AP Calc BC Series — Answers & Worked Solutions

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Chapter 1 — Sequences

1.1 $\dfrac{5}{2}$. Compare degrees (same; ratio of leading coefficients).

1.2 $0$. L'Hôpital: $\frac{1/x}{2x} = \frac{1}{2x^2} \to 0$.

1.3 $e^2$. Standard limit $\left(1+\frac{a}{n}\right)^n \to e^a$.

1.4 Converges to $0$. $\left|\frac{(-1)^n}{n+1}\right| = \frac{1}{n+1} \to 0$, so by squeeze the original $\to 0$.

1.5 $0$. $-\frac{1}{\sqrt{n}} \le \frac{\cos(n^2)}{\sqrt{n}} \le \frac{1}{\sqrt{n}}$, both bounds $\to 0$.

1.6 Diverges. Ratio $\frac{a_{n+1}}{a_n} = \frac{n+1}{3} \to \infty$ — terms grow without bound. Factorial beats exponential.

1.7 $\frac{a_{n+1}}{a_n} = \frac{2}{n+1} < 1$ for $n \ge 2$, so $\{a_n\}$ is decreasing from $n = 2$ onward.

1.8 Monotonicity: $a_1 = 2 < a_2 = \sqrt{8} = 2\sqrt 2 \approx 2.83$. Induction: $a_n > a_{n-1} \Rightarrow 6+a_n > 6+a_{n-1} \Rightarrow a_{n+1} > a_n$.
Bounded above by 3: $a_1 = 2 < 3$. Induction: $a_n < 3 \Rightarrow 6+a_n < 9 \Rightarrow a_{n+1} = \sqrt{6+a_n} < 3$.
MBT → converges. Set $L = \lim a_n$: $L = \sqrt{6+L} \Rightarrow L^2 - L - 6 = 0 \Rightarrow L = 3$ (rejecting $L = -2$).

1.9 $\ln 2$. Rewrite as $\frac{1}{n}\sum_{k=0}^{n}\frac{1}{1+k/n}$, a Riemann sum for $\int_0^1 \frac{dx}{1+x} = \ln 2$.

1.10 $\cos(n\pi) = (-1)^n$, oscillates between $\pm 1$ — diverges.

Chapter 2 — Series Basics, Geometric, p-Series

2.1 Geometric, $a = 4$, $r = 1/3$. Sum $= \dfrac{4}{1 - 1/3} = 6$.

2.2 Geometric starting at $n = 2$: first term $\frac{1}{4}$, $r = \frac{1}{2}$. Sum $= \dfrac{1/4}{1 - 1/2} = \dfrac{1}{2}$.

2.3 $p = 0.99 \le 1$ → diverges.

2.4 $\lim a_n = 3 \ne 0$, n-th term test → diverges.

2.5 Rewrite: $\sum \frac{2 \cdot 2^n}{5^{n-1}} = 4\sum \left(\frac{2}{5}\right)^{n-1}$. First term ($n=1$) is $4$, $r = 2/5$. Sum $= \dfrac{4}{1-2/5} = \dfrac{20}{3}$.

2.6 Telescoping: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$, $S_N = 1 - \frac{1}{N+1} \to 1$.

2.7 $\frac{1}{(2n-1)(2n+1)} = \frac{1}{2}\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right)$, $S_N = \frac{1}{2}\left(1 - \frac{1}{2N+1}\right) \to \frac{1}{2}$.

2.8 $0.\overline{37} = \frac{0.37}{1 - 0.01} = \frac{37}{99}$.

2.9 $3 \cdot 5 + 2 \cdot (-2) = 11$.

2.10 Total distance $= 10 + 2(10 \cdot \frac{2}{3} + 10 \cdot \frac{4}{9} + \cdots) = 10 + 2 \cdot \frac{20/3}{1 - 2/3} = 10 + 40 = 50$ m.

Chapter 3 — Convergence Tests

3.1 $L = \lim \frac{(n+1)^2/2^{n+1}}{n^2/2^n} = \frac{1}{2} < 1$ → converges.

3.2 $L = \lim \frac{3n+1}{4n+2} = \frac{3}{4} < 1$ → converges.

3.3 $\int_2^\infty \frac{dx}{x \ln x} = \ln(\ln x)\big|_2^\infty = \infty$ → diverges.

3.4 $\frac{1}{2^n + n} < \frac{1}{2^n}$, geometric $r = 1/2$ converges → DCT → converges.

3.5 $b_n = \frac{1}{n^2}$, $\lim \frac{(n+\sin n)/(n^3-1)}{1/n^2} = 1$, $\sum 1/n^2$ converges → converges.

