Unit 10 Foundations: Understanding Infinite Sequences and Series (AP Calculus BC)

Defining Convergent and Divergent Infinite Series

Sequences vs. series: what you’re actually studying

A sequence is a list of numbers indexed by an integer (usually starting at n = 1 or n = 0). You can think of it as a function whose input is an integer: for each n you get a term a_n.

An infinite series is what you get when you add the terms of a sequence forever. If the sequence is a_1, a_2, a_3, \dots, then the associated series is

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

The big idea is that a series is not primarily about the individual terms—it’s about the behavior of the running total.

Partial sums: the key mechanism behind convergence

Because you can’t literally “finish” an infinite addition, you study a series through its partial sums. The Nth partial sum is

S_N = \sum_{n=1}^{N} a_n

This creates a new sequence: S_1, S_2, S_3, \dots.

• The sequence a_n tells you the size of the pieces you add.
• The sequence S_N tells you what your total looks like after adding the first N pieces.

That second sequence is what determines whether the infinite sum “settles down.”

Convergent vs. divergent series (core definitions)

An infinite series \sum_{n=1}^{\infty} a_n is **convergent** if the partial sums approach a finite real number. In symbols, if there exists a number L such that

\lim_{N \to \infty} S_N = L

then the series converges and you write

\sum_{n=1}^{\infty} a_n = L

A series is divergent if its partial sums do not approach a finite limit. Divergence can happen in multiple ways:

• S_N grows without bound (to \infty or -\infty).
• S_N oscillates (bounces around) and never settles on one number.
• S_N behaves irregularly and has no limit.

Why this matters in AP Calculus BC: many later tools (Taylor series, power series) are built on knowing when an infinite “sum of approximations” actually produces a meaningful, finite value.

A helpful analogy: walking toward a wall

Imagine you start 1 unit from a wall, then walk half the remaining distance, then half again, etc. Your step sizes form a sequence, and your total distance walked forms partial sums. Even though you take infinitely many steps, you might still approach a finite distance (the wall). That “approach a finite limit” idea is exactly convergence.

Worked example: a convergent series via partial sums

Consider

\sum_{n=1}^{\infty} \frac{1}{2^n}

Look at partial sums:

S_N = \sum_{n=1}^{N} \frac{1}{2^n} = \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^N}

These partial sums increase but by smaller and smaller amounts. In the next section you’ll learn this is a geometric series and the sum is finite. For now, the important point is: you judge convergence by what S_N does as N grows.

Worked example: a divergent series via partial sums

Consider

\sum_{n=1}^{\infty} 1

Here each term is 1, so

S_N = \sum_{n=1}^{N} 1 = N

As N \to \infty, S_N \to \infty, so the series diverges.

Common misconception: “terms go to 0, so the series converges”

It’s necessary for convergence that the terms satisfy

\lim_{n \to \infty} a_n = 0

…but it is not sufficient. You will soon see examples (like the harmonic series) where a_n \to 0 and the series still diverges.

Exam Focus

• Typical question patterns:
  • You’re given a series and asked whether it converges or diverges, often by referencing partial sums conceptually or by applying a specific test.
  • You’re asked to interpret S_N (partial sums) and relate the limit of S_N to the series’ sum.
  • You’re shown a graph/table of partial sums and asked to decide convergence.
• Common mistakes:
  • Confusing the sequence a_n with the partial sum sequence S_N (convergence of the series depends on S_N).
  • Saying “convergent because terms get small” without checking more (terms going to 0 is only a prerequisite).
  • Treating a divergent series as if it has a sum because you can compute a few partial sums.

Geometric Series

What makes a series “geometric”

A geometric sequence is one where each term is obtained by multiplying the previous term by a constant ratio r. If the first term is a, the terms look like

a, ar, ar^2, ar^3, \dots

A geometric series is the sum of those terms:

\sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + ar^3 + \cdots

(You may also see indexing start at n=1; what matters is matching the formula to the first term correctly.)

Why geometric series matter

Geometric series are one of the few infinite series you can sum exactly with an explicit formula. They show up everywhere:

• repeating decimals (like 0.333\dots)
• “half of what’s left” processes (cooling, bouncing balls, iterative algorithms)
• later in AP Calc BC: rewriting functions as power series often begins by converting something into a geometric series

How the convergence works (the crucial condition)

The geometric series converges exactly when the common ratio is between -1 and 1:

|r| < 1

Intuition: if |r| < 1, the terms ar^n shrink to 0 fast enough that the running total levels off. If |r| \ge 1, the terms do not shrink to 0 (or shrink too slowly), so the partial sums cannot settle to a finite limit.

The sum formula (infinite geometric series)

If |r| < 1, then

\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}

Here:

• a is the first term (the term when n=0).
• r is the constant ratio.

