AP Calculus BC Unit 10 Study Guide: Infinite Sequences, Series, Convergence Tests, and Power Series

Sequences: What They Are and How Limits Work

A sequence is a list of numbers written in a specific order. You can think of it as a function whose input is a positive integer %%LATEX0%% (the “term number”) and whose output is the value of the %%LATEX1%%th term, usually written %%LATEX2%%. In this course, %%LATEX3%% is always an integer.

For example, the sequence

$a_n=\frac{1}{n}$

produces the list

$1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots$

Another example (starting at $n=1$) is

$a_n=\frac{n-1}{n}$

which generates

$0,\frac{1}{2},\frac{2}{3},\dots$

Why sequences matter

Sequences are the building blocks for series (sums of infinitely many terms). In AP Calculus BC, nearly every question about infinite series depends on understanding what happens to the underlying terms %%LATEX9%% as %%LATEX10%%. A key idea you’ll use repeatedly is this: if the terms of a series do not go to zero, then the series cannot converge.

Limits of sequences

The limit of a sequence is the value the terms approach as $n$ grows without bound. We write

$\lim_{n\to\infty} a_n=L$

if you can make %%LATEX13%% arbitrarily close to %%LATEX14%% by taking $n$ large enough.

Many sequence limits use the same algebraic strategies you learned for limits of functions: dividing by the highest power, using dominant-term reasoning, or recognizing known limits.

A more formal definition that sometimes appears in explanations is the %%LATEX16%%-%%LATEX17%% definition: a sequence has limit %%LATEX18%% if for any %%LATEX19%% there is an associated positive integer $N$ such that

$|a_n-L|<\varepsilon$

for all $n\ge N$.

Example 1: A rational sequence

Find the limit:

$\lim_{n\to\infty} \frac{3n^2-7}{n^2+5n}$

Reasoning: For large %%LATEX24%%, the highest power dominates. Divide numerator and denominator by %%LATEX25%%.

$\frac{3n^2-7}{n^2+5n}=\frac{3-\frac{7}{n^2}}{1+\frac{5}{n}}$

Now take the limit as $n\to\infty$:

$\lim_{n\to\infty} \frac{3-\frac{7}{n^2}}{1+\frac{5}{n}}=\frac{3-0}{1+0}=3$

So the sequence converges to $3$.

Example 2: A geometric-type sequence

Evaluate:

$\lim_{n\to\infty} \left(\frac{2}{3}\right)^n$

Reasoning: Each term is multiplied by $\frac{2}{3}$ again. Because

$\left|\frac{2}{3}\right|<1$

the powers shrink to $0$.

$\lim_{n\to\infty} \left(\frac{2}{3}\right)^n=0$

Divergence and oscillation

A sequence diverges if it does not approach a single finite number. Divergence can happen in different ways:

• It can grow without bound (for example $a_n=n$).
• It can oscillate without settling (for example $a_n=(-1)^n$).

A common misconception is to think “bounded means convergent.” Not true: %%LATEX37%% is bounded between %%LATEX38%% and $1$ but does not converge.

Exam Focus

• Typical question patterns:
  • Compute $\lim_{n\to\infty} a_n$ for a given formula (often rational, exponential, or involving radicals).
  • Use a sequence limit to decide whether a related series can possibly converge (the nth-term test).
• Common mistakes:
  • Assuming a bounded sequence must converge (it can oscillate).
  • Confusing %%LATEX41%% (terms) with partial sums %%LATEX42%% (sums) once series begin.

Series and Partial Sums: Turning a Sequence into an Infinite Sum

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

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

Equivalently, you can picture it as

$a_1+a_2+a_3+\dots$

where the numbers $a_1,a_2,a_3,\dots$ are the terms of the series.

The key idea: convergence is about partial sums

An infinite sum only makes sense through the concept of partial sums. The $n$th partial sum is

$S_n=\sum_{k=1}^{n} a_k$

As %%LATEX49%% increases, the partial sum includes more and more terms of the series. The series converges if the sequence of partial sums converges to a finite limit %%LATEX50%%:

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

means

$\lim_{n\to\infty} S_n=S$

If $S_n$ does not approach a finite number, the series diverges, which means a divergent series has no sum.

Why “terms go to zero” is necessary but not sufficient

If a series converges, then its terms must approach zero:

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

The contrapositive is a fast divergence check (the nth-term test for divergence): you find the limit of the term sequence.

$\text{If }\lim_{n\to\infty} a_n\ne 0\text{ or does not exist, then }\sum_{n=1}^{\infty} a_n\text{ diverges}$

If the limit equals $0$, the nth-term test does not finish the problem and you must choose a different test.

Be careful: $\lim_{n\to\infty} a_n=0$ does not guarantee convergence. A classic example is the harmonic series:

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

The terms go to zero, but the sum still diverges.