3.6 Ratio $L = \lim \frac{(n+1)!/(2n+2)!}{n!/(2n)!} = \lim \frac{n+1}{(2n+1)(2n+2)} = 0 < 1$ → converges.

3.7 $f(x) = \ln x / x^2$ is decreasing for large $x$. $\int_1^\infty \frac{\ln x}{x^2} dx$ via integration by parts ($u = \ln x$, $dv = x^{-2}dx$) gives $\left[-\frac{\ln x}{x}\right]_1^\infty + \int \frac{dx}{x^2} = 0 + 1 = 1$. Converges.

3.8 Ratio $L = \lim \frac{(n+1)^{n+1}/(n+1)!}{n^n/n!} = \lim \frac{(n+1)^n}{n^n} = e > 1$ → diverges.

3.9 $\arctan n \to \pi/2$, so $\frac{1}{n^2 \arctan n} \sim \frac{2}{\pi n^2}$. Limit comparison with $\frac{1}{n^2}$ → converges.

3.10 Ratio $L = \lim \frac{(2n+2)!/(4^{n+1}((n+1)!)^2)}{(2n)!/(4^n(n!)^2)} = \lim \frac{(2n+1)(2n+2)}{4(n+1)^2} = 1$ — inconclusive. Stirling gives $\frac{(2n)!}{4^n(n!)^2} \sim \frac{1}{\sqrt{\pi n}}$, comparison with $\sum 1/\sqrt{n}$ ($p=1/2$) shows divergence.

Chapter 4 — Alternating Series

4.1 AST converges + $\sum 1/n^2$ converges → absolutely convergent.

4.2 AST converges + $\sum 1/(2n+1)$ diverges (same order as harmonic) → conditionally convergent.

4.3 $b_n = n/(n^2+4)$ decreasing for $n \ge 2$ (derivative argument), $\to 0$ → AST converges. Absolute version $\sum n/(n^2+4)$ matches $\sum 1/n$ in order, diverges → conditionally convergent.

4.4 $\lim \frac{n}{n+1} = 1 \ne 0$, n-th term test → diverges.

4.5 Error $\le b_4 = \frac{1}{4^4} = \frac{1}{256} \approx 0.0039$.

4.6 $\frac{1}{(N+1)!} < 0.0001 \Rightarrow (N+1)! > 10000$. $7! = 5040, 8! = 40320$, so $N+1 \ge 8$, at least 7 terms ($N = 7$).

4.7 $b_n = \ln n / n$. $f(x) = \ln x / x$, $f'(x) = (1 - \ln x)/x^2 < 0$ when $x > e$, so eventually decreasing; $\to 0$. AST converges. Absolute version $\sum \ln n / n$ dominates $\sum 1/n$ (since $\ln n > 1$), diverges → conditionally convergent.

4.8 $b_n = 1/(n \ln n)$ decreasing $\to 0$ → AST converges. Absolute version $\sum 1/(n \ln n)$ diverges by integral test → conditionally convergent.

4.9 $b_n = 1/\sqrt{n^2+1} \to 0$ decreasing → AST converges. Absolute $\sum 1/\sqrt{n^2+1} \sim \sum 1/n$ diverges → conditionally convergent.

4.10 $b_{N+1} = \frac{1}{N+1} < 0.01 \Rightarrow N+1 > 100 \Rightarrow N \ge 100$. At least 100 terms.

Chapter 5 — Power Series

5.1 $L = |x|$, $R = 1$. $x=1$: $\sum 1/(n+1)$ harmonic — diverges. $x=-1$: $\sum (-1)^n/(n+1)$ AST — converges. $[-1, 1)$.

5.2 $L = |x-2|$, $R = 1$. $x=3$: $\sum 1/n^2$ converges; $x=1$: $\sum (-1)^n/n^2$ absolutely converges. $[1, 3]$.

5.3 $L = |2x| / (n+1) \to 0$, so $R = \infty$.

5.4 $L = |x+1|(n+1) \to \infty$ unless $x = -1$. $R = 0$, converges only at $x = -1$.

5.5 $L = |x|$, $R = 1$. $x=1$: $\sum 1/\sqrt n$ diverges; $x=-1$: $\sum (-1)^n/\sqrt n$ AST converges. $[-1, 1)$.

5.6 $L = |x-3|/4$, $R = 4$. $x=7$: $\sum (-1)^n$ diverges; $x=-1$: $\sum 1$ diverges. $(-1, 7)$.

