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AP Calculus BC Practice: Sequences & Series (11 Step-by-Step Explanations)
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AP Calculus BC · Sequences & Series

Sequences & Series: AP Calculus BC Practice Problems

Sequences and series make up one of the largest and most distinctive units on the AP Calculus BC exam. Beyond building Taylor and Maclaurin polynomials, the exam consistently tests how well you understand error — how far a truncated series or polynomial approximation can be from the true value, and how to guarantee that error is small enough. It also tests careful, precise reasoning about which general statements about series are actually always true, since a surprising number of "obviously true" claims about convergence turn out to have counterexamples.

Two error-bounding tools come up constantly. For an alternating series with terms that decrease in size toward zero, the error after truncating is bounded by the size of the very next term left out — this makes alternating series unusually easy to bound. For a general Taylor polynomial, the Lagrange error bound |Rn(x)| \le M|x−a|n+1/(n+1)! (where M bounds the (n+1)th derivative on the relevant interval) works even when the series isn't alternating, at the cost of usually being a looser bound. Both tools show up in "how many terms do I need" and "what's the smallest degree polynomial that guarantees this much accuracy" problems, which require checking successive terms or degrees until the bound finally drops below the target error.

Below are 11 AP Calculus BC-style practice problems covering Taylor polynomial construction, alternating series and Lagrange error bounds, geometric series convergence, choosing a valid series expansion for evaluating a logarithm, and identifying true and false statements about general series properties. Work through each problem, select your answer, and check whether you got it right — every problem includes a complete, step-by-step explanation. Once you've worked through these, keep building your skills with thousands more official-style questions in our free AP Calculus BC QBank.

1
Bounding the error of a truncated Maclaurin series

If the approximate formula cos x = 1−x2/2! is used and |x| < 0.5, then the error is numerically less than

Full Explanation

The full Maclaurin series for cos x is 1−x2/2!+x4/4!−\ldots, an alternating series. Truncating after the x2 term, the error is bounded by the magnitude of the very next term, x4/4!.

Since |x|<0.5, the largest this bound can be is at |x|=0.5: (0.5)4/24 = 0.0625/24 ≈ 0.0026.

The smallest listed value that's still guaranteed to exceed this bound for every valid x is 0.003 (0.001 would fail, since the true error can get as large as about 0.0026).

The correct answer is B.

2
Determining how many series terms guarantee accuracy

The function g(x) = ∑n=0∞ bnxn and g′(x) = −g(x) for all x. If g(0)=1, then how many terms of the series are needed to find g(0.5) correct to three decimal places?

Full Explanation

The condition g′(x)=−g(x) with g(0)=1 identifies g(x) as e−x = \sum(−1)nxn/n!, an alternating series (for x>0).

List the magnitudes of the terms at x=0.5: n=0: 1. n=1: 0.5. n=2: 0.125. n=3: 0.02083. n=4: 0.002604. n=5: 0.0002604.

"Correct to three decimal places" requires error < 0.0005. Using 4 terms (n=0 to 3), the next term (n=4, value 0.002604) still exceeds 0.0005 — not enough. Using 5 terms (n=0 to 4), the next term (n=5, value 0.0002604) is finally under 0.0005.

So 5 terms are needed. The correct answer is C.

3
Sum of a geometric series

The sum of the series ∑n=1∞(2π/π2)n equals

Full Explanation

This is a geometric series with common ratio r = 2\pi/\pi2 = 2/\pi. Since 2/\pi \approx 0.637 < 1, this series converges — ruling out choice D right away.

For a geometric series starting at n=1, the sum is r/(1−r): (2/\pi)/(1−2/\pi).

Multiply numerator and denominator by π to clear the fractions: 2/(\pi−2).

The correct answer is B.

Keep the momentum going.

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4
Error from truncating an alternating series after many terms

When ∑n=1∞ (−1)n−1/(4n−1) is approximated by the sum of its first 200 terms, the error is closest to

Full Explanation

This is an alternating series with terms decreasing to 0, so the error after truncating at 200 terms is bounded by (and closely approximated by) the magnitude of the 201st term.

The 201st term has magnitude 1/(4(201)−1) = 1/803.

1/803 \approx 0.001245.

The closest listed value is 0.0012. The correct answer is B.

5
Minimum Taylor polynomial degree via the Lagrange error bound

You wish to estimate ex, over the interval |x| ≤ 3, with an error less than 0.001. The Lagrange error term suggests that you use a Taylor polynomial at 0 with degree at least

Full Explanation

For f(x)=ex, every derivative is ex itself, and on |x|≤3 the maximum value of any derivative is e3 (at x=3). The Lagrange error bound for a degree-n Taylor polynomial at 0 is |Rn(x)| \le e3\cdot3n+1/(n+1)!.