Telescoping series

A telescoping series is designed so most terms cancel in the partial sums. These are among the most “directly computable” infinite series because you can often find $S_n$ explicitly and then take a limit.

Example: telescoping by decomposition

Consider

$\sum_{n=1}^{\infty} \left(\frac{1}{n}-\frac{1}{n+1}\right)$

Compute partial sums:

$S_n=\sum_{k=1}^{n} \left(\frac{1}{k}-\frac{1}{k+1}\right)$

Write out the first few terms:

$S_n=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots+\left(\frac{1}{n}-\frac{1}{n+1}\right)$

Everything cancels except the first $1$ and the last negative term:

$S_n=1-\frac{1}{n+1}$

Now take the limit:

$\lim_{n\to\infty} S_n=\lim_{n\to\infty} \left(1-\frac{1}{n+1}\right)=1$

So the series converges to $1$.

Exam Focus

• Typical question patterns:
  • Determine whether $\sum a_n$ converges using the nth-term test.
  • Compute the exact sum of a telescoping series by finding $S_n$ and taking a limit.
• Common mistakes:
  • Thinking “terms go to zero” implies convergence.
  • Cancelling incorrectly in telescoping problems (write out several terms to see the pattern).

Geometric Series and p-Series: The Two Core Benchmarks

Two families of series show up constantly because they are easy to classify and they serve as comparison targets.

Geometric series

A geometric series can be written in either of these common forms:

$\sum_{n=0}^{\infty} ar^n$

or

$\sum_{n=1}^{\infty} ar^{n-1}$

where %%LATEX71%% is the first term and %%LATEX72%% is the common ratio.

• It converges if $|r|<1$.
• It diverges if $|r|\ge 1$.

These show up often on the AP exam: you are frequently asked to determine whether a geometric series converges or diverges, and if it converges, to what value.

When it converges, its sum is

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

and the corresponding finite partial sum (sum of the first $n$ terms) is

$S_n=\frac{a(1-r^n)}{1-r}$

Where the finite geometric sum formula comes from (classic derivation)

Let

$S_n=a+ar+ar^2+\dots+ar^{n-1}$

Multiply by $r$:

$rS_n=ar+ar^2+ar^3+\dots+ar^n$

Subtract:

$S_n-rS_n=a-ar^n$

Factor and solve:

$S_n(1-r)=a-ar^n$

$S_n=\frac{a(1-r^n)}{1-r}$

For an infinite geometric series with $|r|<1$, you also use the fact that

$\lim_{n\to\infty} r^n=0$

which leads to the infinite-sum formula

$S=\frac{a}{1-r}$

Example: rewrite as geometric

Evaluate

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

Factor out the constant and recognize a geometric series with first term %%LATEX88%% and ratio %%LATEX89%%:

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

Use the formula starting at %%LATEX91%% by identifying the first term %%LATEX92%%:

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

So the sum is $3\cdot 1=3$.

Example (from benchmark intuition): convergent vs divergent geometric

The series

$\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots$

converges because the ratio is $\frac{1}{2}$, while

$2+2^2+2^3+\dots$

diverges because the ratio is $2$.

Harmonic series and p-series

The harmonic series has the pattern

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

It might look as though it is converging because terms get small, but it diverges because the partial sums do not approach a finite limit.

A p-series has the form

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

It converges if %%LATEX101%% and diverges if %%LATEX102%% (including the harmonic series case %%LATEX103%%). In particular, for %%LATEX104%% it diverges as well.

Example: classify using p-series knowledge

Determine convergence of

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

This is a p-series with

$p=\frac{3}{2}>1$

so it converges.

Exam Focus

• Typical question patterns:
  • Recognize a series as geometric or p-series (possibly after algebraic rewriting).
  • Compute the sum of a convergent geometric series.
• Common mistakes:
  • Using the geometric sum formula when $|r|\ge 1$.
  • Mixing up the p-series cutoff (the boundary is exactly $p=1$).

The Integral Test and Error Bounds: Series as “Continuous” Objects

The integral test connects an infinite series to an improper integral. It is especially useful for terms that come from a nice, decreasing function like %%LATEX109%% or %%LATEX110%%.

Integral Test (what it says)

Suppose %%LATEX111%% where %%LATEX112%% is continuous, positive, and decreasing for %%LATEX113%% (many textbooks state it for %%LATEX114%% as a common starting point). Then

$\sum_{n=N}^{\infty} a_n$

and

$\int_{N}^{\infty} f(x)\,dx$

either both converge or both diverge.

(Original reference image from the earlier notes: https://knowt-user-attachments.s3.amazonaws.com/a56dd38d188547ed9a01c1412cc92d3f.jpeg)

Why it works (intuition)

If %%LATEX117%% is decreasing, the rectangles of width %%LATEX118%% with heights $f(n)$ can be compared to the area under the curve. The integral gives a continuous approximation to the discrete sum.

