Difficulty: Intermediate to Advanced | Prerequisites: Sequences, basic integration, comparison arguments
Series questions on this exam test whether you can determine convergence or divergence (using the divergence test, geometric series test, p-test, telescoping, integral test, and comparison test), and whether you can find exact sums for geometric and telescoping series. Improper integrals extend the same limit-based reasoning to integrals with infinite bounds or discontinuities. Numerical integration (midpoint, trapezoid, Simpson's) asks you to approximate definite integrals and compute the associated error.
Geometric series
A series of the form sum of a * r^n. Converges if and only if |r| < 1, in which case its sum is a / (1 - r).
In simple terms, each term is a fixed multiple of the previous one.
p-series
The series sum of 1/n^p. Converges if p > 1, diverges if p <= 1.
The harmonic series (p = 1) is the classic borderline case that diverges.
Telescoping series
A series whose partial sums collapse because most terms cancel. Typically arises after partial fraction decomposition.
Think of it as a collapsing telescope: write out the first several partial sums and see what cancels.
Divergence test (nth-term test)
If lim a_n is not 0, the series diverges. If lim a_n = 0, the test is inconclusive.
Integral test
If f is positive, continuous, and decreasing on [1, infinity) and f(n) = a_n, then the series and the improper integral of f either both converge or both diverge. The sum of the series does not equal the integral.
Comparison test
If 0 <= a_n <= b_n and sum b_n converges, then sum a_n converges. If a_n >= b_n >= 0 and sum b_n diverges, then sum a_n diverges.
Improper integral
An integral with an infinite limit of integration or an integrand with a discontinuity in the interval. Evaluated as a limit of proper integrals.
Midpoint rule
Approximate the integral using rectangles whose height is the function value at the midpoint of each subinterval: M_n = delta_x * sum of f(x_i*), where x_i* is the midpoint.
Trapezoid rule
Approximate the integral using trapezoids: T_n = (delta_x / 2) * [f(x_0) + 2f(x_1) + ... + 2f(x_{n-1}) + f(x_n)].
Simpson's rule
A weighted average of midpoint and trapezoid approximations: S_n = (delta_x / 3) * [f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + ... + f(x_n)], where n is even.
If lim a_n is not 0, the series diverges.
Problem 9a: sum of (1 + 2k)/(3k + 5). lim = 2/3, which is not 0. Diverges.
Identify the first term a and common ratio r.
Problem 9b: sum from n=0 of 4^(n+1)/5^n = 4 * sum of (4/5)^n = 4 * 1/(1 - 4/5) = 4 * 5 = 20.
Problem 9d: sum of 3e^(-2n) = 3 * sum of (e^(-2))^n = 3 * e^(-2) / (1 - e^(-2)) = 3 / (e^2 - 1).
Problem 9f: sum of 5 * 3^(2n) / 2^n = 5 * sum of (9/2)^n. Since |9/2| > 1, diverges.
Problem 9c: sum of 1/(n(n+2)). Partial fractions: 1/(n(n+2)) = (1/2)[1/n - 1/(n+2)].
Write out partial sums: (1/2)[1 - 1/3 + 1/2 - 1/4 + 1/3 - 1/5 + ...]. Most terms cancel.
Sum = (1/2)(1 + 1/2) = 3/4.
Problem 9e: sum of 5 / sqrt(4n^5) = (5/2) * sum of 1/n^(5/2). Since p = 5/2 > 1, converges. (Exact sum not available in closed form.)
11a: sum from n=1 of 2^(n+1)/5^n = 2 * sum of (2/5)^n = 2 * (2/5)/(1 - 2/5) = 2 * (2/3) = 4/3.
11b: sum from n=0 of 3^(4n)/4^(3n) = sum of (81/64)^n. Since 81/64 > 1, diverges.
Problem 12 asks you to determine whether each series is geometric and, if so, find its sum.
12a: 2 + 3 + 6 + 10 + ... Ratios: 3/2, 6/3 = 2, 10/6 = 5/3. Not constant. Not geometric.
12b: 3 - 6 + 12 - 24 + ... First term a = 3, ratio r = -2. Since |r| = 2 > 1, the series diverges.
12c: 3 + 2 + 4/3 + 8/9 + ... First term a = 3, ratio r = 2/3. Sum = 3 / (1 - 2/3) = 9.
12d: 1/2 + 3/4 + 7/8 + 15/16 + ... Terms approach 1, so lim a_n = 1, which is not 0. Diverges. Not geometric (ratios are not constant).
