AP Calculus BC Infinite Sequences and Series — Worked Answer Explanations

Unit 10 · 12 questions explained

1. Question 1 · Easy

   Does the geometric series $\sum _{n=0}^{\infty }{\left(\frac{2}{3}\right)}^n$ converge? If so, find its sum.

   • A
     Diverges

     Why not A: Incorrectly applies the divergence test; $a_n\to 0$ and $|r|=2/3<1$, so the series converges.
   • B
     Converges to $\frac{2}{3}$

     Why not B: Returns the common ratio $r$ rather than the sum formula $a/(1-r)$.
   • C
     Converges to $3$Correct
   • D
     Converges to $2$

     Why not D: Uses sum $=1/(1-r)$ but writes $1/(1-2/3)=3$ correctly — actually this gives $3$; D's answer $2$ comes from using $1/(2/3)$ instead.
   Explanation

   Geometric series: ${\sum }_{n=0}^{\infty }{r}^n=\frac{a}{1-r}$ for $|r|<1$. Here $a=1$ (first term at $n=0$) and $r=2/3$. Sum $=\frac{1}{1-2/3}=\frac{1}{1/3}=3$.

   Key takeaway

   Geometric series converges iff $|r| < 1$; sum $= a/(1-r)$ where $a$ is the first term.

2. Question 2 · Easy

   Does the $p$-series $\sum _{n=1}^{\infty }\frac{1}{n^{3/2}}$ converge or diverge?

   • A
     Diverges, because terms go to $0$

     Why not A: Confuses necessary condition (terms $\to 0$) with sufficient condition for convergence; many divergent series have terms $\to 0$.
   • B
     Diverges, because $p=3/2>1$

     Why not B: Inverts the $p$-series rule: $p>1$ gives convergence, not divergence.
   • C
     Converges, because $p=3/2>1$Correct
   • D
     Converges, because $p=3/2<2$

     Why not D: States an incorrect convergence criterion; the only condition is $p>1$, not $p<2$.
   Explanation

   $p$-series $\sum 1/n^p$ converges if and only if $p>1$. Here $p=3/2>1$, so the series converges.

   Key takeaway

   $p$-series test: $\sum 1/n^p$ converges if $p > 1$, diverges if $p \leq 1$; memorize the boundary $p = 1$ (harmonic series diverges).

3. Question 3 · Easy

   Find the interval of convergence of the power series $\sum _{n=0}^{\infty }\frac{x^n}{n+1}$.

   • A
     $(-1,1)$

     Why not A: Finds the radius of convergence $R=1$ but doesn't check endpoints; $x=1$ gives $\sum 1/(n+1)$ (diverges) and $x=-1$ gives $\sum (-1{)}^n/(n+1)$ (converges by alternating series).
   • B
     $[-1,1]$

     Why not B: Includes both endpoints; but $x=1$ gives the harmonic series $\sum 1/(n+1)$ which diverges.
   • C
     $[-1,1)$Correct
   • D
     $(-\infty ,\infty )$

     Why not D: Concludes the series converges everywhere; the Ratio Test gives radius $R=1$, not $\infty$.
   Explanation

   Ratio Test: ​xn/(n+1)xn+1/(n+2)​​=∣x∣⋅n+2n+1​→∣x∣. Converges when $|x|<1$. Endpoints: $x=1$: $\sum 1/(n+1)$ diverges (harmonic). $x=-1$: $\sum (-1{)}^n/(n+1)$ converges (alternating series test). Interval: $[-1,1)$.

   Key takeaway

   Always test endpoints separately after finding the radius; one endpoint may converge (alternating series) while the other diverges (harmonic).

4. Question 4 · Easy

   Use the Ratio Test to determine whether $\sum _{n=1}^{\infty }\frac{n!}{2^n}$ converges or diverges.