Remainder (error) estimate for the Integral Test

If you approximate the series by the partial sum $S_n$, the remainder is

$R_n=\sum_{k=n+1}^{\infty} a_k$

For the integral test conditions (positive, decreasing), you can bound the error using integrals:

$\int_{n+1}^{\infty} f(x)\,dx\le R_n\le\int_{n}^{\infty} f(x)\,dx$

This is powerful on exams when you’re asked how many terms are needed so that the approximation is within a given tolerance.

Example: using the integral test for convergence

Test

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

Let

$f(x)=\frac{1}{x\ln x}$

for %%LATEX125%%. It is positive and decreasing for large enough %%LATEX126%%. Consider

$\int_{2}^{\infty} \frac{1}{x\ln x}\,dx$

Use substitution

$u=\ln x$

so

$du=\frac{1}{x}dx$

Then

$\int_{2}^{\infty} \frac{1}{x\ln x}\,dx=\int_{\ln 2}^{\infty} \frac{1}{u}\,du$

But

$\int_{\ln 2}^{\infty} \frac{1}{u}\,du$

diverges, so the series diverges.

Example: terms needed for a p-series approximation

How large must $n$ be so that

$\sum_{k=1}^{n} \frac{1}{k^2}$

approximates

$\sum_{k=1}^{\infty} \frac{1}{k^2}$

within error less than $0.01$?

Here %%LATEX136%% and %%LATEX137%%. Use

$R_n\le\int_{n}^{\infty} \frac{1}{x^2}\,dx$

Compute the integral:

$\int_{n}^{\infty} \frac{1}{x^2}\,dx=\left[-\frac{1}{x}\right]_{n}^{\infty}=\frac{1}{n}$

To guarantee error less than $0.01$, require

$\frac{1}{n}<0.01$

So %%LATEX142%%. Taking %%LATEX143%% is sufficient.

Exam Focus

• Typical question patterns:
  • Decide convergence or divergence by comparing a series to an improper integral.
  • Use the remainder bound to find how many terms are needed for a given accuracy.
• Common mistakes:
  • Forgetting the conditions (positive and decreasing) before applying the integral test.
  • Using the %%LATEX144%% and %%LATEX145%% bounds backwards.

Comparison Tests: Direct and Limit Comparison

Most series you meet are not exactly geometric or p-series, so you often prove convergence by comparing them to something you already understand.

Direct Comparison Test

Assume

$a_n\ge 0$

and

$b_n\ge 0$

• If %%LATEX148%% for all sufficiently large %%LATEX149%% and %%LATEX150%% converges, then %%LATEX151%% converges.
• If %%LATEX152%% for all sufficiently large %%LATEX153%% and %%LATEX154%% diverges, then %%LATEX155%% diverges.

Intuition: if the terms of one series are always less than the terms of a convergent benchmark series (eventually), the smaller one cannot “build up” enough to diverge. Similarly, if the terms are larger than a divergent benchmark (eventually), it cannot converge.

Example: direct comparison (convergence)

Test

$\sum_{n=1}^{\infty} \frac{1}{n^2+5}$

For all $n$,

$n^2+5\ge n^2$

So

$\frac{1}{n^2+5}\le\frac{1}{n^2}$

And since

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

converges (p-series with $p=2$), the given series converges by comparison.

Example: direct comparison (divergence)

Test

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

For %%LATEX163%%, %%LATEX164%%, so

$\frac{1}{n-1}\ge\frac{1}{n}$

Since

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

diverges, the given series diverges.

Limit Comparison Test

Direct comparison can be tricky when inequalities aren’t obvious. The limit comparison test avoids this by comparing ratios.

Let

$a_n>0$

and

$b_n>0$

and suppose

$\lim_{n\to\infty} \frac{a_n}{b_n}=c$

where

$0<c<\infty$

Then %%LATEX171%% and %%LATEX172%% either both converge or both diverge.

(Original reference image from the earlier notes: https://knowt-user-attachments.s3.amazonaws.com/d13ec47c1a4b4f9a9cd9a57b6db59660.jpeg)

Example: limit comparison with a rational expression

Test

$\sum_{n=1}^{\infty} \frac{4n+1}{n^2-3n}$

For large $n$, the expression behaves like

$\frac{4n}{n^2}=\frac{4}{n}$

Choose

$b_n=\frac{1}{n}$

Compute the limit:

$\lim_{n\to\infty} \frac{\frac{4n+1}{n^2-3n}}{\frac{1}{n}}=\lim_{n\to\infty} \frac{n(4n+1)}{n^2-3n}$

Simplify:

$\frac{n(4n+1)}{n(n-3)}=\frac{4n+1}{n-3}$

Then

$\lim_{n\to\infty} \frac{4n+1}{n-3}=4$

Since the limit is finite and positive, the series behaves like $\sum \frac{1}{n}$, which diverges. Therefore the given series diverges.

