Calculus 2 Quiz: Representing Series As Power Series
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Representing Series As Power SeriesQuestion 1 of 7

The power series ∑n=0∞(−1)nx2n+12n+1\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} represents which of the following functions?

arctan⁡(x)\arctan(x) for ∣x∣≤1|x| \leq 1
sin⁡(x)\sin(x) for all real xx
ln⁡(1+x)\ln(1+x) for −1<x≤1-1 < x \leq 1
11+x2\frac{1}{1+x^2} for ∣x∣<1|x| < 1
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Calculus 2 Quiz

Calculus 2 Quiz: Representing Series As Power Series

Practice Representing Series As Power Series in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Representing Series As Power Series, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The power series ∑n=0∞(−1)nx2n+12n+1\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} represents which of the following functions?

  1. arctan⁡(x)\arctan(x) for ∣x∣≤1|x| \leq 1 (correct answer)
  2. sin⁡(x)\sin(x) for all real xx
  3. ln⁡(1+x)\ln(1+x) for −1<x≤1-1 < x \leq 1
  4. 11+x2\frac{1}{1+x^2} for ∣x∣<1|x| < 1
Explanation: The given series ∑n=0∞(−1)nx2n+12n+1=x−x33+x55−x77+⋯\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots is the Taylor series for arctan⁡(x)\arctan(x). Choice B is incorrect because sin⁡(x)=∑n=0∞(−1)nx2n+1(2n+1)!\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} (with factorial in denominator). Choice C is wrong because ln⁡(1+x)=∑n=1∞(−1)n+1xnn\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n} (different pattern). Choice D represents the derivative of arctan⁡(x)\arctan(x), not the function itself.

Question 2

Consider the power series representation of 11−x−x2\frac{1}{1-x-x^2}. If we write this as ∑n=0∞anxn\sum_{n=0}^{\infty} a_n x^n, which recurrence relation do the coefficients ana_n satisfy?

  1. an=an−1+an−2a_n = a_{n-1} + a_{n-2} for n≥2n \geq 2 with a0=0,a1=1a_0 = 0, a_1 = 1
  2. an=an−1−an−2a_n = a_{n-1} - a_{n-2} for n≥2n \geq 2 with a0=1,a1=1a_0 = 1, a_1 = 1
  3. an=an−1+an−2a_n = a_{n-1} + a_{n-2} for n≥2n \geq 2 with a0=1,a1=1a_0 = 1, a_1 = 1 (correct answer)
  4. an=2an−1+an−2a_n = 2a_{n-1} + a_{n-2} for n≥2n \geq 2 with a0=1,a1=1a_0 = 1, a_1 = 1
Explanation: When you encounter a function like 11−x−x2\frac{1}{1-x-x^2} and need to find its power series coefficients, the key insight is to use the relationship between the function and its series representation to derive a recurrence relation. Starting with 11−x−x2=∑n=0∞anxn\frac{1}{1-x-x^2} = \sum_{n=0}^{\infty} a_n x^n, multiply both sides by the denominator: (1−x−x2)∑n=0∞anxn=1(1-x-x^2) \sum_{n=0}^{\infty} a_n x^n = 1. Expanding the left side gives you ∑n=0∞anxn−∑n=0∞anxn+1−∑n=0∞anxn+2=1\sum_{n=0}^{\infty} a_n x^n - \sum_{n=0}^{\infty} a_n x^{n+1} - \sum_{n=0}^{\infty} a_n x^{n+2} = 1. Reindexing the shifted series: ∑n=0∞anxn−∑n=1∞an−1xn−∑n=2∞an−2xn+2=1\sum_{n=0}^{\infty} a_n x^n - \sum_{n=1}^{\infty} a_{n-1} x^n - \sum_{n=2}^{\infty} a_{n-2} x^{n+2} = 1. For this equation to hold, the coefficient of each power of xx on the left must match the right side. The constant term gives a0=1a_0 = 1, the coefficient of xx gives a1−a0=0a_1 - a_0 = 0, so a1=1a_1 = 1. For n≥2n \geq 2, we get an−an−1−an−2=0a_n - a_{n-1} - a_{n-2} = 0, which rearranges to an=an−1+an−2a_n = a_{n-1} + a_{n-2}. Option A has the wrong initial conditions (a0=0a_0 = 0), while option B uses subtraction instead of addition in the recurrence. Option D incorrectly doubles the an−1a_{n-1} term. Option C correctly captures both the recurrence relation an=an−1+an−2a_n = a_{n-1} + a_{n-2} and the initial conditions a0=1,a1=1a_0 = 1, a_1 = 1. Study tip: When finding recurrence relations from generating functions, always multiply by the denominator and match coefficients systematically. The Fibonacci-like recurrence here connects to the famous sequence!