   • A
     Converges by Ratio Test

     Why not A: Computes $|a_{n+1}/a_n|=(n+1)/2\to \infty >1$ and misreads the Ratio Test: limit $>1$ means divergence, not convergence.
   • B
     Diverges by Ratio Test (limit $=\infty >1$)Correct
   • C
     Inconclusive (limit $=1$)

     Why not C: Computes $a_{n+1}/a_n$ incorrectly; the ratio grows without bound, not to $1$.
   • D
     Converges by comparison with $\sum 1/2^n$

     Why not D: $n!$ grows faster than $2^n$, so $n!/2^n>1/2^n$; this comparison would show a lower bound, not convergence.
   Explanation

   $\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{2^{n+1}}\cdot \frac{2^n}{n!}=\frac{n+1}{2}\to \infty >1$. Since the ratio exceeds $1$, the Ratio Test concludes divergence.

   Key takeaway

   Ratio Test: compute $\lim|a_{n+1}/a_n|$; $< 1$ converges, $> 1$ diverges, $= 1$ inconclusive; $n!$ always dominates exponentials.

5. Question 5 · Easy

   Find the Maclaurin series for $f(x)=e^{-x}$ up to the $x^3$ term.

   • A
     $1-x+\frac{x^2}{2}-\frac{x^3}{6}+\cdots$Correct
   • B
     $1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots$

     Why not B: Gives the Maclaurin series for $e^{+x}$, forgetting that $e^{-x}$ alternates in sign.
   • C
     $-1+x-\frac{x^2}{2}+\frac{x^3}{6}+\cdots$

     Why not C: Applies signs incorrectly: the constant term is $e^0=1$, not $-1$.
   • D
     $x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$

     Why not D: Gives the series for $\ln (1+x)$, not $e^{-x}$.
   Explanation

   Substitute $-x$ into $e^x={\sum }_{n=0}^{\infty }\frac{x^n}{n!}$: $e^{-x}={\sum }_{n=0}^{\infty }\frac{(-x{)}^n}{n!}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots =1-x+\frac{x^2}{2}-\frac{x^3}{6}+\cdots$

   Key takeaway

   Maclaurin series for $e^{-x}$: substitute $-x$ into the $e^x$ series; signs alternate starting with $+1$.

6. Question 6 · Easy

   Does the alternating series $\sum _{n=1}^{\infty }\frac{(-1{)}^{n+1}}{n^2}$ converge?

   • A
     No, because terms are not all positive.

     Why not A: The Alternating Series Test allows alternating signs; the requirement is $|a_n|$ decreasing to zero.
   • B
     Yes, absolutely (since $\sum 1/n^2$ converges).Correct
   • C
     Conditionally convergent only, not absolutely.

     Why not C: $\sum 1/n^2$ is a convergent $p$-series ($p=2>1$), so the series is absolutely convergent, not merely conditional.
   • D
     Yes, but only conditionally (alternating series test).

     Why not D: Ignores that $\sum |a_n|=\sum 1/n^2$ converges; the series is absolutely, not merely conditionally, convergent.
   Explanation

   ${\sum }_{n=1}^{\infty }|a_n|={\sum }_{n=1}^{\infty }1/n^2$, a convergent $p$-series with $p=2>1$. Absolute convergence implies convergence. The series converges absolutely.

   Key takeaway

   Absolute convergence: if $\sum|a_n|$ converges, then $\sum a_n$ converges; always check for absolute convergence before applying the weaker alternating series test.

7. Question 7 · Medium

   Find the Taylor series for $\ln (1+x)$ centered at $x=0$ (Maclaurin series).