Exam Focus

• Typical question patterns:
  • Pick a known benchmark (often $\frac{1}{n^p}$) and justify comparison.
  • Use limit comparison for rational functions or expressions with roots.
• Common mistakes:
  • Using direct comparison with the inequality in the wrong direction.
  • Getting a limit of %%LATEX182%% or %%LATEX183%% in limit comparison and concluding “both converge or both diverge” (that conclusion only works when the limit is a positive finite constant).

Alternating Series, Absolute vs Conditional Convergence

Not all convergent series have positive terms. Alternating signs can create cancellation that makes a series converge even when the corresponding positive-term series diverges.

Alternating Series Test (Leibniz Test)

An alternating series typically looks like

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

or

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

or

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

where

$b_n\ge 0$

The series converges if:

1. %%LATEX188%% is decreasing for all sufficiently large %%LATEX189%% (many notes phrase this as %%LATEX190%% for all %%LATEX191%% once you’re far enough out), and
2. $\lim_{n\to\infty} b_n=0$.

Why it works: the partial sums zig-zag and get trapped in a shrinking interval.

Example: alternating harmonic series

Test

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

Here

$b_n=\frac{1}{n}$

decreases and goes to $0$, so the series converges by the alternating series test.

Absolute vs conditional convergence

Given a series $\sum a_n$:

• It is absolutely convergent if $\sum |a_n|$ converges.
• It is conditionally convergent if %%LATEX198%% converges but %%LATEX199%% diverges.

This is sometimes summarized as an “absolute convergence theorem”: if a series converges absolutely, then it converges.

Example: conditional convergence

We already know

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

converges. But

$\sum_{n=1}^{\infty} \left|\frac{(-1)^{n-1}}{n}\right|=\sum_{n=1}^{\infty} \frac{1}{n}$

diverges. So the alternating harmonic series is conditionally convergent.

Alternating Series Estimation Theorem (error bound)

When summing an alternating series with a finite number of terms, the partial sum will not equal the full infinite sum. If an alternating series converges by the alternating series test, then the error when approximating by $S_n$ is at most the next term:

$|R_n|=\left|\sum_{k=n+1}^{\infty} a_k\right|\le b_{n+1}$

In practical terms: if you sum the first 10 terms, the error is less than the absolute value of the 11th term (assuming the hypotheses are met).

Example: how many terms for an alternating approximation

How many terms of

$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2}$

are needed so that the error is less than $0.001$?

Here

$b_n=\frac{1}{n^2}$

We want

$b_{n+1}<0.001$

So

$\frac{1}{(n+1)^2}<0.001$

That means

$(n+1)^2>1000$

So

$n+1>\sqrt{1000}\approx 31.62$

and %%LATEX211%% works. Using the first 31 terms guarantees error less than %%LATEX212%%.

Exam Focus

• Typical question patterns:
  • Decide convergence of an alternating series and classify absolute vs conditional.
  • Use the alternating series error bound to choose $n$ for a required accuracy.
• Common mistakes:
  • Forgetting to check that $b_n$ decreases (or is eventually decreasing).
  • Applying the alternating error bound to a series that is not alternating or not decreasing.

Ratio and Root Tests: Best for Factorials and Exponentials

Some series involve factorials, exponentials, or products where comparison to p-series is awkward. The ratio test and root test handle these efficiently.

Ratio Test

For a series %%LATEX215%% (often applied to %%LATEX216%% when signs alternate), compute

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

• If $L<1$, the series converges absolutely.
• If %%LATEX219%% (or %%LATEX220%%), the series diverges.
• If $L=1$, the test is inconclusive.

(Original reference image from the earlier notes: https://knowt-user-attachments.s3.amazonaws.com/d875fe737ef44e8e84f0f937b60cb06e.jpeg)

Example: factorial series (diverges)

Test

$\sum_{n=1}^{\infty} \frac{n!}{3^n}$

Let

$a_n=\frac{n!}{3^n}$

Then

$\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)!}{3^{n+1}}\cdot\frac{3^n}{n!}=\frac{n+1}{3}$

So

$L=\lim_{n\to\infty} \frac{n+1}{3}=\infty$

Since $L>1$, the series diverges.

Example: ratio test convergence

Test

$\sum_{n=1}^{\infty} \frac{3^n}{n!}$

Let

$a_n=\frac{3^n}{n!}$

Then

$\left|\frac{a_{n+1}}{a_n}\right|=\frac{3^{n+1}}{(n+1)!}\cdot\frac{n!}{3^n}=\frac{3}{n+1}$

Thus

$L=\lim_{n\to\infty} \frac{3}{n+1}=0$

So the series converges absolutely.