A quick warning: if your series starts at n=1, the first term is ar, not a. You can either rewrite it to start at n=0 or use the finite-sum formula and take a limit, but rewriting is usually easiest.

Where the formula comes from (mechanism, not magic)

Consider the partial sum

S_N = a + ar + ar^2 + \cdots + ar^N

Multiply by r:

rS_N = ar + ar^2 + \cdots + ar^{N+1}

Subtract:

S_N - rS_N = a - ar^{N+1}

Factor:

S_N(1-r) = a(1-r^{N+1})

So

S_N = \frac{a(1-r^{N+1})}{1-r}

If |r| < 1, then r^{N+1} \to 0, giving

\lim_{N \to \infty} S_N = \frac{a}{1-r}

Seeing this derivation once helps you avoid a common mistake: trying to apply \frac{a}{1-r} when |r| \ge 1 (where the limit step fails).

Worked example 1: summing a geometric series

Find the sum:

\sum_{n=0}^{\infty} 3\left(\frac{2}{5}\right)^n

1) Identify a and r.

The first term (when n=0) is

a = 3\left(\frac{2}{5}\right)^0 = 3

The ratio is

r = \frac{2}{5}

2) Check convergence:

|r| = \left|\frac{2}{5}\right| < 1

So it converges.

3) Apply the sum formula:

\sum_{n=0}^{\infty} 3\left(\frac{2}{5}\right)^n = \frac{3}{1-\frac{2}{5}} = \frac{3}{\frac{3}{5}} = 5

Worked example 2: recognizing a disguised geometric series

Determine whether the series converges, and if so find its sum:

\sum_{n=1}^{\infty} \frac{7}{4}\left(-\frac{1}{2}\right)^{n-1}

This is geometric because the power is linear in n.

• First term occurs at n=1:

a = \frac{7}{4}\left(-\frac{1}{2}\right)^{0} = \frac{7}{4}

• Common ratio:

r = -\frac{1}{2}

Convergence check:

|r| = \frac{1}{2} < 1

Sum:

\sum_{n=1}^{\infty} \frac{7}{4}\left(-\frac{1}{2}\right)^{n-1} = \frac{\frac{7}{4}}{1-\left(-\frac{1}{2}\right)} = \frac{\frac{7}{4}}{\frac{3}{2}} = \frac{7}{6}

Repeating decimals connection (why this is “real”)

A repeating decimal is an infinite geometric series. For instance,

0.333\dots = 0.3 + 0.03 + 0.003 + \cdots

This is geometric with a = 0.3 and r = 0.1, so

0.333\dots = \frac{0.3}{1-0.1} = \frac{0.3}{0.9} = \frac{1}{3}

This is a concrete example of an infinite sum producing a finite, exact value.

Exam Focus

• Typical question patterns:
  • Identify whether a series is geometric, find r, decide convergence using |r| < 1, and compute the sum.
  • Rewrite an expression (often involving powers like k^{n} or k^{n-1}) into geometric form.
  • Use geometric series to model or interpret a repeating process (including repeating decimals).
• Common mistakes:
  • Using \frac{a}{1-r} without checking |r| < 1.
  • Misidentifying the first term a because the index starts at n=1 instead of n=0.
  • Confusing the ratio r with the first term (always compute ratio as “next term divided by previous term”).

The nth Term Test for Divergence

What the test says (and what it does not say)

The nth term test for divergence (also called the divergence test) gives a quick way to prove a series diverges.

If

\lim_{n \to \infty} a_n \ne 0

or if the limit does not exist, then the series

\sum_{n=1}^{\infty} a_n

diverges.

This is a one-way test:

• If terms do not go to 0, the series definitely diverges.
• If terms do go to 0, the test tells you nothing (the series might converge or diverge).

Why it’s true (intuition)

A convergent series has partial sums that approach a finite number. That means adding more terms eventually changes the total by smaller and smaller amounts. If the terms don’t shrink to 0, then each new term keeps making a noticeable change in the partial sums, preventing them from settling.

More formally, if S_N \to L, then the difference between consecutive partial sums must go to 0:

S_{N+1} - S_N = a_{N+1}

If S_N approaches a limit, then S_{N+1} - S_N \to 0, so necessarily a_{N+1} \to 0.

How to use it effectively

1) Identify the general term a_n.
2) Compute \lim_{n \to \infty} a_n (or analyze whether it exists).
3) If the limit is not 0 (or doesn’t exist), you can immediately conclude divergence.
4) If the limit equals 0, stop using this test and switch to another idea (geometric, p-series, comparison, etc.).

Worked example 1: limit not zero

Determine whether the series diverges:

\sum_{n=1}^{\infty} \frac{n}{n+1}

Compute the term limit:

\lim_{n \to \infty} \frac{n}{n+1} = \lim_{n \to \infty} \frac{1}{1+\frac{1}{n}} = 1

Since the term limit is 1 (not 0), the series diverges by the nth term test.