   • A
     $\sum _{n=1}^{\infty }\frac{x^n}{n}$

     Why not A: Gives all positive terms; the series for $\ln (1+x)$ must alternate.
   • B
     $\sum _{n=1}^{\infty }\frac{(-1{)}^{n+1}{x}^n}{n}$Correct
   • C
     $\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^n}{n!}$

     Why not C: Gives the Maclaurin series for $e^{-x}$, not $\ln (1+x)$.
   • D
     $\sum _{n=1}^{\infty }\frac{(-1{)}^n{x}^n}{n}$

     Why not D: Off by sign: at $n=1$, gives $-x$ instead of $+x$ (the first term of $\ln (1+x)$ is $+x$).
   Explanation

   Integrate the geometric series $\frac{1}{1+x}={\sum }_{n=0}^{\infty }(-x{)}^n={\sum }_{n=0}^{\infty }(-1{)}^n{x}^n$. $\ln (1+x)={\int }_0^x\frac{1}{1+t}dt={\sum }_{n=0}^{\infty }(-1{)}^n\frac{x^{n+1}}{n+1}={\sum }_{n=1}^{\infty }\frac{(-1{)}^{n+1}{x}^n}{n}$ for $x\in (-1,1]$.

   Key takeaway

   Derive $\ln(1+x)$ series by integrating $1/(1+x) = \sum(-x)^n$; the resulting alternating series converges on $(-1, 1]$.

8. Question 8 · Medium

   Find the Lagrange error bound for approximating $\sin (0.1)$ using the third-degree Taylor polynomial about $x=0$.

   • A
     $\frac{(0.1{)}^4}{4!}$Correct
   • B
     $\frac{(0.1{)}^3}{3!}$

     Why not B: Uses the last included term's bound instead of the first omitted term's bound ($n+1=4$).
   • C
     $\frac{(0.1{)}^5}{5!}$

     Why not C: Uses degree $5$ instead of $4$; the third-degree polynomial omits terms starting at degree $4$.
   • D
     $\frac{(0.1{)}^4}{4}$

     Why not D: Writes $4$ in the denominator instead of $4!=24$.
   Explanation

   Third-degree Maclaurin polynomial for $\sin x$: $x-x^3/3!$. The error is bounded by $\frac{|f^{(4)}(c)|\cdot |x{|}^4}{4!}$ for some $c$ between $0$ and $x$. Since $|f^{(4)}|=|{\sin }^{(4)}(c)|=|\sin c|\le 1$, the bound is $\frac{(0.1{)}^4}{4!}=\frac{10^{-4}}{24}$.

   Key takeaway

   Lagrange error bound: $|R_n(x)| \leq \dfrac{M|x-a|^{n+1}}{(n+1)!}$ where $M$ bounds $|f^{(n+1)}|$; use degree $n+1$ (the first omitted degree).

9. Question 9 · Hard

   The Maclaurin series for $\frac{1}{1-x}$ is ${\sum }_{n=0}^{\infty }{x}^n$. Use this to find a series for $\frac{x}{(1-x{)}^2}$.

   • A
     $\sum _{n=1}^{\infty }n{x}^n$Correct
   • B
     $\sum _{n=0}^{\infty }{x}^{n+1}$

     Why not B: Returns the series for $x/(1-x)$ (integrating or shifting the geometric series without differentiating).
   • C
     $\sum _{n=0}^{\infty }(n+1)x^n$

     Why not C: Gives the series for $d/dx[1/(1-x)]=1/(1-x{)}^2=\sum (n+1)x^n$ but forgets the extra factor of $x$.
   • D
     $\sum _{n=0}^{\infty }n{x}^{n-1}$

     Why not D: This is the derivative of $\sum x^n$, i.e., $\sum n{x}^{n-1}=1/(1-x{)}^2$, but not multiplied by $x$.
   Explanation

   Differentiate $\frac{1}{1-x}={\sum }_{n=0}^{\infty }{x}^n$: $\frac{1}{(1-x{)}^2}={\sum }_{n=1}^{\infty }n{x}^{n-1}$. Multiply both sides by $x$: $\frac{x}{(1-x{)}^2}={\sum }_{n=1}^{\infty }n{x}^n$.

