Unit 10: Infinite Sequences and Series

Sequences: Definitions, Notation, and Limits

A sequence is an infinite succession (list) of numbers arranged in a specific order, usually generated by a rule. In calculus, you typically write a sequence as $\{a_n\}$ (read “the sequence $a_n$”), where $a_n$ is the $n$th term and $n$ is always an integer (usually a positive integer).

Formally, a sequence is a function whose domain is the positive integers. The key idea is that sequences let you study what happens “in the long run” as $n$ becomes very large.

A common example is

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

Beginning with $n=1$, the terms are $0,\frac{1}{2},\frac{2}{3},\dots$

Why sequences matter

Sequences are the foundation for infinite series. A series is built by adding up terms of a sequence. Before you can decide whether an infinite sum makes sense, you need language for whether the terms approach a value, approach zero, oscillate, or blow up.

Common sequence behaviors

A convergent sequence is one where the limit exists and is a finite real number. If that limit exists and equals $L$, you write

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

A divergent sequence is one where the limit does not exist or is infinite (goes to $\infty$ or $-\infty$).

An oscillating sequence is a common kind of divergence where the terms bounce between values and never settle, such as

$a_n = (-1)^n$

The official (epsilon) definition of sequence limit

A sequence has a limit $L$ if for any $\varepsilon>0$ there is an associated positive integer $N$ such that

$|a_n-L|<\varepsilon$

for all

$n\ge N$

If so, the sequence converges to $L$.

How you actually evaluate limits of sequences

Many limits of sequences can be treated like limits of functions by thinking of $n$ as “very large.” In AP Calculus, you can safely use the same limit tools you learned for functions as long as you reason correctly.

Common strategies include using known limits (like $\frac{1}{n}\to 0$), dominant-term reasoning (especially for rational expressions in $n$), and the Squeeze Theorem (trapping a sequence between two others with the same limit).

Example 1 (dominant terms): Find

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

Divide numerator and denominator by $n^2$:

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

As $n\to\infty$, the fractions go to $0$, so

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

Example 2 (Squeeze idea): Consider

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

Because $\frac{1}{n}\to 0$ and $\lim_{x\to 0}\sin x = 0$, you get

$\lim_{n\to\infty} \sin\left(\frac{1}{n}\right) = 0$

A crucial necessary condition for series later

If you plan to add terms of a sequence as a series, a non-negotiable fact is:

If $\sum a_n$ converges, then

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

This does not mean that if $a_n\to 0$ then the series converges. That false reversal causes lots of wrong answers (for example, the harmonic series has terms going to zero but still diverges).

Exam Focus

• Typical question patterns:
  • Evaluate $\lim_{n\to\infty} a_n$ for an explicit formula $a_n$.
  • Decide whether a sequence converges or diverges, sometimes with oscillation.
  • Use the fact “series convergent implies terms go to zero” as a quick divergence check.
• Common mistakes:
  • Assuming $a_n\to 0$ guarantees $\sum a_n$ converges.
  • Thinking you “can’t” take limits because $n$ is discrete (you can).
  • Treating an oscillating sequence as “no limit” without checking if it actually approaches a single value.
  • Losing track of dominant powers when simplifying rational expressions in $n$.

Infinite Series: Partial Sums, Convergence, and First Examples

An infinite series is an expression of the form

$a_1+a_2+a_3+\cdots$

In sigma notation, the associated series of a sequence $\{a_n\}$ is

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

An “infinite sum” only makes sense if the running totals settle toward a finite number.

Partial sums are the definition

The rigorous meaning of

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

comes from partial sums:

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

If the sequence of partial sums $\{S_N\}$ converges to a limit $S$, meaning

$\lim_{N\to\infty} S_N = S$

then the series converges and $S$ is called the **sum** of the series. If $\{S_N\}$ diverges, then the series diverges and has no sum.

(Heuristically, as $N$ increases you include more and more terms, so when the partial sums approach a limit, that limit is the value the infinite sum is defined to have.)

The nth-term test for divergence

The simplest test is the nth-term test:

If

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

or the limit does not exist, then

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

diverges.

This is only a one-way test: if $a_n\to 0$, you still need more information.

Geometric series (finite and infinite)

A geometric series has terms with a constant ratio. Two common index forms are:

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

or

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

These show up often on the AP exam: you’re often asked whether the series converges or diverges, and if it converges, to what value.