Question 3

The function F(x)=∫0xsin⁡ttdtF(x) = \int_0^x \frac{\sin t}{t} dt can be represented as a power series. Which of the following correctly gives the first four non-zero terms of this series?

  1. x−x33!+x55!−x77!+⋯x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots
  2. x−x33⋅3!+x55⋅5!−x77⋅7!+⋯x - \frac{x^3}{3 \cdot 3!} + \frac{x^5}{5 \cdot 5!} - \frac{x^7}{7 \cdot 7!} + \cdots (correct answer)
  3. x−x44⋅3!+x66⋅5!−x88⋅7!+⋯x - \frac{x^4}{4 \cdot 3!} + \frac{x^6}{6 \cdot 5!} - \frac{x^8}{8 \cdot 7!} + \cdots
  4. 1−x22⋅3!+x44⋅5!−x66⋅7!+⋯1 - \frac{x^2}{2 \cdot 3!} + \frac{x^4}{4 \cdot 5!} - \frac{x^6}{6 \cdot 7!} + \cdots
Explanation: When you encounter a function defined as an integral that needs to be expressed as a power series, the key strategy is to find the power series of the integrand first, then integrate term by term. Start with the power series for sin⁡t\sin t: sin⁡t=t−t33!+t55!−t77!+⋯\sin t = t - \frac{t^3}{3!} + \frac{t^5}{5!} - \frac{t^7}{7!} + \cdots To find the series for sin⁡tt\frac{\sin t}{t}, divide each term by tt: sin⁡tt=1−t23!+t45!−t67!+⋯\frac{\sin t}{t} = 1 - \frac{t^2}{3!} + \frac{t^4}{5!} - \frac{t^6}{7!} + \cdots Now integrate this series term by term from 0 to xx: F(x)=∫0x(1−t23!+t45!−t67!+⋯ )dtF(x) = \int_0^x \left(1 - \frac{t^2}{3!} + \frac{t^4}{5!} - \frac{t^6}{7!} + \cdots\right) dt F(x)=x−x33⋅3!+x55⋅5!−x77⋅7!+⋯F(x) = x - \frac{x^3}{3 \cdot 3!} + \frac{x^5}{5 \cdot 5!} - \frac{x^7}{7 \cdot 7!} + \cdots This confirms answer B is correct. Answer A forgets to account for the division by tt in the original integrand - it's just the antiderivative of sin⁡t\sin t. Answer C incorrectly increases all the powers by 1 beyond what integration requires, likely from confusion about the integration process. Answer D starts with a constant term of 1, which is impossible since F(0)=0F(0) = 0 (the integral from 0 to 0 is zero). Remember: when integrating power series term by term, increase each power by 1 and divide by the new power. This technique works whenever you have a function defined by an integral with a series-expandable integrand.

Question 4

If h(x)=∑n=0∞x3n+13n+1h(x) = \sum_{n=0}^{\infty} \frac{x^{3n+1}}{3n+1}, then h′(x)h'(x) can be expressed in terms of which elementary function?