   Key takeaway

   Differentiate a known power series to get new series; then multiply by $x$ or $x^2$ to shift or scale as needed.

10. Question 10 · Hard

    Determine the radius of convergence of $\sum _{n=0}^{\infty }\frac{(x-3{)}^n}{4^n(n+1{)}^2}$.

    • A
      $R=1$

      Why not A: Ignores the $4^n$ in the denominator and finds $R=1$ from $|x-3|<1$.
    • B
      $R=4$Correct
    • C
      $R=16$

      Why not C: Squares the $4^n$ denominator factor erroneously, getting $R=4^2=16$.
    • D
      $R=\infty$

      Why not D: Concludes that $(n+1{)}^2$ in the denominator guarantees convergence for all $x$; misapplies the comparison test.
    Explanation

    Ratio Test: ​(x−3)n/[4n(n+1)2](x−3)n+1/[4n+1(n+2)2]​​=4∣x−3∣​⋅(n+2n+1​)2→4∣x−3∣​. Converges when $|x-3|/4<1$, i.e., $|x-3|<4$. Radius $R=4$.

    Key takeaway

    Ratio Test for power series about $x = a$: the limit gives $|x-a|/R$; set $< 1$ to find $R$; the $(n+1)^2$ factor vanishes in the ratio limit.

11. Question 11 · Hard

    Use the Maclaurin series $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$ to find $\underset{x\to 0}{\lim }\frac{\sin x-x}{x^3}$.

    • A
      $-\frac{1}{6}$Correct
    • B
      $0$

      Why not B: Substitutes $x=0$ in $\sin x-x=0$ and concludes $0/0=0$ without computing the leading term.
    • C
      $\frac{1}{6}$

      Why not C: Correct magnitude but wrong sign; the leading term of $\sin x-x$ is $-x^3/6$.
    • D
      $1$

      Why not D: Computes $\lim (\sin x)/x^3=\lim 1/x^2=\infty$ and then makes an error, or confuses with $\lim \sin x/x=1$.
    Explanation

    $\sin x-x=\left(x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots \right)-x=-\frac{x^3}{6}+O(x^5)$. $\frac{\sin x-x}{x^3}=\frac{-x^3/6+O(x^5)}{x^3}=-\frac{1}{6}+O(x^2)\to -\frac{1}{6}$.

    Key takeaway

    Series expand the numerator, cancel, and read off the limit from the leading remaining term; this avoids repeated L'Hôpital applications.

12. Question 12 · Hard

    The Taylor polynomial of degree $4$ for $f(x)=\cos x$ about $x=0$ is $P_4(x)=1-\frac{x^2}{2}+\frac{x^4}{24}$. Use the Lagrange error bound to estimate $|\cos (0.5)-P_4(0.5)|$.

    • A
      $\frac{(0.5{)}^5}{120}$Correct
    • B
      $\frac{(0.5{)}^4}{24}$

      Why not B: Uses the last included term's bound; Lagrange bound uses the first omitted term (degree $5$).
    • C
      $\frac{(0.5{)}^6}{720}$

      Why not C: Uses degree $6$ error bound instead of degree $5$; degree $4$ polynomial omits degree $5$ first.
    • D
      $\frac{(0.5{)}^5}{5}$

      Why not D: Uses $5$ in the denominator instead of $5!=120$.
    Explanation

    The first omitted term in $P_4$ is degree $5$: $f^{(5)}(x)=\sin x$ for $\cos x$ (5th derivative of $\cos x$). $|f^{(5)}(c)|=|\sin c|\le 1$. Lagrange bound: $|R_4(0.5)|\le \frac{1\cdot (0.5{)}^5}{5!}=\frac{(0.5{)}^5}{120}$.

    Key takeaway

    Lagrange error for $P_n$: use $(n+1)$th derivative bound divided by $(n+1)!$; for $P_4$ and $\cos x$, the 5th derivative is $\sin x$ with $|\sin c| \leq 1$.