Root Test

Compute

$L=\lim_{n\to\infty} \sqrt[n]{|a_n|}$

• If $L<1$, the series converges absolutely.
• If $L>1$, it diverges.
• If $L=1$, inconclusive.

The root test is especially useful when terms look like something “to the $n$” power.

Example: root test inconclusive

Test

$\sum_{n=1}^{\infty} \left(\frac{n}{n+1}\right)^n$

Let

$a_n=\left(\frac{n}{n+1}\right)^n$

Then

$\sqrt[n]{a_n}=\frac{n}{n+1}$

So

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

The root test is inconclusive here; you would need a different approach.

Exam Focus

• Typical question patterns:
  • Apply the ratio test to factorial or exponential expressions.
  • Use the ratio test or root test to find the radius of convergence of a power series.
• Common mistakes:
  • Forgetting absolute values in the ratio or root test when signs alternate.
  • Concluding convergence or divergence when $L=1$ (the test is inconclusive).

Power Series: Infinite Polynomials and Their Convergence

A power series centered at $c$ looks like an infinite polynomial:

$\sum_{n=0}^{\infty} a_n(x-c)^n$

Here %%LATEX243%% are constants and %%LATEX244%% is the variable.

Why power series matter

Power series are the bridge between infinite series and functions. In BC calculus, you use them to represent functions (sometimes exactly), approximate function values with polynomials, differentiate and integrate functions via their series, and evaluate otherwise difficult limits and integrals.

Convergence behavior: radius and interval

A key fact (used routinely in AP problems) is that a power series has one of three convergence patterns:

• It converges only at $x=c$.
• It converges for all real $x$.
• It converges for %%LATEX247%% and diverges for %%LATEX248%%, and you must test endpoints %%LATEX249%% and %%LATEX250%% separately.

The number %%LATEX251%% is the radius of convergence, and the set of all %%LATEX252%% values where the series converges is the interval of convergence.

A common bookkeeping issue is that some power series are written starting at %%LATEX253%% rather than %%LATEX254%%, which changes the first term but does not change the overall convergence strategy.

Special cases to remember:

• If the series converges only for %%LATEX255%%, then %%LATEX256%%.
• If the series converges for all real %%LATEX257%%, then %%LATEX258%%.

When centered at %%LATEX259%%, the interval is typically written %%LATEX260%% before checking endpoints.

(Original reference image from the earlier notes: https://knowt-user-attachments.s3.amazonaws.com/a40b3988ed984aa69bd06871e6bcb23e.jpeg)

Finding the radius of convergence (usually ratio test)

Most AP BC problems use the ratio test because it handles powers cleanly.

Example: find radius and interval of convergence

Find the interval of convergence of

$\sum_{n=1}^{\infty} \frac{(x-2)^n}{n}$

Let

$a_n=\frac{(x-2)^n}{n}$

Apply the ratio test:

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

Take the limit:

$L=\lim_{n\to\infty} |x-2|\cdot\frac{n}{n+1}=|x-2|$

Convergence requires $L<1$, so

$|x-2|<1$

This gives the open interval $1<x<3$.

Now test endpoints.

At $x=1$, the series becomes

$\sum_{n=1}^{\infty} \frac{(1-2)^n}{n}=\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$

This converges by the alternating series test.

At $x=3$, the series becomes

$\sum_{n=1}^{\infty} \frac{(3-2)^n}{n}=\sum_{n=1}^{\infty} \frac{1}{n}$

This diverges (harmonic series).

So the interval of convergence is %%LATEX272%% and the radius of convergence is %%LATEX273%%.

Convergence is about the full series, not just $|x-c|<R$

A frequent error is to stop after finding $R$ and forget to test endpoints. Endpoints can converge, diverge, or (less commonly) one of each.

Exam Focus

• Typical question patterns:
  • Use the ratio test to find $R$ and then test endpoints to get the interval.
  • Determine for which $x$ a given power series defines a function.
• Common mistakes:
  • Not testing endpoints.
  • Algebra mistakes when simplifying the ratio test (especially handling powers of $x-c$).

Differentiating and Integrating Power Series (Term-by-Term)

One of the most useful properties of power series is that, within their interval of convergence, they behave like polynomials: you can differentiate and integrate term-by-term.

The core rules

If

$f(x)=\sum_{n=0}^{\infty} a_n(x-c)^n$

then for $x$ in the interval of convergence,

$f'(x)=\sum_{n=1}^{\infty} na_n(x-c)^{n-1}$

and

$\int f(x)\,dx=C+\sum_{n=0}^{\infty} \frac{a_n}{n+1}(x-c)^{n+1}$

Why this matters

This is the engine behind many AP BC tasks: generating a new series from a known one, finding a series for a derivative or antiderivative, representing functions as power series, and approximating integrals using Maclaurin or Taylor expansions.