A geometric series converges if and only if

$|r|<1$

It diverges if

$|r|>1$

and when

$r=1$

the partial sums grow linearly.

For instance,

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

converges, while

$2+2^2+2^3+\cdots$

diverges.

Finite geometric sum formula (first $n$ terms): If the first term is $a$ and the common ratio is $r$, then the sum of the first $n$ terms is

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

Why this works (algebra trick): Write

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

Multiply by $r$ and subtract:

$rS_n=ar+ar^2+\cdots+ar^n$

Subtracting gives

$S_n-rS_n=a-ar^n$

so

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

and dividing by $1-r$ produces the formula above.

Infinite geometric sum: If $|r|<1$ then

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

and the infinite sum is

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

Example 1 (geometric): Find

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

Here $a=5$ and $r=\frac{2}{3}$, and since $\left|\frac{2}{3}\right|<1$ it converges:

$\sum_{n=0}^{\infty} 5\left(\frac{2}{3}\right)^n=\frac{5}{1-\frac{2}{3}}=15$

Common misconception: Students sometimes apply the geometric sum formula even when $|r|\ge 1$.

Telescoping series

A telescoping series is one where algebraic rewriting creates massive cancellation in partial sums.

Example 2 (telescoping): Evaluate

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

Partial sums:

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

Writing out terms shows everything cancels except the first and last negative term, giving

$S_N=1-\frac{1}{N+1}$

Then

$\lim_{N\to\infty} S_N=1$

So the series converges to $1$.

Exam Focus

• Typical question patterns:
  • Use partial sums to justify a telescoping sum.
  • Identify a geometric series embedded in an expression and compute its sum.
  • Apply the nth-term test quickly to prove divergence.
• Common mistakes:
  • Forgetting that the nth-term test only proves divergence, not convergence.
  • Using $\frac{a}{1-r}$ when the series starts at $n=1$ without adjusting the first term.
  • Missing cancellation in telescoping series because you do not write out several terms.
  • Applying geometric convergence rules without checking $|r|<1$.

Benchmark Series: Harmonic and p-Series (and Why They’re So Useful)

Some series appear so often that they become “benchmarks” you compare other series to.

The harmonic series

The harmonic series is

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

It diverges. This is surprising at first because

$\frac{1}{n}\to 0$

but the partial sums still grow without bound (very slowly, but unbounded).

p-series

A p-series is

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

It has a complete convergence rule:

• Converges if $p>1$
• Diverges if $p\le 1$

In particular, it diverges for the range

$0<p<1$

and the harmonic series is the special case $p=1$.

Why p-series matter

Many complicated-looking terms behave “like” $\frac{1}{n^p}$ for large $n$. If you can compare your series to a p-series, you can decide convergence without needing an exact sum.

Example 1: Determine whether

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

converges.

This is a p-series with

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

so it converges.

Example 2: Determine whether

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

converges.

This is a classic borderline type. A standard approach uses the integral test (see later). The key warning is that “a little smaller than $\frac{1}{n}$” is not automatically enough to guarantee convergence.

Exam Focus

• Typical question patterns:
  • Recognize a p-series and decide convergence from $p$.
  • Compare a new series to $\sum \frac{1}{n}$ or $\sum \frac{1}{n^p}$.
  • Use “harmonic diverges” to justify divergence of a related series.
• Common mistakes:
  • Thinking any $\sum \frac{1}{n^p}$ converges because terms go to zero.
  • Confusing the condition $p>1$ with $p\ge 1$.
  • Treating logarithms as powers (they grow much more slowly than any positive power).

Convergence Tests for Series with Positive Terms

Many important tests assume $a_n>0$ (at least eventually). When all terms are positive, partial sums increase, so the only question is whether they level off (converge) or grow without bound (diverge).

Direct comparison test

Suppose

$0\le a_n\le b_n$

for all sufficiently large $n$.

• If $\sum b_n$ converges, then $\sum a_n$ converges.
• If $\sum a_n$ diverges, then $\sum b_n$ diverges.

Example 1 (comparison to p-series): Test

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

Since

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

we have

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

and $\sum \frac{1}{n^2}$ converges, so the given series converges.

A common mistake is to flip the inequality the wrong way. With denominators, a bigger denominator means a smaller fraction.

Limit comparison test

The direct comparison test can be hard because you must find a clean inequality. The limit comparison test avoids that by comparing ratios.

Let $a_n>0$ and $b_n>0$ and compute

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

If

$0<L<\infty$

then $\sum a_n$ and $\sum b_n$ either both converge or both diverge.