Worked example 2: limit does not exist

Determine whether the series diverges:

\sum_{n=1}^{\infty} (-1)^n

Here a_n = (-1)^n oscillates between -1 and 1, so

\lim_{n \to \infty} a_n

does not exist. Therefore the series diverges by the nth term test.

A classic “trap”: limit equals zero but still diverges

Consider

\sum_{n=1}^{\infty} \frac{1}{n}

Here

\lim_{n \to \infty} \frac{1}{n} = 0

So the nth term test cannot conclude anything. In fact, the series diverges (this is the harmonic series, discussed next). This is why you should remember: “terms to 0” is required, but not enough.

Exam Focus

• Typical question patterns:
  • Quick check: compute \lim_{n \to \infty} a_n and decide if divergence is guaranteed.
  • “Explain your reasoning”: justify divergence by explicitly referencing the nth term test.
  • Mixed problems: first apply nth term test; if inconclusive, move to another recognized series type.
• Common mistakes:
  • Claiming convergence when \lim_{n \to \infty} a_n = 0 (the test does not work that direction).
  • Computing the wrong limit because a_n was misread (for example, confusing \frac{n}{n+1} with \frac{1}{n+1}).
  • Forgetting that “limit does not exist” also triggers divergence.

Harmonic Series and p-Series

The harmonic series: the most important divergent example

The harmonic series is

\sum_{n=1}^{\infty} \frac{1}{n}

Its terms go to 0, but the series diverges. Conceptually, it diverges because the terms decrease too slowly: even though you add tiny amounts, there are infinitely many of them and they accumulate without bound.

A common intuition (often illustrated in calculus courses) is grouping terms to show that the partial sums eventually exceed any bound. You don’t need the full proof for many AP problems, but you do need the fact: the harmonic series diverges.

p-series: a whole family you can classify instantly

A p-series is a series of the form

\sum_{n=1}^{\infty} \frac{1}{n^p}

where p is a real constant.

The p-series test tells you exactly when it converges:

• If p > 1, the series converges.
• If p \le 1, the series diverges.

This is one of the most high-value “recognition” results in Unit 10 because it lets you decide convergence without heavy computation.

Why the boundary is at p = 1 (intuition)

As p increases, terms like \frac{1}{n^p} drop off faster. Faster decay makes it more plausible that the total added amount stays finite.

• At p = 1 you get the harmonic series, which diverges.
• For p > 1, the terms shrink quickly enough for convergence.
• For 0 < p < 1, the terms shrink too slowly, leading to divergence.

You don’t need to re-prove this each time; you recognize the pattern and apply the classification.

Notation and recognition table

Many problems are really about spotting that something is (or can be rewritten as) a p-series.

Worked example 1: identify a p-series

Determine convergence:

\sum_{n=1}^{\infty} \frac{1}{n^{5}}

This is a p-series with p = 5. Since p > 1, it converges.

Worked example 2: rewrite to reveal the p-series

Determine convergence:

\sum_{n=1}^{\infty} \frac{1}{n\sqrt{n}}

Rewrite the term:

\frac{1}{n\sqrt{n}} = \frac{1}{n \cdot n^{1/2}} = \frac{1}{n^{3/2}}

So the series is

\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}

This is a p-series with p = \frac{3}{2}, and since p > 1, it converges.

Worked example 3: harmonic series recognition

Determine convergence:

\sum_{n=1}^{\infty} \frac{10}{n}

This is a constant multiple of the harmonic series. The term \frac{10}{n} is still of the form \frac{C}{n}, which corresponds to p = 1. Constant multiples do not change convergence behavior, so it diverges.

What can go wrong: confusing p-series with “anything rational”

A common error is to see a fraction with n in it and assume p-series. But p-series require a very specific structure: a constant numerator and a denominator that is exactly n^p.

For example:

• \sum \frac{1}{n+1} is not literally a p-series, though it behaves like one and also diverges (it’s essentially harmonic after an index shift).
• \sum \frac{1}{n(n+1)} is not a p-series (and in fact converges, but you’d need other tools to show it).

Within “fundamentals,” your job is to correctly identify true p-series and the harmonic series, and to know their convergence behavior.

Exam Focus

• Typical question patterns:
  • Classify a given series as harmonic or p-series and state convergence based on the value of p.
  • Rewrite an expression (using exponent rules) to reveal \frac{1}{n^p}.
  • Use harmonic series divergence as a key fact when a series is a constant multiple or an index shift of \sum \frac{1}{n}.
• Common mistakes:
  • Misidentifying p when radicals are involved (for example, forgetting that \sqrt{n} = n^{1/2}).
  • Thinking that any series with terms approaching 0 converges (harmonic series is the counterexample you must remember).
  • Treating \sum \frac{1}{n^p} as convergent for all positive p (the cutoff is p > 1, not p > 0).