A subtle but important point: differentiating or integrating a power series keeps the same center $c$ and typically the same radius of convergence, though endpoint behavior can change.

Example: differentiate a power series and discuss convergence

Given

$f(x)=\sum_{n=0}^{\infty} \frac{(x-1)^n}{2^n}$

Differentiate term-by-term:

$f'(x)=\sum_{n=1}^{\infty} n\frac{(x-1)^{n-1}}{2^n}$

For convergence, note the original is geometric:

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

So it converges when

$\left|\frac{x-1}{2}\right|<1$

That is

$|x-1|<2$

So the radius is %%LATEX289%% and the differentiated series converges at least on %%LATEX290%% (endpoint behavior can be checked if needed).

Exam Focus

• Typical question patterns:
  • Find a power series for a new function by differentiating or integrating a known series.
  • Use a derived power series to approximate values or compute coefficients.
• Common mistakes:
  • Differentiating the index incorrectly (watch the starting index shift from %%LATEX291%% to %%LATEX292%%).
  • Forgetting to update the power of $x-c$ after differentiation or integration.

Taylor Polynomials and Taylor Series: Building Functions from Derivatives

A Taylor polynomial approximates a function near a point %%LATEX294%% using information from derivatives at %%LATEX295%%. The Taylor series is what you get if you keep adding higher-degree terms forever.

Taylor polynomial (degree $n$)

If %%LATEX297%% can be differentiated %%LATEX298%% times at %%LATEX299%%, the degree-%%LATEX300%% Taylor polynomial about $x=a$ is

$P_n(x)=\sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k$

Equivalently, written out:

$P_n(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\dots+\frac{f^{(n)}(a)}{n!}(x-a)^n$

(Original reference image from the earlier notes: https://knowt-user-attachments.s3.amazonaws.com/4742db23289d4067951dd1f440ba5b82.jpeg)

Taylor series

The Taylor series is the infinite extension:

$\sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}(x-a)^k$

If the Taylor series converges to $f(x)$ on an interval, then you have a powerful representation of the function as an infinite polynomial.

Maclaurin series

A Maclaurin series is just a Taylor series centered at $0$:

$\sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}x^k$

Why Taylor series matter in BC

Taylor series unify many topics: polynomial approximations (for estimation), creating new series from known ones through algebra/differentiation/integration, and evaluating tricky limits by replacing functions with their series near a point.

Worked example: build a Taylor polynomial from derivatives

Find the degree-3 Taylor polynomial for %%LATEX308%% about %%LATEX309%%.

Compute derivatives:

$f(x)=\ln x$

$f'(x)=\frac{1}{x}$

$f''(x)=-\frac{1}{x^2}$

$f'''(x)=\frac{2}{x^3}$

Evaluate at $x=1$:

$f(1)=0$

$f'(1)=1$

$f''(1)=-1$

$f'''(1)=2$

Now plug into the Taylor polynomial:

$P_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)}{2!}(x-1)^2+\frac{f'''(1)}{3!}(x-1)^3$

So

$P_3(x)=(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3$

A common mistake is forgetting the factorial in the denominator.

Exam Focus

• Typical question patterns:
  • Construct a Taylor polynomial from a table of derivative values at a point.
  • Recognize the general term $\frac{f^{(k)}(a)}{k!}(x-a)^k$ and match coefficients.
• Common mistakes:
  • Confusing the expansion point %%LATEX322%% (using %%LATEX323%% instead of $(x-a)^k$).
  • Dropping factorials or miscomputing derivative values.

Common Maclaurin Series You Must Know (and How to Generate More)

AP Calculus BC expects you to know several “parent” Maclaurin series. You then build many other series from them using substitution, algebra, differentiation, and integration.

The geometric-series template

The geometric series

$\sum_{n=0}^{\infty} x^n$

converges for $|x|<1$ and sums to

$\sum_{n=0}^{\infty} x^n=\frac{1}{1-x}$

This identity is the starting point for many derived series. Replacing %%LATEX328%% with another expression (like %%LATEX329%% or $3x$) is one of the fastest ways to generate new power series.

Exponential, sine, and cosine

These are fundamental Maclaurin series:

$e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$

$\sin x=\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$

$\cos x=\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$

These converge for all real $x$.

Logarithm and arctangent

A key derived series (obtainable from geometric series by integration) is:

$\ln(1+x)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^n}{n}$

This converges for $-1<x\le 1$.

Another common one is:

$\arctan x=\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}$

This converges for $-1\le x\le 1$.