Example 2: Test

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

For large $n$ this behaves like

$\frac{3n}{n^2}=\frac{3}{n}$

Choose

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

Then

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

Since $0<L<\infty$ and $\sum \frac{1}{n}$ diverges, the given series diverges.

Integral test

If $f$ is positive, continuous, and decreasing for

$x\ge 1$

and

$a_n=f(n)$

then

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

and

$\int_1^{\infty} f(x)\,dx$

either both converge or both diverge.

Example 3: Test

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

Let

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

Then

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

Substitute

$u=\ln x$

so

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

to get

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

This converges, so the series converges.

Ratio test

The ratio test is powerful when factorials, exponentials, or repeated products appear.

For $a_n>0$, compute

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

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

Example 4: Test

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

Let

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

Then

$\frac{a_{n+1}}{a_n}=\frac{n+1}{2n}$

So

$L=\lim_{n\to\infty}\frac{n+1}{2n}=\frac{1}{2}$

Therefore the series converges.

Root test

The root test is often useful when terms look like something to the $n$th power.

Compute

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

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

Example 5: Test

$\sum_{n=1}^{\infty} \left(\frac{3n}{5n+1}\right)^n$

Let

$a_n=\left(\frac{3n}{5n+1}\right)^n$

Then

$\sqrt[n]{a_n}=\frac{3n}{5n+1}$

So

$L=\lim_{n\to\infty}\frac{3n}{5n+1}=\frac{3}{5}$

Since $\frac{3}{5}<1$, the series converges.

Exam Focus

• Typical question patterns:
  • Decide which test to use based on the form (powers, factorials, rational functions, logs).
  • Use comparison/limit comparison to a p-series.
  • Use ratio test with factorials or exponentials.
  • Use the integral test on classic forms like $\frac{1}{n(\ln n)^p}$.
• Common mistakes:
  • Using the integral test when $f(x)$ is not decreasing or not positive.
  • Forgetting that ratio/root tests can be inconclusive at $L=1$.
  • Setting up limit comparison with a poor choice of $b_n$ (choose the dominant behavior).
  • Thinking “smaller than a divergent series” is enough to prove convergence (it is not).

Alternating Series, Absolute vs Conditional Convergence

Not all series have positive terms. Many important series alternate signs, such as

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

where $b_n>0$.

Alternating series test (Leibniz test)

An alternating series

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

converges if all of the following are true (at least eventually):

1. The terms alternate in sign (the factor $(-1)^{n+1}$ enforces this).
2. $b_n$ is decreasing.
3. $\lim_{n\to\infty} b_n = 0$.

Example 1: Test

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

Here $b_n=\frac{1}{n}$ decreases and goes to $0$, so the series converges.

Absolute convergence vs conditional convergence

Given a series $\sum a_n$, consider the series of absolute values $\sum |a_n|$.

• Absolutely convergent means $\sum |a_n|$ converges. Then $\sum a_n$ also converges.
• Conditionally convergent means $\sum a_n$ converges, but $\sum |a_n|$ diverges.

This is sometimes summarized as the Absolute Convergence Theorem:

If

$\sum |a_n|$

converges, then

$\sum a_n$

converges.

Example 2 (conditional): The series

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

converges by the alternating series test, but

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

diverges, so it is conditionally convergent.

Example 3 (absolute): Test

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

Check absolute convergence:

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

Use the ratio test with $a_n=\frac{n}{3^n}$:

$\frac{a_{n+1}}{a_n}=\frac{n+1}{3n}$

So

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

Since $\frac{1}{3}<1$, the absolute value series converges, so the original series converges absolutely.

Alternating series error estimate (remainder bound)

If an alternating series converges by the alternating series test, then the error when approximating the infinite sum by the $N$th partial sum is bounded by the next term.

Let

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

and let $S_N$ be the partial sum through $n=N$. Then the remainder

$R_N=S-S_N$

satisfies

$|R_N|\le b_{N+1}$

Equivalently, if you sum the first 10 terms, the error is less than the absolute value of the 11th term.

Example 4 (error control): How many terms of

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

are needed to approximate the sum within $0.001$?

We need

$b_{N+1}=\frac{1}{N+1}\le 0.001$

So

$N+1\ge 1000$

and

$N\ge 999$

Using the first 999 terms guarantees error at most $0.001$.