12e: 0.2 + 0.02 + 0.002 + 0.0002 + ... First term a = 0.2, ratio r = 0.1. Sum = 0.2 / (1 - 0.1) = 2/9.
These are the kinds of conceptual traps that appear on exams. Know the reasoning, not just the answer.
13a: "If lim a_n = 0, then sum a_n converges." False. Counterexample: the harmonic series sum 1/n. lim 1/n = 0, but the series diverges.
13b: "If sum a_n converges, then lim a_n = 0." True. This is the contrapositive of the divergence test.
13c: "If sum a_n and sum b_n both converge, then sum (a_n +/- b_n) converge." True. Convergent series can be added or subtracted term by term.
13d: "sum b (constant) always diverges for any constant b." False if b = 0 (the series sums to 0). True for any b that is not 0 (partial sums grow without bound).
13e: "sum (-1)^(2n)/n is convergent by the alternating series test." False. (-1)^(2n) = 1 for all n, so this is just the harmonic series sum 1/n, which diverges. It is not alternating.
13f: "The sequence {-1} (the constant -1) converges to -1." True. Every term is -1, so the limit is -1.
13g: "If sum a_n satisfies the integral test and f(n) = a_n, then sum a_n = integral from 1 to infinity of f(x) dx." False. The integral test says they share the same convergence behaviour, but the sum and the integral generally have different values.
Replace infinity with a variable b, evaluate the proper integral, then take the limit as b -> infinity.
Problem 14: integral from 3 to infinity of (3 + cos(3x))/e^(3x) dx. Since 2/e^(3x) <= (3 + cos(3x))/e^(3x) <= 4/e^(3x) and integral of 4/e^(3x) converges, this integral converges by comparison.
Problem 17a: integral from 1 to infinity of 2/(3x+1)^3 dx. Substitute u = 3x+1: (2/3) * integral of u^(-3) du = (2/3)[-1/(2u^2)] = -1/(3u^2). Evaluate from 4 to infinity: 0 - (-1/48) = 1/48.
Problem 17c: integral from 0 to infinity of x^3 * e^(-2x^4) dx. Substitute u = x^4: (1/4) * integral of e^(-2u) du = (1/4)[-1/(2) * e^(-2u)] from 0 to infinity = 1/8.
Problem 17b: integral from 0 to 1 of (1+x)/sqrt(x) dx. The integrand has a discontinuity at x = 0. Take lim as a -> 0+ of the integral from a to 1. Integral = 2sqrt(x) + (2/3)x^(3/2) from a to 1 = (2 + 2/3) - (2sqrt(a) + ...) -> 8/3.
Problem 17d: integral from 1 to 2 of 1/(2-x) dx. Discontinuity at x = 2. Integral = -ln|2-x| from 1 to b as b -> 2-. This gives -ln(2-b) + ln(1), which -> infinity. Diverges.
18a: "If f is continuous on [a,b], then integral of f(x) dx is the area under f." False. Only if f >= 0. If f takes negative values, the integral gives signed area.
18b: "integral of 1/(x^3 + sin(x)) dx from 1 to infinity converges by comparison to integral of 1/x^3 dx." True. For large x, x^3 + sin(x) >= x^3 - 1 > (1/2)x^3, so the integrand is bounded above by 2/x^3.
18c: "If f(x) > 0 and integral f(x) converges, then integral 1/f(x) converges." False. Take f(x) = 1/x^2; its integral converges, but 1/f(x) = x^2 and the integral of x^2 diverges.
18g: "If integral f(x) converges, then integral a*f(x) converges for any a > 0." True. Scalar multiples preserve convergence.
18h: "If integral f(x) converges, then integral xf(x) converges." False. Take f(x) = 1/x^2; xf(x) = 1/x, whose integral diverges.
Problem 16 uses f(x) = cos(x^2) on [0, 1] with n = 4 subdivisions. delta_x = (1 - 0)/4 = 1/4.
Subinterval endpoints: x_0 = 0, x_1 = 1/4, x_2 = 1/2, x_3 = 3/4, x_4 = 1.
Midpoints: 1/8, 3/8, 5/8, 7/8.
M_4 = (1/4)[cos((1/8)^2) + cos((3/8)^2) + cos((5/8)^2) + cos((7/8)^2)].
T_4 = (1/8)[cos(0) + 2cos((1/4)^2) + 2cos((1/2)^2) + 2cos((3/4)^2) + cos(1)].
S_4 = (1/12)[cos(0) + 4cos(1/16) + 2cos(1/4) + 4cos(9/16) + cos(1)].
