \qquad a_n > 0.$$

The signs flip every term. That oscillation is what saves these series: contributions partially cancel each other out, slowing the partial sums down enough to converge in cases where the corresponding all-positive series would diverge.

The Alternating Harmonic Series

The cleanest example is the alternating harmonic series:

$$\sum_{n=1}^{\infty} \rac{(-1)^{n+1}}{n} = 1 - \rac{1}{2} + \rac{1}{3} - \rac{1}{4} + \rac{1}{5} - \cdots$$

Same magnitudes as the harmonic series; same $a_n = 1/n$. But where the harmonic series adds those magnitudes and grows without bound, the alternating version subtracts every other one. Watch what happens.

Partial sums of the harmonic series $\sum 1/n$ (top) climb without bound, while the alternating harmonic $\sum (-1)^{n+1}/n$ (bottom) oscillates and tightens around $\ln 2 \approx 0.693$. Same terms in absolute value; the alternating signs are doing all the work.

Key insight

The alternating harmonic series converges, even though the (regular) harmonic series $\sum 1/n$ diverges. Adding alternating signs is enough to take a divergent series and force it to converge, provided the magnitudes shrink to zero.

The Alternating Series Test

Alternating Series Test (AST)

Let $a_n > 0$. The alternating series

$$\sum_{n=1}^{\infty} (-1)^n a_n \qquad\ ext{and}\qquad \sum_{n=1}^{\infty} (-1)^{n+1} a_n$$

converge if both of the following conditions are met:

$\displaystyle\lim_{n \to \infty} a_n = 0$, the terms shrink to zero, and
$a_{n+1} \leq a_n$, the magnitudes are decreasing (eventually).

Plain English: if the terms of an alternating series decrease in absolute value to zero, the series converges. That's the whole test.

Watch out, convergence only

The AST can only be used to argue that a series converges. It can never be used to argue divergence. If the AST conditions are not met, you have learned nothing about whether the series diverges; you have only learned that this test did not apply. To check for divergence, proceed to the $n$th Term Test: if $\lim a_n \neq 0$, the series diverges.

Alternating Series Error Bound

Once you know an alternating series converges, the partial sums sweep past the true sum on alternate sides, with each pass tighter than the last. The next term you didn't add tells you exactly how far off you could possibly be.

Alternating Series Error Bound (Remainder)

If a convergent alternating series satisfies $a_{n+1} \leq a_n$, then the magnitude of the remainder $R_N$ from using $s_N$ to approximate the true sum $S$ is no larger than the magnitude of the first omitted term:

$$|S - s_N| \;\leq\; |a_{N+1}|.$$

In words: the error from stopping at $s_N$ is at most the size of the first term you didn't include.

Stop at N5

s_N

0.78333

S = ln 2

0.69315

|S − s_N|

0.09019

|a_N+1| (bound)

0.16667

Drag $N$. The vertical bracket is the actual error; the right-hand bar is the guaranteed bound $|a_{N+1}|$. The error is always smaller.

Partial sums of the alternating harmonic series oscillate around $S = \ln 2$, getting tighter with each step. The error $|S - s_N|$ is bracketed in blue; the bound $|a_{N+1}| = 1/(N+1)$ is shown to the right for comparison. Stopping earlier means a larger guaranteed error, but it never exceeds the first omitted term.

This is unusually generous. Most error estimates in calculus are loose; this one tells you the exact ceiling. To guarantee error below some target $\varepsilon$, just choose $N$ large enough that $1/(N+1) < \varepsilon$, done.

On the Exam

Free-response questions often ask you to estimate $S$ with $s_N$ and bound the error. The answer is mechanical: state that the series satisfies the AST hypotheses (alternating, $a_n \to 0$, decreasing), then write $|S - s_N| \leq |a_{N+1}|$ and plug in. Skipping the hypothesis check costs points.

10.8

Absolute & Conditional Convergence

The alternating harmonic series converges. Its all-positive cousin, the harmonic series, diverges. Same magnitudes, different fate. The signs are doing the work, and that distinction has a name.

Given a series $\sum a_n$, we can ask two different questions: does $\sum a_n$ converge, and does $\sum |a_n|$ converge? The answers don't have to agree, and the gap between them is exactly what absolute and conditional convergence are for.

Definition · Absolute vs. Conditional Convergence

Let $\sum a_n$ be an infinite series.

$\sum a_n$ is absolutely convergent when $\sum |a_n|$ converges.

$\sum a_n$ is conditionally convergent when $\sum a_n$ converges, but $\sum |a_n|$ diverges.

Absolute convergence is the stronger condition: the series converges even if you strip away every minus sign. Conditional convergence is fragile by comparison, the convergence depends entirely on the cancellation between positive and negative terms.

The Absolute Convergence Theorem

Absolute Convergence Theorem

$$\ ext{If } \sum_{n=1}^{\infty} |a_n| \ ext{ converges, then } \sum_{n=1}^{\infty} a_n \ ext{ converges.}$$

In words: absolute convergence implies convergence.

Read the direction carefully. The implication runs one way. If the series of absolute values converges, the original series is guaranteed to converge. The converse is not true: $\sum a_n$ can converge while $\sum |a_n|$ diverges. That is precisely the conditional case, the alternating harmonic series.