Exam Focus

• Typical question patterns:
  • Determine whether an alternating series converges, and whether it is absolute or conditional.
  • Use ratio/root to check absolute convergence first.
  • Use the alternating series remainder bound to find the number of terms needed for a desired accuracy.
• Common mistakes:
  • Forgetting to verify $b_n$ is decreasing for the alternating series test.
  • Assuming “alternating” automatically means convergent.
  • Confusing conditional vs absolute convergence (always check $\sum |a_n|$).
  • Using $b_N$ instead of the theorem’s “next term” $b_{N+1}$.

Power Series: Building Functions from Infinite Polynomials

A power series centered at $x=a$ has the form

$\sum_{n=0}^{\infty} c_n(x-a)^n$

where $c_n$ are constants. It is like an infinite-degree polynomial.

Why power series matter

Power series turn difficult function questions into algebra. If a function can be represented as a power series, you can differentiate and integrate term-by-term (within its interval of convergence), approximate values, and estimate errors.

Convergence depends on $x$

Unlike a number series, a power series may converge for some $x$ values and diverge for others. There is always an interval of convergence, typically

$|x-a|<R$

where $R$ is the radius of convergence.

• If $|x-a|<R$, the series converges (often absolutely).
• If $|x-a|>R$, the series diverges.
• If $|x-a|=R$, you must test endpoints separately.

Special cases worth knowing:

• If the series converges only when $x=a$, then $R=0$.
• If the series converges absolutely for all real $x$, then $R=\infty$.

Also, pay attention to where the index starts. Starting at $n=0$ versus $n=1$ changes the first term but does not change convergence behavior; you may need to rewrite a series to match a known form.

Finding the radius and interval of convergence (ratio test)

Let

$a_n(x)=c_n(x-a)^n$

Compute

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

and solve

$L<1$

for $x$.

A common “shortcut” form (when the needed limit exists) is that the radius can be found by solving

$\lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right||x-a|<1$

Example 1 (interval of convergence): Find the interval of convergence of

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

Let

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

Then

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

So

$L=|x-2|$

Ratio test gives convergence for

$|x-2|<1$

which is $1<x<3$. Now test endpoints:

• At $x=1$ the series becomes $\sum \frac{(-1)^n}{n}$, which converges by the alternating series test.
• At $x=3$ the series becomes $\sum \frac{1}{n}$, which diverges (harmonic).

So the interval of convergence is

$[1,3)$

Term-by-term differentiation and integration

If a power series converges on $|x-a|<R$, then inside that interval you may differentiate and integrate term-by-term, producing new power series with the same radius of convergence.

If

$f(x)=\sum_{n=0}^{\infty} c_n(x-a)^n$

then for $|x-a|<R$,

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

and

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

Example 2 (differentiate a power series): Suppose

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

This is geometric with

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

for

$|x|<1$

Differentiate term-by-term:

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

Differentiate the closed form:

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

Therefore, for $|x|<1$,

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

Exam Focus

• Typical question patterns:
  • Use the ratio test to find radius and interval of convergence.
  • Test endpoint convergence using alternating, p-series, comparison, etc.
  • Differentiate/integrate a power series and state the new interval of convergence.
• Common mistakes:
  • Forgetting endpoint testing.
  • Misapplying the ratio test algebra when powers involve $x-a$.
  • Assuming the interval of convergence changes after differentiation or integration (radius stays the same).
  • Losing track of whether the series starts at $n=0$ or $n=1$.

Taylor and Maclaurin Series: Turning Derivatives into Polynomials (and Beyond)

A Taylor series represents a function as an infinite power series centered at $x=a$, built from the function’s derivatives at $a$.

The Taylor series formula

If $f$ is infinitely differentiable near $x=a$, its Taylor series is

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

The special case $a=0$ is called a Maclaurin series:

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

Taylor polynomials

The Taylor polynomial of degree $n$ (also written as $P_n(x)$) about $x=a$ is the finite truncation:

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

Equivalently, written out in expanded form,

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

Standard Maclaurin series you should know

These are commonly used (often provided):

Geometric (for $|x|<1$):

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

Exponential (all real $x$):

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

Sine (all real $x$):

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

Cosine (all real $x$):

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

Natural log (for $-1<x\le 1$):

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

Arctangent (for $|x|\le 1$):

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

Building new Taylor series from known ones

On the AP exam you often start from a known series and apply substitution, multiplication/division by simple functions, and term-by-term differentiation or integration.