  1. 13(11−x3)\frac{1}{3}\left(\frac{1}{1-x^3}\right) for ∣x∣<1|x| < 1
  2. 11+x3\frac{1}{1+x^3} for ∣x∣<1|x| < 1
  3. 13(11+x3)\frac{1}{3}\left(\frac{1}{1+x^3}\right) for ∣x∣<1|x| < 1
  4. 11−x3\frac{1}{1-x^3} for ∣x∣<1|x| < 1 (correct answer)
Explanation: Taking the derivative of h(x)=∑n=0∞x3n+13n+1h(x) = \sum_{n=0}^{\infty} \frac{x^{3n+1}}{3n+1} term by term: h′(x)=∑n=0∞ddx(x3n+13n+1)=∑n=0∞(3n+1)x3n3n+1=∑n=0∞x3nh'(x) = \sum_{n=0}^{\infty} \frac{d}{dx}\left(\frac{x^{3n+1}}{3n+1}\right) = \sum_{n=0}^{\infty} \frac{(3n+1)x^{3n}}{3n+1} = \sum_{n=0}^{\infty} x^{3n}. Now, ∑n=0∞x3n=∑n=0∞(x3)n\sum_{n=0}^{\infty} x^{3n} = \sum_{n=0}^{\infty} (x^3)^n, which is a geometric series with first term 1 and common ratio x3x^3. For ∣x3∣<1|x^3| < 1 (i.e., ∣x∣<1|x| < 1), this sums to 11−x3\frac{1}{1-x^3}. Choice A includes an unnecessary factor of 13\frac{1}{3}, choice B has the wrong sign in the denominator, and choice C has both the wrong sign and an unnecessary factor.

Question 5

The function g(x)=ln⁡(1+x3)xg(x) = \frac{\ln(1+x^3)}{x} has a power series representation for x≠0x \neq 0 near x=0x = 0. What is the radius of convergence of this series?

  1. R=13R = \frac{1}{3}
  2. R=1R = 1 (correct answer)
  3. R=3R = 3
  4. The series does not converge for any x≠0x \neq 0
Explanation: When finding the radius of convergence for a power series representation of a function, you need to identify where the function itself is analytic (has no singularities). The radius of convergence extends from the center of the series to the nearest singularity. To find the power series for g(x)=ln⁡(1+x3)xg(x) = \frac{\ln(1+x^3)}{x}, start with the known series ln⁡(1+u)=u−u22+u33−u44+⋯\ln(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} + \cdots for ∣u∣<1|u| < 1. Substituting u=x3u = x^3: ln⁡(1+x3)=x3−x62+x93−x124+⋯\ln(1+x^3) = x^3 - \frac{x^6}{2} + \frac{x^9}{3} - \frac{x^{12}}{4} + \cdots Dividing by xx: g(x)=x2−x52+x83−x114+⋯g(x) = x^2 - \frac{x^5}{2} + \frac{x^8}{3} - \frac{x^{11}}{4} + \cdots The original series ln⁡(1+u)\ln(1+u) converges for ∣u∣<1|u| < 1, so ln⁡(1+x3)\ln(1+x^3) converges when ∣x3∣<1|x^3| < 1, which means ∣x∣<1|x| < 1. Since dividing by xx doesn't change the radius of convergence, g(x)g(x) has radius R=1R = 1. The correct answer is B. Choice A (R=13R = \frac{1}{3}) incorrectly assumes the cube root relationship applies inversely. Choice C (R=3R = 3) mistakenly multiplies instead of taking the cube root. Choice D is wrong because the series clearly converges for values like x=0.5x = 0.5. Study tip: The radius of convergence is determined by the nearest singularity to the expansion point. For composite functions like f(xn)f(x^n), if f(u)f(u) has radius RfR_f, then f(xn)f(x^n) has radius Rf1/nR_f^{1/n}.

Question 6

If g(x)=∫0xt21+t4dtg(x) = \int_0^x \frac{t^2}{1+t^4} dt, which power series represents g(x)g(x) near x=0x = 0?