Check successive degrees: at n=12, the bound is about 0.00514 (too large). At n=13, about 0.00110 (still just over 0.001). At n=14, about 0.00022 (finally under 0.001).

The smallest degree that guarantees error under 0.001 is 14.

The correct answer is D.

6
Third-degree Taylor polynomial for a trig function

The third-degree Taylor polynomial P3(x) for cos x about π/4 is

Full Explanation

Evaluate cos x and its first three derivatives at π/4: cos(\pi/4)=\sqrt2/2, −sin(\pi/4)=−\sqrt2/2, −cos(\pi/4)=−\sqrt2/2, sin(\pi/4)=\sqrt2/2.

Build the polynomial: P3(x) = \sqrt2/2 + (−\sqrt2/2)(x−\pi/4) + (−\sqrt2/2)/2!\,(x−\pi/4)2 + (\sqrt2/2)/3!\,(x−\pi/4)3.

Factor out \sqrt2/2 = 1/\sqrt2 from every term: (1/\sqrt2)[1−(x−\pi/4)−(1/2)(x−\pi/4)2+(1/6)(x−\pi/4)3].

The correct answer is A.

6 down, 5 to go — over halfway there.

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7
Building a Taylor polynomial from given derivative values

Let k be a function for which all derivatives exist at x=2. If k(2)=k′(2)=k″(2)=k′′′(2)=8, which third-degree polynomial best approximates k there?

Full Explanation

The third-degree Taylor polynomial centered at x=2 is P3(x) = k(2)+k′(2)(x−2)+[k″(2)/2!](x−2)2+[k′′′(2)/3!](x−2)3.

Substitute all four given values (each equal to 8): 8+8(x−2)+(8/2)(x−2)2+(8/6)(x−2)3.

Simplify the coefficients: 8/2=4 and 8/6=4/3.

P3(x) = 8+8(x−2)+4(x−2)2+(4/3)(x−2)3. The correct answer is B.

8
Identifying the false statement about series

Which of the following statements about series is false?

Full Explanation

Choice A claims the sum starting from k=1 always equals the sum starting from any later index m. This is false in general — shifting the starting index drops the terms u1 through um−1, which changes the sum unless those dropped terms happen to add to 0.

Choice B is true: multiplying every term of a divergent series by a nonzero constant can't make it converge. Choice C is true: a convergent series plus a divergent series is always divergent (if the sum converged, subtracting the convergent piece would force the divergent piece to converge too, a contradiction).

Choice D is true — it's the standard divergence test (necessary, though not sufficient, condition for convergence).

The correct answer is A.

9
Which statement about series is always true

Which of the following statements is always true?

Full Explanation

Choice A is false in general: \sum(−1)n/\sqrt{n} converges (by the alternating series test), but squaring the terms gives \sum1/n, the divergent harmonic series.

Choice B is only guaranteed for alternating series with decreasing terms — it's not a universal property of every series. Choice C is false: the harmonic series \sum1/n has terms approaching 0 but still diverges.

Since every one of (A), (B), and (C) fails as a blanket statement, none of them are always true.

The correct answer is D.

9 down, 2 to go — almost there.

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10
Choosing a valid series expansion to compute a logarithm

Which of the following series can be used to compute ln(1.3)?

Full Explanation

To compute ln(1.3) with a series, the series needs to actually converge to ln(x) at x=1.3.

Choice B fails immediately: ln x isn't even defined (nor differentiable) at x=0, so no Taylor series can be built there. Choice A gives a series for ln(x+1), which at x=1.3 would compute ln(2.3), not ln(1.3). Choice D similarly targets ln(x+1) evaluated near x=2.3, not what's needed.

The Taylor series for ln x centered at x=1 has radius of convergence 1, meaning it converges for x in (0,2) — and 1.3 falls right inside that interval. Evaluating this series at x=1.3 gives exactly ln(1.3).

The correct answer is C.

11
Identifying a false claim about power series arithmetic

Let f(x) = ∑n=0∞anxn and g(x) = ∑n=0∞bnxn. Suppose both series converge for |x| < R. Let x0 be a number such that |x0| < R. Which statement is false?

Full Explanation

Choices A, B, and C describe standard, well-established properties of power series within their radius of convergence: they can be added termwise, they're continuous, and they can be differentiated termwise.

Choice D proposes a shortcut for multiplying two power series: pairing up the nth coefficients directly (anbn) and using x02n. This is not how power series multiplication actually works.

The correct product f(x)g(x) is given by the Cauchy product, whose nth coefficient is a full convolution sum \sum_{k=0}^{n}a_kb_{n−k} — not simply a_nb_n. Choice D's formula is a fabricated shortcut, not a real theorem.

The correct answer is D.

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