Example 1 (substitution): Find a power series for

$\frac{1}{1+x^2}$

Start with

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

Replace $x$ by $-x^2$:

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

The condition $|x|<1$ still results from $|-x^2|<1$.

Example 2 (integration to get arctan): Integrate term-by-term from $0$ to $x$:

$\int_0^x \frac{1}{1+t^2}\,dt=\int_0^x\sum_{n=0}^{\infty}(-1)^n t^{2n}\,dt$

This produces

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

(valid at least for $|x|<1$, and in this case it also converges at the endpoints).

Exam Focus

• Typical question patterns:
  • Write the first few nonzero terms of a Taylor/Maclaurin series for a function.
  • Create a new series from a known one via substitution and term-by-term calculus.
  • Identify the interval of convergence after manipulating a series.
• Common mistakes:
  • Losing factorial structure or sign alternation in sine/cosine/exponential series.
  • Substituting into a known series but not updating the convergence condition.
  • Confusing a Taylor polynomial (finite) with the Taylor series (infinite).
  • Making index-shift errors after differentiating/integrating.

Taylor Polynomial Approximation and Error (Remainder) Estimates

A Taylor polynomial approximates a function near the center $a$. On exams and in applications, you must control how good the approximation is.

The remainder

If

$f(x)=P_N(x)+R_N(x)$

then $R_N(x)$ is the **remainder** (error) after degree $N$.

Two main BC error tools are:

1. Alternating-series error bound (when the Taylor series is alternating with decreasing term magnitudes)
2. Lagrange error bound (general bound using the next derivative)

Alternating series remainder for a Taylor series

If a Taylor/Maclaurin series alternates and its term magnitudes decrease to $0$, then truncating gives an error bounded by the first omitted term:

$|R_N(x)|\le b_{N+1}(x)$

This matches the broader alternating-series error bound: the error is less than the absolute value of the next term.

Example 1 (cosine approximation): Approximate $\cos(0.2)$ using the Maclaurin polynomial through the $x^4$ term, and bound the error.

Maclaurin series:

$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$

Use

$P_4(x)=1-\frac{x^2}{2}+\frac{x^4}{24}$

Evaluate:

$P_4(0.2)=1-\frac{(0.2)^2}{2}+\frac{(0.2)^4}{24}$

Compute:

$(0.2)^2=0.04$

$(0.2)^4=0.0016$

So

$P_4(0.2)=1-0.02+0.000066666\ldots$

Thus

$P_4(0.2)\approx 0.980066666\ldots$

Error bound is the next term magnitude:

$\left|\frac{x^6}{6!}\right|$

At $x=0.2$:

$\left|\frac{(0.2)^6}{720}\right|$

Since

$(0.2)^6=0.000064$

the error is at most

$\frac{0.000064}{720}\approx 8.888\times 10^{-8}$

Lagrange error bound

The Lagrange form of the remainder gives a general bound:

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

where $M$ is a number such that

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

for all $t$ between $a$ and $x$.

A practical note: if you are finding an $n$th-degree Taylor polynomial and the series terms are decreasing in magnitude, a good approximation to the error is often the next nonzero term, but the Lagrange bound is the general guarantee.

Example 2 (Lagrange bound with $e^x$): Approximate $e^{0.1}$ by the Maclaurin polynomial of degree 2, and bound the error.

For $f(x)=e^x$,

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

So

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

For the remainder, $N=2$ so use the third derivative. All derivatives are $e^x$, so

$f^{(3)}(t)=e^t$

On $[0,0.1]$ the maximum is at $t=0.1$, so

$M=e^{0.1}$

Then

$|R_2(0.1)|\le\frac{e^{0.1}}{3!}(0.1)^3$

So

$|R_2(0.1)|\le\frac{e^{0.1}}{6000}$

A common mistake is to set $M=f^{(N+1)}(a)$ automatically. $M$ must bound the derivative on the whole interval between $a$ and $x$.

Exam Focus

• Typical question patterns:
  • Use a Taylor polynomial to approximate a function value and give an error bound.
  • Decide how many terms are needed to guarantee a desired accuracy.
  • Choose alternating-series error vs Lagrange error depending on the situation.
• Common mistakes:
  • Using the alternating error bound when the terms are not decreasing in magnitude.
  • Forgetting factorials in the next omitted term.
  • Choosing $M$ incorrectly (must be a maximum bound on the interval).