  1. ∑n=0∞(−1)nx4n+3\sum_{n=0}^{\infty} (-1)^n x^{4n+3}
  2. ∑n=0∞(−1)nx4n+24n+2\sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{4n+2}
  3. ∑n=0∞(−1)nx4n+34n+3\sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+3}}{4n+3} (correct answer)
  4. ∑n=1∞(−1)n+1x4n−14n−1\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{4n-1}}{4n-1}
Explanation: When you encounter a definite integral with a variable upper limit, you're looking at a situation where you can find the power series by first expanding the integrand, then integrating term by term. Start with the integrand t21+t4\frac{t^2}{1+t^4}. Since 11+u=∑n=0∞(−1)nun\frac{1}{1+u} = \sum_{n=0}^{\infty} (-1)^n u^n for ∣u∣<1|u| < 1, you can substitute u=t4u = t^4 to get: 11+t4=∑n=0∞(−1)n(t4)n=∑n=0∞(−1)nt4n\frac{1}{1+t^4} = \sum_{n=0}^{\infty} (-1)^n (t^4)^n = \sum_{n=0}^{\infty} (-1)^n t^{4n} Therefore: t21+t4=t2∑n=0∞(−1)nt4n=∑n=0∞(−1)nt4n+2\frac{t^2}{1+t^4} = t^2 \sum_{n=0}^{\infty} (-1)^n t^{4n} = \sum_{n=0}^{\infty} (-1)^n t^{4n+2} Now integrate term by term from 0 to xx: g(x)=∫0x∑n=0∞(−1)nt4n+2dt=∑n=0∞(−1)n∫0xt4n+2dt=∑n=0∞(−1)nx4n+34n+3g(x) = \int_0^x \sum_{n=0}^{\infty} (-1)^n t^{4n+2} dt = \sum_{n=0}^{\infty} (-1)^n \int_0^x t^{4n+2} dt = \sum_{n=0}^{\infty} (-1)^n \frac{x^{4n+3}}{4n+3} This matches answer choice C. Looking at the wrong answers: A lacks the denominator (4n+3)(4n+3) that comes from integration. B has the wrong exponent in the numerator (4n+24n+2 instead of 4n+34n+3) and would result from forgetting to integrate. D starts the sum at n=1n=1 and uses different indexing that doesn't match our derivation. Study tip: When integrating power series term by term, remember that ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1} - both the exponent and denominator increase by 1. Always start with the geometric series formula when you see expressions like 11+u\frac{1}{1+u}.

Question 7

Consider the function f(x)=2x1−3x2f(x) = \frac{2x}{1-3x^2}. Which power series correctly represents f(x)f(x) for appropriate values of xx?

  1. ∑n=0∞2x(3x2)n=2x+6x3+18x5+54x7+⋯\sum_{n=0}^{\infty} 2x(3x^2)^n = 2x + 6x^3 + 18x^5 + 54x^7 + \cdots (correct answer)
  2. ∑n=0∞2(3x2)n=2+6x2+18x4+54x6+⋯\sum_{n=0}^{\infty} 2(3x^2)^n = 2 + 6x^2 + 18x^4 + 54x^6 + \cdots
  3. ∑n=0∞(2x)(3x)n=2x+6x2+18x3+54x4+⋯\sum_{n=0}^{\infty} (2x)(3x)^n = 2x + 6x^2 + 18x^3 + 54x^4 + \cdots
  4. ∑n=1∞2x(3x2)n−1=2x+6x3+18x5+54x7+⋯\sum_{n=1}^{\infty} 2x(3x^2)^{n-1} = 2x + 6x^3 + 18x^5 + 54x^7 + \cdots
Explanation: To represent f(x)=2x1−3x2f(x) = \frac{2x}{1-3x^2} as a power series, we factor out 2x2x and use the geometric series formula. We have f(x)=2x⋅11−3x2f(x) = 2x \cdot \frac{1}{1-3x^2}. Since 11−u=∑n=0∞un\frac{1}{1-u} = \sum_{n=0}^{\infty} u^n for ∣u∣<1|u| < 1, we get 11−3x2=∑n=0∞(3x2)n\frac{1}{1-3x^2} = \sum_{n=0}^{\infty} (3x^2)^n. Therefore, f(x)=2x∑n=0∞(3x2)n=∑n=0∞2x(3x2)nf(x) = 2x \sum_{n=0}^{\infty} (3x^2)^n = \sum_{n=0}^{\infty} 2x(3x^2)^n. Choice B omits the factor of xx, choice C incorrectly uses 3x3x instead of 3x23x^2, and choice D uses an unnecessarily complicated indexing that doesn't match the standard form.