One-way street

Convergence of $\sum a_n$ does not imply convergence of $\sum |a_n|$. The alternating harmonic series $\sum (-1)^{n+1}/n$ converges to $\ln 2$, yet $\sum |(-1)^{n+1}/n| = \sum 1/n$ diverges. That asymmetry is why conditional convergence has its own name.

Why the theorem matters

It turns a hard question into an easier one. Showing $\sum a_n$ converges directly often means wrestling with cancellation between positive and negative terms; showing $\sum |a_n|$ converges lets you use every positive-term tool you already have, the integral test, $p$-series, comparison tests, and (next) the Ratio Test. Confirm absolute convergence, and the original series comes along for the ride.

Two examples to keep straight

$\sum (-1)^{n+1} / n^2$ is absolutely convergent: $\sum 1/n^2$ is a convergent $p$-series ($p = 2$), so the original converges by the Absolute Convergence Theorem.

$\sum (-1)^{n+1} / n$ is conditionally convergent: it converges by the AST, but $\sum 1/n$ diverges. Strip the signs and the series falls apart.

10.9

The Ratio Test

Most of the tests so far have a sweet spot, a $p$-series, a geometric series, a positive-decreasing function. The Ratio Test takes a different angle. Instead of comparing $\sum a_n$ to a known parent series, it asks how fast each term grows relative to the previous one. If consecutive terms shrink fast enough, the series converges; if they grow, it diverges.

The motivation comes straight from the geometric series. For $\sum b r^n$, every term is a fixed ratio $r$ times the one before it: $a_{n+1}/a_n = r$. We already know geometric series converge exactly when $|r| < 1$. The Ratio Test generalizes that idea to series whose ratio isn’t constant, but approaches a constant in the limit.

Setup · Reading $r$ off a Geometric Series

Consider $\displaystyle\sum_{n=1}^{\infty} b r^n$. The $n$th term is $a_n = b r^n$, so the next term is $a_{n+1} = b r^{n+1}$. Form the ratio:

$$\rac{a_{n+1}}{a_n} \;=\; \rac{b r^{n+1}}{b r^n} \;=\; r.$$

For a true geometric series the ratio is exactly $r$ for every $n$. The Ratio Test says: if the ratio of consecutive terms approaches a limit $L$ in absolute value, then $L$ plays the role of $|r|$.

The Test

The Ratio Test

Let $\sum a_n$ be a series with nonzero terms, and define

$$L \;=\; \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|.$$

• If $L < 1$, the series converges absolutely.
• If $L > 1$ (or $L = \infty$), the series diverges.
• If $L = 1$, the test is inconclusive.

Read the first case carefully: when $L < 1$, the Ratio Test gives you absolute convergence, not just convergence. That is the theorem from the previous section paying off, the Ratio Test actually shows $\sum |a_n|$ converges, and then the Absolute Convergence Theorem hands you convergence of $\sum a_n$ for free.

Watching the Ratio

Here’s the series $\displaystyle\sum_{n=1}^{\infty} n^2 x^n$. Its ratio is

$$\left|\frac{a_{n+1}}{a_n}\right| = \left(\frac{n+1}{n}\right)^{\!2} \cdot |x| \;\longrightarrow\; |x|.$$

So $L = |x|$. Drag $x$ across the boundary at $|x| = 1$ and watch the dots settle to a different limit each time. The dashed horizontal line is $L$; the solid line at height $1$ is the boundary the test cares about.

x0.50

L = lim |a_n+1/a_n|

0.500

Compare to 1

L < 1

Verdict

converges absolutely

The ratio $|a_{n+1}/a_n|$ for $\sum n^2 x^n$, plotted against $n$. Each dot is a single ratio; the dashed line is the limit $L = |x|$. Drag $x$ across the boundary at $1$ to see the verdict flip from absolutely convergent, to inconclusive, to divergent.

When to reach for it

Recognizing a Ratio Test problem

Use the Ratio Test when $n$ appears in multiple bases in the series, typically exponentials and factorials. Things like $\sum \frac{2^n}{n!}$, $\sum \frac{n!}{n^n}$, or $\sum \frac{(-3)^n}{n \cdot 5^n}$. The ratio causes most of those bases to telescope, leaving something tractable.

Conversely, do not reach for the Ratio Test on plain $p$-series like $\sum 1/n^p$. The ratio comes out to $\big(n/(n+1)\big)^p \to 1$, and the test is inconclusive. Recognize the structure and use the $p$-series test instead.

$L = 1$ tells you nothing

A common mistake: concluding divergence (or convergence) from $L = 1$. The Ratio Test simply does not apply there, you have learned nothing about the series. Switch tools: $p$-series, integral test, or comparison are usually the next moves.

On the Exam

Free-response problems that introduce a power series or a series with factorials almost always want the Ratio Test. The expected work: write $\big|a_{n+1}/a_n\big|$, simplify, take the limit, compare to $1$, and state the conclusion, and when $L < 1$, the conclusion is absolute convergence, not just convergence. Get the wording right.