5.7 Center 0; $R \ge 4$ (since $x=4$ converges) and $R \le 5$ (since $x=-5$ diverges), so $R \in [4, 5)$.

• $x = 2$: $|x|=2 < 4$, must converge
• $x = -3$: $|x|=3 < 4$, must converge
• $x = 5$: $|x|=5 > R$, must diverge
• $x = -7$: $|x|=7 > 5 > R$, must diverge

5.8 $L = |x+2|/3$, $R = 3$. $x=1$: $\sum 1/(n+1)$ harmonic diverges; $x=-5$: $\sum (-1)^n/(n+1)$ AST converges. $[-5, 1)$.

5.9 $L = x^2 \cdot \lim \frac{n}{n+1} = x^2$. $R^2 = 1 \Rightarrow R = 1$. At $x = \pm 1$: both give $\sum (-1)^n/n$, conditionally convergent. $[-1, 1]$.

5.10 $g$ converges when $|x^2| < 3 \Rightarrow |x| < \sqrt{3}$. $R_g = \sqrt 3$.

Chapter 6 — Taylor / Maclaurin

6.1 $e^{2x} = \sum \frac{(2x)^n}{n!} = \sum \frac{2^n x^n}{n!}$.

6.2 $\sin(3x) = \sum \frac{(-1)^n (3x)^{2n+1}}{(2n+1)!} = \sum \frac{(-1)^n 3^{2n+1} x^{2n+1}}{(2n+1)!}$.

6.3 $P_4(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24}$.

6.4 $\ln x$ at $a=1$: $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}(x-1)^n$.

6.5 $f^{(n)}(x) = (-1)^n n! / x^{n+1}$, $f^{(n)}(2) = (-1)^n n! / 2^{n+1}$. Coefficient $c_n = (-1)^n / 2^{n+1}$.
$$\frac{1}{x} = \sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}}(x-2)^n$$

6.6 $f^{(n)}(1) = e$ for all $n$. $T(x) = e \sum \frac{(x-1)^n}{n!} = e \cdot e^{x-1} = e^x$. ✓

6.7 $P_4(0.2) = 1 - 0.02 + \frac{0.0016}{24} \approx 0.98007$. Matches $\cos(0.2) \approx 0.98007$ to 5 decimals.

6.8 $c_3 = f^{(3)}(2)/3!$ → $f^{(3)}(2) = 30$.

6.9 $\sin$ at $a = \pi$: $\sin\pi = 0, \cos\pi = -1, -\sin\pi = 0, -\cos\pi = 1, \ldots$. First three nonzero terms: $-(x-\pi) + \frac{(x-\pi)^3}{6} - \frac{(x-\pi)^5}{120}$.

6.10 $f(x) = (1+x)^{1/2}$: $f(0)=1, f'(0)=1/2, f''(0)=-1/4, f'''(0)=3/8$.
$P_3(x) = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16}$.

Chapter 7 — Common Maclaurin & Operations

7.1 $e^{-x} = \sum \frac{(-x)^n}{n!} = \sum \frac{(-1)^n x^n}{n!}$.

7.2 $\cos(x^2) = \sum \frac{(-1)^n (x^2)^{2n}}{(2n)!} = \sum \frac{(-1)^n x^{4n}}{(2n)!}$.

7.3 $\frac{x}{1-x} = x \sum x^n = \sum_{n=1}^\infty x^n$.

7.4 Differentiate $\frac{1}{1-x} = \sum x^n$: $\frac{1}{(1-x)^2} = \sum_{n=1}^\infty n x^{n-1} = \sum_{n=0}^\infty (n+1) x^n$.

7.5 $1 - \cos x = \frac{x^2}{2} - \frac{x^4}{24} + \cdots$, divide by $x^2$: $\frac{1}{2} - \frac{x^2}{24} + \cdots \to \frac{1}{2}$.

7.6 $e^x - 1 - x = \frac{x^2}{2} + \frac{x^3}{6} + \cdots$, divide by $x^2$: $\frac{1}{2} + \frac{x}{6} \to \frac{1}{2}$.

7.7 $\sin(x^2) = x^2 - \frac{x^6}{6} + \cdots$, integrate: $\int_0^1 = \frac{1}{3} - \frac{1}{42} + \cdots$. First two terms: $\frac{1}{3} - \frac{1}{42} = \frac{13}{42} \approx 0.310$.

7.8 $e^x = \sum x^n/n!$ at $x = -1$: $\sum (-1)^n/n! = e^{-1} = 1/e$.

7.9 $-\ln(1-x) = \sum x^n/n$ for $|x|<1$. At $x = 1/2$: $\sum 1/(n \cdot 2^n) = -\ln(1/2) = \ln 2$.

7.10 $\arctan x = x - x^3/3 + x^5/5 - \cdots$. $P_5(0.1) = 0.1 - 0.001/3 + 0.00001/5 \approx 0.0996687$. AST error $\le b_4 = (0.1)^7/7 \approx 1.4\times 10^{-9}$.