Compute: cos(0) = 1, cos(1/16) = 0.9981, cos(1/4) = 0.9689, cos(9/16) = 0.8316, cos(1) = 0.5403.
S_4 = (1/12)[1 + 4(0.9981) + 2(0.9689) + 4(0.8316) + 0.5403] = (1/12)[1 + 3.9924 + 1.9378 + 3.3264 + 0.5403] = (1/12)(10.7969) = 0.8997 (approximately).
E_n = Exact - Approximation = 0.9045 - 0.8997 = 0.0048.
Numerical integration is how engineers and scientists handle integrals that have no closed-form antiderivative. Simpson's rule in particular is used in signal processing, physics simulations, and anywhere a function is known only at discrete sample points.
"lim a_n = 0 means the series converges." This is the single most common error in this unit. The harmonic series is the standard counterexample.
Students confuse the integral test conclusion with equality: the integral test says the series and the integral share the same convergence status, not the same value.
(-1)^(2n) = 1 for every n. This catches students who assume any (-1)^(something) expression is alternating.
For geometric series, students sometimes misidentify the first term or the ratio. Always write the series in standard form a * r^n and read off a and r carefully.
In Simpson's rule, the coefficient pattern is 1, 4, 2, 4, 2, ..., 4, 1. Students sometimes get the alternating 4s and 2s wrong or use the wrong delta_x multiplier (it is delta_x / 3, not delta_x / 2).
Expect at least one "identify whether this series is geometric" question. Write out the first few terms and check the ratio.
True/false questions on series (like Problem 13) test conceptual understanding. Always have a counterexample ready.
Improper integrals with comparison are common: know the standard comparisons (1/x^p, e^(-x)).
Simpson's rule is often asked as a compute-and-round problem. Carry enough decimal places.
True or False: If lim a_n = 0, the series sum a_n must converge. (False.)
What is the sum of the geometric series 5 + 5/3 + 5/9 + 5/27 + ...? (Answer: 5 / (1 - 1/3) = 15/2.)
True or False: The series sum 1/n^(3/2) converges. (True; p = 3/2 > 1.)
Fill in the blank: In Simpson's rule with n subdivisions, the multiplier in front of the bracket is ____. (Answer: delta_x / 3.)
True or False: integral from 1 to infinity of e^(-x) dx converges. (True; it equals 1/e.)
Q: Determine whether sum from k=1 to infinity of (1 + 2k)/(3k + 5) converges or diverges.
A: lim (1 + 2k)/(3k + 5) = 2/3, which is not 0. By the divergence test, the series diverges.
Q: Find the sum of sum from n=0 to infinity of 4^(n+1)/5^n.
A: Rewrite as 4 * sum of (4/5)^n = 4 * 1/(1 - 4/5) = 4 * 5 = 20.
Q: Find the sum of sum from n=1 to infinity of 1/(n(n+2)).
A: Partial fractions: (1/2)[1/n - 1/(n+2)]. Telescoping gives (1/2)(1 + 1/2) = 3/4.
Q: Evaluate integral from 1 to infinity of 2/(3x+1)^3 dx.
A: Substitute u = 3x + 1. Integral becomes (2/3) * integral from 4 to infinity of u^(-3) du = (2/3)[1/(2u^2)] from 4 to infinity = (2/3)(1/32) = 1/48.
Q: Set up the trapezoid rule approximation for integral from 0 to 1 of cos(x^2) dx with n = 4.
A: delta_x = 1/4. T_4 = (1/8)[cos(0) + 2cos(1/16) + 2cos(1/4) + 2cos(9/16) + cos(1)].
Q: Is the statement "sum (-1)^(2n)/n converges by the alternating series test" true or false?
A: False. (-1)^(2n) = 1 for all n, so this is the harmonic series 1/n, which diverges. It is not alternating.
The integral test connects series to improper integrals directly: if you can evaluate the improper integral, you know whether the series converges (though not its sum). This bridges the material from the first half of the course (integration techniques) to the second half (series).
Numerical integration connects to differential equations because many ODEs have solutions that cannot be written in closed form, so engineers approximate them numerically using the same quadrature ideas (midpoint, trapezoid, Simpson's).
Geometric series reappear later in Taylor and Maclaurin series, where 1/(1 - x) = sum x^n is the starting point for generating power series representations of other functions.
Geometric series, p-series, harmonic series, telescoping series, divergence test, nth-term test, integral test, comparison test, partial sums, improper integral, convergent integral, divergent integral, midpoint rule, trapezoid rule, Simpson's rule, numerical approximation, error in numerical integration, Calculus II, MA-UY 1124, MA-UY 1424, NYU Courant