Source: AP Calculus BC Mock 3 (College Board 2008 PT)
Tags: series, geometric series, Taylor series, Maclaurin series, power series, radius of convergence, interval of convergence, ratio test, Lagrange error bound, convergence tests, AP Calculus BC
Difficulty: Advanced Prerequisites: Solid grasp of derivatives (all orders), integration, and basic sequences. This is the most BC-specific topic on the exam.
Infinite series questions make up roughly 25 to 30% of the BC exam and are the main content that distinguishes BC from AB. You need to evaluate geometric series, recognise standard Maclaurin series (e^x, sin x, cos x), find the radius and interval of convergence for power series, determine convergence or divergence using a range of tests, and apply the Lagrange error bound. Series questions appear in both multiple-choice sections and typically in one full free-response question.
Memorize the five standard Maclaurin series and the geometric series formula. For convergence, the ratio test handles most power series. At the endpoints, check separately using a different test. The Lagrange error bound is the go-to tool for bounding approximation error.
Geometric series
A series of the form sum of a*r^n from n=0 to infinity. It converges when |r| < 1, and the sum is a/(1 - r). Think of it as the simplest infinite series: each term is a fixed multiple of the previous one.
Maclaurin series
A Taylor series centred at x = 0. The Maclaurin series of f is sum of f^(n)(0)/n! * x^n. In simple terms, it is the polynomial of infinite degree that matches the function and all its derivatives at x = 0.
Taylor series
The series sum of f^(n)(a)/n! * (x - a)^n, which represents f(x) near x = a. A Taylor series centred at a = 0 is called a Maclaurin series.
Power series
A series of the form sum of a_n * (x - c)^n. It converges within a radius R of its centre c and may or may not converge at the endpoints. Think of it as a polynomial that might go on forever.
Radius of convergence
The value R such that a power series converges for |x - c| < R and diverges for |x - c| > R. It can be 0, a positive number, or infinity.
Interval of convergence
The set of all x-values for which a power series converges. It is the open interval (c - R, c + R) plus whichever endpoints (if any) also give convergence when tested separately.
Ratio test
For a series of positive terms, compute L = lim(n->inf) |a_{n+1}/a_n|. If L < 1 the series converges absolutely; if L > 1 it diverges; if L = 1 the test is inconclusive.
Lagrange error bound
The error in using a degree-n Taylor polynomial is at most |f^(n+1)(z)| / (n+1)! * |x - a|^(n+1), where z is some value between x and a. In practice, you bound |f^(n+1)(z)| by its maximum on the interval.
p-series
A series of the form sum of 1/n^p. It converges when p > 1 and diverges when p <= 1. The harmonic series (p = 1) diverges; 1/n^2 converges.
Sum from n=0 to infinity of (-2/3)^n is geometric with a = 1 and r = -2/3. Since |r| = 2/3 < 1, it converges. Sum = 1/(1 - (-2/3)) = 1/(5/3) = 3/5.
Any series that can be written as a constant times r^n is geometric. If |r| >= 1, it diverges.
The series 1 + x^2/2! + x^4/4! + x^6/6! + ... + x^(2n)/(2n)! + ... uses only even powers with all positive signs. This is the Maclaurin series for cosh(x) = (e^x + e^(-x))/2.
Comparison to known series: e^x = 1 + x + x^2/2! + x^3/3! + ..., cos(x) = 1 - x^2/2! + x^4/4! - ..., sin(x) = x - x^3/3! + x^5/5! - .... The given series has all positive terms with even powers, matching cosh(x).
If f^(n)(0) = (-1)^n * (n+1)/[(n+2)*2^n], then the Maclaurin series is sum of [f^(n)(0)/n!] * x^n = sum of (-1)^n * (n+1)/[(n+2)*2^n * n!] * x^n.
Compute the first few terms: n=0 gives 1/(21) = 1/2; n=1 gives (-1)2/(321) = -1/3; n=2 gives (1)3/(44*2) = 3/32. Match these against the answer choices.
The Taylor polynomial P(x) = (x-3)^2 - (x-3)^4/2! + (x-3)^6/3! - ... + (-1)^(n+1) * (x-3)^(2n)/n! - ... + (x-3)^100/50!. This is centred at a = 3.
The coefficient of (x-3)^30 is (-1)^(15+1)/15! = 1/15!. Since the Taylor coefficient of (x-a)^k is f^(k)(a)/k!, we get f^(30)(3)/30! = 1/15!, so f^(30)(3) = 30!/15!.
For sum of (x-3)^(2n)/n, use the ratio test on the absolute values. The ratio simplifies to |x-3|^2 * n/(n+1), which tends to |x-3|^2. For convergence: |x-3|^2 < 1, so |x-3| < 1. The radius is 1.
Interval: 2 < x < 4. At x = 2: sum of (-1)^(2n)/n = sum of 1/n, which is the harmonic series (diverges). At x = 4: sum of 1/n (diverges). So the interval is 2 < x < 4, open at both ends.
I. Sum of 1/sqrt(n): this is a p-series with p = 1/2 < 1. It diverges.
II. Sum of 3^n/n!: apply the ratio test: 3^(n+1)/(n+1)! / [3^n/n!] = 3/(n+1) -> 0 < 1. It converges.
III. Sum of (e/pi)^n: this is geometric with r = e/pi, approximately 0.865 < 1. It converges.
So II and III converge.