Working with Power Series as Models: Center Shifts, Multiplication, and Differential Equations

Power series are a flexible “language” for representing and manipulating functions.

Shifting the center and rewriting to match known series

A series centered at $a$ uses powers of $(x-a)$. Often you rewrite a function to match a known geometric or known Maclaurin series.

Example 1 (shift via substitution): Starting with

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

find a series for

$\frac{1}{2-x}$

Rewrite:

$\frac{1}{2-x}=\frac{1}{2}\cdot\frac{1}{1-\frac{x}{2}}$

Now apply geometric series:

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

Convergence requires

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

so

$|x|<2$

Multiplying a power series by a polynomial

If

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

then

$x^k\sum_{n=0}^{\infty} c_n x^n=\sum_{n=0}^{\infty} c_n x^{n+k}$

Example 2: Find a power series for

$\frac{x}{1-x}$

Using

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

multiply by $x$:

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

(valid for $|x|<1$).

Power series and differential equations (concept level)

In BC, you may see that a function defined by a power series can be differentiated and substituted into relationships. While full power-series solutions of differential equations are beyond AP, the key takeaways are:

• Power series let you build functions that satisfy certain derivative patterns.
• Term-by-term differentiation makes it possible to match coefficients and identify a function.

This shows up in problems where you are given a series for $f(x)$ and asked for series for $f'(x)$, $\int f(x)dx$, or a related expression. You can also approximate an integral by replacing the integrand with its Maclaurin series and integrating term-by-term (within the interval of convergence).

Exam Focus

• Typical question patterns:
  • Rewrite an expression to fit a known geometric or known Maclaurin series.
  • Shift/scale a series by substituting $bx$ or $(x-a)$.
  • Multiply/divide by simple terms and rewrite with clean sigma notation.
  • Differentiate/integrate a given power series and use it to approximate values.
• Common mistakes:
  • Forgetting to transform the convergence interval after substituting $bx$.
  • Incorrect index shifting (starting at $n=0$ vs $n=1$).
  • Treating the new interval of convergence as automatic rather than derived from the original condition.

Strategies for Choosing a Convergence Test (How to Think, Not Just What to Memorize)

Convergence tests can feel like a menu where you guess randomly. A better approach is to read the structure of $a_n$ and let that structure suggest a tool.

Recognize “signature” term patterns

• Looks like $\frac{1}{n^p}$: use p-series, comparison, or limit comparison.
• Has factorials $n!$: ratio test is usually best.
• Has something to the $n$th power: root test or ratio test.
• Has logs like $\ln n$: comparison or integral test often works.
• Alternating signs: check absolute convergence first; if not, alternating series test.
• Difference of fractions: try algebra for telescoping.

Why “inconclusive” happens

Some tests are designed for specific growth rates. Ratio/root tests are tuned to exponential (geometric-like) decay and often give

$L=1$

for polynomial-type series (like p-series). That is why you also need comparison and integral tests.

Worked “test selection” examples

Example 1: Decide convergence of

$\sum_{n=1}^{\infty} \frac{n^3}{5^n}$

The exponential $5^n$ dominates polynomial $n^3$, so use the ratio test. Let

$a_n=\frac{n^3}{5^n}$

Then

$\frac{a_{n+1}}{a_n}=\frac{(n+1)^3}{5n^3}$

So

$L=\lim_{n\to\infty}\frac{(n+1)^3}{5n^3}=\frac{1}{5}$

It converges.

Example 2: Decide convergence of

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

This is a classic integral test case. Let

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

Then

$\int_2^{\infty}\frac{1}{x\ln x}\,dx$

Substitute

$u=\ln x$

so

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

to get

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

This diverges, so the series diverges.

A common mistake is to compare $\frac{1}{n\ln n}$ directly to $\frac{1}{n}$ and conclude “smaller so convergent.” Smaller than a divergent series tells you nothing; you need to be smaller than a convergent benchmark.

Exam Focus

• Typical question patterns:
  • “Does it converge? Which test?” with justification.
  • Mixed-feature series (factorials and powers, logs and powers, alternating plus rational terms).
  • Endpoint convergence of power series requiring multiple different tests.
• Common mistakes:
  • Picking a test that cannot apply (comparison needs nonnegative terms; alternating test needs alternating structure and decreasing terms).
  • Treating “smaller than harmonic” as automatically convergent.
  • Not stating reasoning clearly (free response often expects the test name and its conditions).
  • Forgetting that the nth-term test is only for divergence, not convergence.