How to create new series from known ones

You’ll repeatedly do these moves:

1. Substitute: replace %%LATEX339%% with some expression (like %%LATEX340%% or $(x-3)$). This changes the interval of convergence accordingly.
2. Multiply or divide by powers of $x$: shifts powers and coefficients.
3. Differentiate or integrate term-by-term: generates series for related functions.
4. Combine series: add or subtract series to represent more complicated functions.

Example: build a series for a rational function

Find a power series for

$\frac{1}{1+x}$

Start from

$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^n$

Replace %%LATEX345%% by %%LATEX346%%:

$\frac{1}{1-(-x)}=\frac{1}{1+x}=\sum_{n=0}^{\infty} (-x)^n=\sum_{n=0}^{\infty} (-1)^n x^n$

Convergence requires %%LATEX348%%, so %%LATEX349%%.

Example: build a series for a related logarithm

Find a series for $\ln(1-x)$.

Use %%LATEX351%% and substitute %%LATEX352%%:

$\ln(1-x)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}(-x)^n}{n}$

Simplify:

$\ln(1-x)=-\sum_{n=1}^{\infty} \frac{x^n}{n}$

The convergence interval transforms from %%LATEX355%% for %%LATEX356%% into %%LATEX357%% for %%LATEX358%%.

Exam Focus

• Typical question patterns:
  • Start from a known Maclaurin series and derive a new one by substitution and algebra.
  • State the interval of convergence after transforming the series.
• Common mistakes:
  • Transforming the series but forgetting to transform the convergence interval.
  • Losing an index shift (for example confusing a series starting at %%LATEX359%% versus %%LATEX360%%).

Taylor Remainder (Error) and When a Taylor Series Equals the Function

Taylor polynomials are approximations. To use them responsibly, you need a way to talk about the error.

Lagrange form of the remainder

If %%LATEX361%% is the degree-%%LATEX362%% Taylor polynomial for %%LATEX363%% centered at %%LATEX364%%, then the remainder is

$R_n(x)=f(x)-P_n(x)$

A standard bound used in AP Calculus is the Lagrange error bound:

$|R_n(x)|\le\frac{M|x-a|^{n+1}}{(n+1)!}$

where $M$ is a number such that

$|f^{(n+1)}(t)|\le M$

for all %%LATEX369%% between %%LATEX370%% and $x$.

(Original reference image from the earlier notes: https://knowt-user-attachments.s3.amazonaws.com/10726dbb765d4c64bd3b4029f9428e93.jpeg)

A practical rule-of-thumb that appears often in class notes: if you are finding an $n$th-degree Taylor polynomial and the associated Taylor series behaves like a decreasing alternating series at your evaluation point, then a good approximation to the error bound is the next nonzero term (because the alternating-series bound applies in that situation).

Worked example: error bound for $e^x$

Approximate $e^{0.1}$ using the degree-2 Maclaurin polynomial and bound the error.

For %%LATEX375%%, all derivatives are %%LATEX376%%. The degree-2 Maclaurin polynomial is

$P_2(x)=1+x+\frac{x^2}{2}$

So

$P_2(0.1)=1+0.1+\frac{0.1^2}{2}=1.105$

Now bound the error. Here %%LATEX379%%, %%LATEX380%%, and we need an upper bound %%LATEX381%% on %%LATEX382%% for %%LATEX383%% between %%LATEX384%% and %%LATEX385%%. Since %%LATEX386%% increases, the maximum occurs at %%LATEX387%%, so you can take %%LATEX388%%. For a simple numerical bound without knowing %%LATEX389%% exactly, you can use %%LATEX390%%.

Using $M=e$ gives

$|R_2(0.1)|\le\frac{e|0.1|^3}{3!}=\frac{e\cdot 0.001}{6}$

This is approximately $0.000453$.

When does the Taylor series equal the function?

In AP BC, you generally rely on known facts: for functions like %%LATEX394%%, %%LATEX395%%, and %%LATEX396%%, the Taylor series converges to the function for all real %%LATEX397%%. For other functions, the Taylor series might converge only on an interval, and sometimes it converges but not to the original function outside a certain region.

Practically, on the AP exam, you are usually told (or expected to know from the standard list) when a series representation is valid and on what interval.

Exam Focus

• Typical question patterns:
  • Use the Lagrange remainder bound to guarantee an approximation is within a tolerance.
  • Use alternating-series error bounds for alternating Taylor series (like %%LATEX398%% and %%LATEX399%%) when applicable.
• Common mistakes:
  • Using %%LATEX400%% incorrectly (it must bound the absolute value of the next derivative on the entire interval between %%LATEX401%% and $x$).
  • Forgetting that Taylor error depends on $|x-a|^{n+1}$ and factorial growth; missing parentheses or powers changes everything.

Using Series to Evaluate Limits and Compute Approximations

A major payoff of Taylor and Maclaurin series is that they let you replace complicated functions with simpler polynomial-like expressions near a point.