For sum of a_n*(x-2)^n with a_0 = 5 and a_n = [(2n+1)/(3n-1)] * a_{n-1}, the ratio |a_n/a_{n-1}| -> 2/3 as n -> infinity. The radius of convergence is 1/(2/3) = 3/2.
If f^(4)(x) = e^(sin x) and we use a third-degree Taylor polynomial about x = 0 to approximate f on [0, 1], the error is at most max|f^(4)(z)| / 4! * |x|^4 for z in [0, 1].
Since e^(sin x) is maximised at x = pi/2, but on [0,1] the maximum of e^(sin z) is e^(sin 1) = e^0.8415 approximately 2.32. So error <= 2.32/24 * 1 approximately 0.097.
For g(x) = [cos(2x) - 1]/x^2 when x is not 0: start with cos(2x) = 1 - (2x)^2/2! + (2x)^4/4! - (2x)^6/6! + .... Then cos(2x) - 1 = -4x^2/2 + 16x^4/24 - 64x^6/720 + .... Dividing by x^2: g(x) = -2 + 2x^2/3 - 4x^4/45 + ....
The constant term is -2, so g(0) = -2 (confirmed by L'Hopital's rule as well).
\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}, \quad |r|<1e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}, \quad \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}, \quad \cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}\text{Ratio test: } L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|, \quad R = \frac{1}{L}|R_n(x)| \le \frac{\max|f^{(n+1)}(z)|}{(n+1)!}\,|x-a|^{n+1}Students often forget that the geometric series formula sum = a/(1-r) uses the FIRST TERM as "a," not the base of the exponent. If the series starts at n = 0, the first term is a*r^0 = a.
When checking endpoints for interval of convergence, students sometimes assume both endpoints behave the same way. Always test each endpoint separately: one may converge and the other diverge.
A common error with the ratio test is concluding something when L = 1. The ratio test is inconclusive at L = 1; you must use a different test.
Students mix up the Maclaurin series for cos(x) and cosh(x). Both use even powers, but cos(x) has alternating signs while cosh(x) = (e^x + e^(-x))/2 has all positive signs.
⚠️ Recognising a given series as a known function (e^x, sin, cos, cosh, 1/(1-x)) is tested every year. Memorize the five standard series.
⚠️ Finding the interval of convergence is a near-certain question. Use the ratio test for the interior, then test endpoints individually.
⚠️ The Lagrange error bound appears in the FRQ section. You need to bound the (n+1)th derivative, not compute it exactly.
⚠️ Reading a Taylor polynomial to extract f^(n)(a) uses the relationship: coefficient of (x-a)^n = f^(n)(a)/n!. Rearrange to find f^(n)(a).
⚠️ The p-series and geometric series tests are the quickest ways to settle convergence of simple series.
True or false: The sum of (1/2)^n from n = 0 to infinity equals 2. (True, a = 1, r = 1/2, sum = 2)
Fill in the blank: The Maclaurin series for e^x has a general term of ______. (x^n / n!)
True or false: Sum of 1/sqrt(n) converges. (False, it is a p-series with p = 1/2 < 1)
Fill in the blank: If the ratio test gives L for a power series, then R = ______. (1/L)
True or false: The interval of convergence always includes both endpoints. (False, you must check each endpoint separately)
Q: What is the value of the sum from n = 0 to infinity of (-2/3)^n?
A: 3/5. Geometric series with a = 1, r = -2/3. Sum = 1/(1+2/3) = 1/(5/3) = 3/5.
Q: The Maclaurin series 1 + x^2/2! + x^4/4! + ... + x^(2n)/(2n)! + ... represents which function?
A: (e^x + e^(-x))/2, also written cosh(x). It is the even-power, all-positive-sign version of the exponential series.
Q: For sum of (x-3)^(2n)/n with radius of convergence 1, what is the interval of convergence?
A: 2 < x < 4 (open at both ends). Both endpoints give the harmonic series, which diverges.
Q: Which of these converge: I. sum of 1/sqrt(n), II. sum of 3^n/n!, III. sum of (e/pi)^n?
A: II and III only. I is a divergent p-series; II converges by ratio test; III is a convergent geometric series.
Q: If the 4th derivative of f is e^(sin x), what is the Lagrange error bound for the 3rd-degree Taylor polynomial on [0, 1]?
A: Approximately 0.097. Max of e^(sin z) on [0,1] is about 2.32; bound = 2.32/24 approximately 0.097.
Series connect back to derivatives (Taylor coefficients come from derivatives at the centre), to integration (term-by-term integration of power series), and to differential equations (power series solutions). L'Hopital's rule can often confirm the value a series sums to, and the Fundamental Theorem underpins term-by-term integration of series.
geometric series, Maclaurin series, Taylor series, Taylor polynomial, power series, radius of convergence, interval of convergence, ratio test, root test, comparison test, p-series, harmonic series, Lagrange error bound, Taylor remainder, cosh, convergence tests, absolute convergence, conditional convergence, AP Calculus BC series