Infinite sequences and series (BC) practice questions (AP Calculus)

Sample questions

All 5 free AP Calculus questions on this unit, from the free AP Calculus BC Practice Exam. Each answer is behind its own toggle; the whole paper, its PDF and the full answer key are on the exam page.

1. Question 12 marks · Series convergence
   The interval of convergence of the power series $\displaystyle\sum_{n=1}^{\infty} \frac{(x - 2)^n}{n \cdot 3^n}$ is
   • A.
     $(-1, 5)$
   • B.
     $[-1, 5)$
   • C.
     $(-1, 5]$
   • D.
     $[-1, 5]$
   Show answerFrom the free AP Calculus BC Practice Exam →
2. Question 22 marks · Series error bounds
   The series $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2} = 1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} + \cdots$ converges to a sum $S$. If the first four terms are used to approximate $S$, the alternating series error bound guarantees that the error is at most
   • A.
     $\dfrac{1}{16}$
   • B.
     $\dfrac{1}{25}$
   • C.
     $\dfrac{1}{36}$
   • D.
     $\dfrac{1}{5}$
   Show answerFrom the free AP Calculus BC Practice Exam →
3. Question 32 marks · Series error bounds
   A function $f$ has derivatives of all orders, and $\left|f^{(4)}(x)\right| \leq 6$ for all $x$ in $[0, 1]$. If the third-degree Taylor polynomial for $f$ about $x = 0$ is used to approximate $f\!\left(\tfrac{1}{2}\right)$, the Lagrange error bound guarantees that the error is at most
   • A.
     $\dfrac{1}{16}$
   • B.
     $\dfrac{1}{32}$
   • C.
     $\dfrac{1}{64}$
   • D.
     $\dfrac{3}{8}$
   Show answerFrom the free AP Calculus BC Practice Exam →
4. Question 42 marks · Taylor series
   The third-degree Taylor polynomial for $f(x) = \ln x$ about $x = 1$ is
   • A.
     $(x - 1) - \dfrac{(x-1)^2}{2} + \dfrac{(x-1)^3}{3}$
   • B.
     $(x - 1) - \dfrac{(x-1)^2}{2!} + \dfrac{(x-1)^3}{3!}$
   • C.
     $(x - 1) + \dfrac{(x-1)^2}{2} + \dfrac{(x-1)^3}{3}$
   • D.
     $1 + (x - 1) - \dfrac{(x-1)^2}{2} + \dfrac{(x-1)^3}{3}$
   Show answerFrom the free AP Calculus BC Practice Exam →
5. Question 59 marks · Taylor series
   Series. Let $f$ be the function defined by $f(x) = e^{-x^2}$, and let $g$ be the function defined by $g(x) = \displaystyle\int_0^x f(t)\,dt$. You may use the fact that the Maclaurin series for $e^u$ is $\displaystyle\sum_{n=0}^{\infty} \frac{u^n}{n!}$, which converges for all real $u$.
   (a)

   Write the first four nonzero terms and the general term of the Maclaurin series for $f$. Use the series to find $f^{(6)}(0)$.

   [3]
   (b)

   Write the first four nonzero terms and the general term of the Maclaurin series for $g$.

   [2]
   (c)

   Use the first two nonzero terms of the series in part (b) to approximate $g\!\left(\tfrac{1}{2}\right)$. Show that this approximation differs from the true value of $g\!\left(\tfrac{1}{2}\right)$ by less than $\dfrac{1}{300}$.

   [2]
   (d)

   Find $\displaystyle\lim_{x \to 0} \frac{g(x) - x}{x^3}$.

   [2]
   Show answerFrom the free AP Calculus BC Practice Exam →

Key terms

The definitions and formulas AP Calculus expects for infinite sequences and series (bc), typeset as they appear on the paper.

Convergence of a series

A series converges when its sequence of partial sums has a finite limit. If the terms do not tend to zero the series diverges (nth-term test), but terms tending to zero does not guarantee convergence.

$$\lim_{n \to \infty} a_n \ne 0 \;\Rightarrow\; \sum a_n \text{ diverges}$$

Geometric series

Converges when the common ratio has absolute value less than 1, to a closed-form sum. It also gives the Maclaurin series for $\frac{1}{1 - x}$.

$$\sum_{n=0}^{\infty} a r^n = \frac{a}{1 - r}, \quad |r| < 1$$

p-series and the integral test

The $p$-series converges when $p > 1$ and diverges otherwise (the harmonic series, $p = 1$, diverges). The integral test compares a positive decreasing series with the matching improper integral.

$$\sum_{n=1}^{\infty} \frac{1}{n^p} \text{ converges} \iff p > 1$$

Comparison and ratio tests

Direct and limit comparison compare a series with a known $p$-series or geometric series. The ratio test gives absolute convergence when the limit of the ratio of successive terms is below 1, and is the usual tool for radius of convergence.

$$\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| = L: \quad L < 1 \text{ converges}, \; L > 1 \text{ diverges}$$

Alternating series test and error bound

An alternating series whose terms decrease in absolute value to zero converges, and the error after $n$ terms is at most the size of the first omitted term.

$$|S - S_n| \le a_{n+1}$$

Taylor polynomial and Taylor series

The polynomial that matches $f$ and its first $n$ derivatives at $x = a$; letting $n$ grow gives the Taylor series. A Maclaurin series is the case $a = 0$.

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

Lagrange error bound

The remainder after a degree-$n$ Taylor polynomial is bounded by the maximum size $M$ of the $(n+1)$th derivative on the interval. State what $M$ is and where it comes from.

$$|R_n(x)| \le \frac{M}{(n+1)!}\,|x - a|^{n+1}$$

Maclaurin series to memorise

The four series the exam assumes, with their intervals of convergence: $e^x$, $\sin x$ and $\cos x$ converge for all $x$; $\frac{1}{1-x}$ for $|x| < 1$.

$$e^x = \sum \frac{x^n}{n!}, \quad \sin x = \sum \frac{(-1)^n x^{2n+1}}{(2n+1)!}, \quad \cos x = \sum \frac{(-1)^n x^{2n}}{(2n)!}, \quad \frac{1}{1-x} = \sum x^n$$

Frequently asked questions

Other AP Calculus topics

• Limits and continuity
• Differentiation: definition and basic rules
• Composite, implicit and inverse functions
• Contextual applications of differentiation
• Analytical applications of differentiation
• Integration and accumulation of change
• Differential equations
• Applications of integration
• Parametric, polar and vector functions (BC)

All AP Calculus topics · AP Calculus exam guide · All free practice exams