The core strategy for limits

When %%LATEX404%% is close to %%LATEX405%% (or close to another center), replace functions with their truncated series and simplify. The lowest-power nonzero term usually determines the limit.

This is especially useful for indeterminate forms like $\frac{0}{0}$ where repeated L’Hôpital’s Rule would be slow.

Example: classic series-based limit

Evaluate

$\lim_{x\to 0} \frac{\sin x-x}{x^3}$

Use the Maclaurin series

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots$

Then

$\sin x-x=-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots$

Divide by $x^3$:

$\frac{\sin x-x}{x^3}=-\frac{1}{3!}+\frac{x^2}{5!}-\dots$

As $x\to 0$, the higher-power terms vanish, so the limit is

$-\frac{1}{6}$

A common mistake is stopping too early: if you approximate %%LATEX414%%, you would incorrectly get %%LATEX415%%. You need enough terms to capture the first non-canceling behavior.

Approximating definite integrals with series

If you have a power series for $f(x)$, you can integrate term-by-term to approximate integrals that have no elementary antiderivative.

Example: integrate a series to approximate an area

Approximate

$\int_{0}^{1/2} \frac{1}{1-x^2}\,dx$

Use the geometric identity

$\frac{1}{1-u}=\sum_{n=0}^{\infty} u^n$

with $u=x^2$:

$\frac{1}{1-x^2}=\sum_{n=0}^{\infty} x^{2n}$

This is valid for %%LATEX421%%, which includes %%LATEX422%%. Integrate term-by-term:

$\int_{0}^{1/2} \frac{1}{1-x^2}\,dx=\sum_{n=0}^{\infty} \int_{0}^{1/2} x^{2n}\,dx$

Compute each integral:

$\int_{0}^{1/2} x^{2n}\,dx=\left[\frac{x^{2n+1}}{2n+1}\right]_{0}^{1/2}=\frac{(1/2)^{2n+1}}{2n+1}$

So

$\int_{0}^{1/2} \frac{1}{1-x^2}\,dx=\sum_{n=0}^{\infty} \frac{(1/2)^{2n+1}}{2n+1}$

For a numerical approximation, take a few terms:

$\frac{(1/2)^1}{1}+\frac{(1/2)^3}{3}+\frac{(1/2)^5}{5}+\dots$

That is

$\frac{1}{2}+\frac{1}{24}+\frac{1}{160}+\dots$

You can bound the error depending on the series type (often geometric-type tail bounds or alternating bounds if alternating).

Real-world interpretation: why series approximations are useful

In science and engineering, measuring devices and computations often work best with polynomials. Taylor polynomials let you approximate non-polynomial behavior using a small number of arithmetic operations, while also controlling error.

Exam Focus

• Typical question patterns:
  • Evaluate a limit by replacing functions with their series and simplifying.
  • Approximate a value (function or integral) using a Taylor polynomial and give an error bound.
• Common mistakes:
  • Using too few terms so that leading terms cancel and you lose the true dominant behavior.
  • Forgetting the interval where a substituted series is valid (for example using a geometric-series expansion outside $|x|<1$).

Putting It Together: Choosing the Right Convergence Test

By the time you reach the AP exam, the hard part is often not executing a test but choosing it efficiently. Different tests shine for different structures.

How to decide (conceptual guide)

If you see:

• A clear geometric form %%LATEX429%% (or %%LATEX430%%): use geometric convergence and sum if asked.
• A pure $\frac{1}{n^p}$ form: use p-series.
• A rational function in %%LATEX432%%: use limit comparison with the dominant power of %%LATEX433%%.
• An alternating sign with decreasing magnitude: use the alternating series test, then check absolute convergence if asked.
• Factorials or products like $n!$: use the ratio test.
• Terms like %%LATEX435%% or %%LATEX436%%: use the root test or ratio test.
• A function-like term $f(n)$ where integrals are easy and terms are positive and decreasing: use the integral test.

Important warning: “inconclusive” is a real outcome

Ratio and root tests often give $L=1$. That is not failure, it is information that you must switch strategies (often to comparison, p-series recognition, or alternating series tools).

Example: ratio test inconclusive, use p-series

Consider

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

Ratio test gives

$\left|\frac{a_{n+1}}{a_n}\right|=\frac{n}{n+1}$

and

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

Inconclusive. You must recognize it as the harmonic series (p-series with $p=1$), which diverges.

Exam Focus

• Typical question patterns:
  • “Determine whether the series converges” where multiple tests could work but one is simplest.
  • “Justify your answer” requiring a named test and correct conditions.
• Common mistakes:
  • Forcing an inappropriate test (for example integral test on a messy expression when limit comparison is simpler).
  • Treating “inconclusive” as “diverges” or “converges.”