Chapter 8 — Lagrange Error

8.1 $f^{(4)} = \sin x$, $|f^{(4)}| \le 1$, $M=1$. $|R_3| \le \frac{1}{4!}(0.4)^4 = \frac{0.0256}{24} \approx 1.07 \times 10^{-3}$.

8.2 $f^{(5)} = e^x$, on $[0, 0.3]$ bounded by $e^{0.3} < 1.4$, take $M=1.4$. $|R_4| \le \frac{1.4}{5!}(0.3)^5 \approx 2.83 \times 10^{-5}$.

8.3 $f^{(7)} = -\sin x$, $M = 1$. $|R_6| \le \frac{1}{7!}(0.5)^7 \approx 1.55 \times 10^{-6}$.

8.4 Estimate $e^1$, $M = 3$. $\frac{3}{(n+1)!} < 10^{-4} \Rightarrow (n+1)! > 30000$. $8! = 40320$ ✓, $7! = 5040$ ✗. At least degree 7.

8.5 $M = 1$. $\frac{1}{(n+1)!} < 0.001 \Rightarrow (n+1)! > 1000$. $7! = 5040$ ✓, $6! = 720$ ✗. At least degree 6.

8.6 $|f^{(n+1)}(t)| = |\sin t \text{ or } \cos t| \le 1$. $|R_n(x)| \le \frac{|x|^{n+1}}{(n+1)!} \to 0$ for any fixed $x$. So the Taylor series converges to $\sin x$ for every $x$.

8.7 $|R_5(1)| \le \frac{5}{6!}(1)^6 = \frac{5}{720} = \frac{1}{144} \approx 6.94 \times 10^{-3}$.

8.8 $\ln(1+x)$, $f^{(n+1)}(x) = (-1)^n n! / (1+x)^{n+1}$. On $[0, 0.2]$: $|f^{(n+1)}| \le n!$. Error $\le \frac{n!}{(n+1)!}(0.2)^{n+1} = \frac{(0.2)^{n+1}}{n+1}$. Need $< 0.001$: $n=3$: $0.0016/4 = 4\times 10^{-4}$ ✓; $n=2$: $0.008/3 \approx 2.7\times 10^{-3}$ ✗. At least degree 3.

8.9 $\cos(\pi/4) = \sqrt{2}/2 \approx 0.7071$.
$P_6(\pi/4) = 1 - \frac{(\pi/4)^2}{2} + \frac{(\pi/4)^4}{24} - \frac{(\pi/4)^6}{720} \approx 0.70710$.
Error $\le \frac{1}{8!}(\pi/4)^8 \approx 2.32 \times 10^{-6}$.

8.10
(a) $f^{(n)}(0) = 2^n$. $P_3(x) = 1 + 2x + 2x^2 + \frac{4 x^3}{3}$.
(b) $P_3(0.1) = 1 + 0.2 + 0.02 + 0.001333 \approx 1.221333$.
(c) $f^{(4)}(t) = 16 e^{2t}$, on $[0, 0.1]$: $\le 16 e^{0.2} < 19.6$. Take $M = 19.6$.
$|R_3| \le \frac{19.6}{4!}(0.1)^4 \approx 8.17 \times 10^{-5}$.

Cross-Chapter Challenges

X.1 Absolute version: ratio $\lim \frac{(n+1)/e^{n+1}}{n/e^n} = \frac{1}{e} < 1$ → absolutely convergent.

X.2 $\sin x$ at $x = \pi/4$: $\sin(\pi/4) = \sqrt{2}/2$.

X.3 $\frac{x}{1+x^2} = x \sum (-x^2)^n = \sum (-1)^n x^{2n+1}$, $|x|<1$, $R=1$. Endpoints $x=\pm 1$: $\sum (\pm 1)(-1)^n$, diverge. Interval $(-1, 1)$.

X.4 $\frac{\sin t}{t} = 1 - \frac{t^2}{6} + \frac{t^4}{120} - \cdots$. Integrate: $f(x) = x - \frac{x^3}{18} + \frac{x^5}{600} - \cdots$. Radius $\infty$ (matches $\sin t / t$).

X.5 $y = \sum c_n x^n$, $y' = \sum (n+1) c_{n+1} x^n$. Equation $y' = y \Rightarrow (n+1) c_{n+1} = c_n \Rightarrow c_{n+1} = c_n/(n+1)$. With $c_0 = 1$: $c_n = 1/n!$. So $y = \sum x^n/n! = e^x$.

End. Total: 80 main + 5 challenge = 85 problems.