AP Calculus BC Quiz

AP Calculus BC Quiz: Alternating Series Error Bound

Practice Alternating Series Error Bound in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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For $\sum_{k=1}^{\infty}(-1)^{k+1}\frac{1}{k^3}$ , what is the maximum error when approximating the sum by $S_{10}$ ?

What this quiz covers

This quiz focuses on Alternating Series Error Bound, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

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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.

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Question 1

For $\sum_{k=1}^{\infty}(-1)^{k+1}\frac{1}{k^3}$ , what is the maximum error when approximating the sum by $S_{10}$ ?

$\le \dfrac{1}{10^3}$
$\le \dfrac{1}{11^3}$ (correct answer)
$\le \dfrac{1}{12^3}$
$\le \displaystyle\sum_{k=11}^{\infty}\dfrac{1}{k^3}$
$\le \dfrac{1}{2\cdot 11^3}$

Explanation: The skill being tested here is the alternating series error bound, which provides a way to estimate the error when approximating an infinite alternating series with a partial sum. For an alternating series that satisfies the conditions of the alternating series test—terms alternating in sign, decreasing in absolute value, and approaching zero—the error in using the partial sum S_n is less than the absolute value of the next term, a_{n+1}. This bound works because the remainder after n terms is bracketed between zero and the first omitted term, ensuring the actual sum lies within S_n ± |a_{n+1}|. In this case, for the series ∑ (-1)^{k+1}/k^3 approximated by S_10, the next term is 1/11^3, so the error is at most 1/11^3. A tempting distractor like ≤ 1/10^3 fails because it uses the last included term instead of the next one, which underestimates the bound since the error is actually smaller than the next term, not the previous. A transferable strategy for error bounds in alternating series is to always identify the first omitted term after verifying the series meets the convergence criteria.

Question 2

For $\sum_{n=1}^{\infty}(-1)^{n}\frac{3}{\sqrt{n}}$ , what is the maximum error when approximating with $50$ terms?

$\le \dfrac{3}{\sqrt{50}}$
$\le \dfrac{3}{\sqrt{51}}$ (correct answer)
$\le \dfrac{3}{\sqrt{52}}$
$\le \dfrac{3}{2\sqrt{51}}$
$\le \left|\dfrac{3}{\sqrt{51}}-\dfrac{3}{\sqrt{50}}\right|$

Explanation: This question tests the alternating series error bound for $\sum_{n=1}^{\infty}(-1)^{n}\frac{3}{\sqrt{n}}$ using 50 terms. When we stop at $n=50$ , the first omitted term is at $n=51$ . The absolute value of this term is $\left|(-1)^{51}\frac{3}{\sqrt{51}}\right| = \frac{3}{\sqrt{51}}$ . Choice E, $\left|\frac{3}{\sqrt{51}}-\frac{3}{\sqrt{50}}\right|$ , incorrectly suggests using the difference between consecutive terms rather than the next term itself. The key principle for alternating series error bounds is that the error is bounded by the absolute value of the first term not included in your partial sum.

Question 3

Consider $\sum_{k=0}^{\infty}(-1)^k\frac{4}{(k+2)^2}$ . What is the alternating-series error bound after $n$ terms?

$\dfrac{4}{(n+2)^2}$
$\dfrac{4}{(n+1)^2}$
$\dfrac{4}{(n+3)^2}$ (correct answer)
$\dfrac{4}{n+3}$
$\dfrac{2}{(n+3)^2}$

Explanation: The skill being tested here is the alternating series error bound, which provides an upper limit on the approximation error for convergent alternating series. For ∑_{k=0}^∞ (-1)^k (4/(k+2)^2), terms alternate with decreasing sizes to zero. After n terms, meaning up to k=n, the error bound is 4/(n+3)^2, the magnitude of the term at k=n+1. This bound is valid as the remainder is an alternating series itself, with absolute value less than its first term due to monotonic decrease. A tempting distractor like 4/(n+2)^2 fails by using the last included term instead of the omitted one. Always identify the general term and ensure the error bound is the absolute value of the (n+1)th term for transferable strategy across alternating series.

Question 4

If $\sum_{n=1}^{\infty}(-1)^{n}\frac{5}{\sqrt{n}}$ is approximated by $S_{25}$ , what is the alternating-series error bound?

$\le \frac{5}{\sqrt{26}}$ (correct answer)
$\le \frac{5}{\sqrt{25}}$
$\le \frac{5}{26}$
$\le \frac{5}{\sqrt{24}}$
$\le \frac{5}{2\sqrt{26}}$

Explanation: This problem asks for the alternating series error bound when approximating $\sum_{n=1}^{\infty}(-1)^{n}\frac{5}{\sqrt{n}}$ by $S_{25}$ . The Alternating Series Error Bound theorem tells us that when we truncate a convergent alternating series at the nth term, the error is at most the absolute value of the (n+1)th term. Since we're using $S_{25}$ (the sum of the first 25 terms), the error is bounded by the 26th term: $|a_{26}| = \frac{5}{\sqrt{26}}$ . Choice B ( $\le \frac{5}{\sqrt{25}}$ ) incorrectly uses the last included term rather than the first omitted term, a common mistake. Remember that the error bound for alternating series is always the absolute value of the first term you leave out, not the last term you include.

Question 5

For $S=\sum_{k=2}^{\infty}(-1)^k\frac{3}{k\ln k}$ , what is the maximum error when using $S_5$ ?

$\left|S-S_5\right|\le \frac{3}{6\ln 6}$ (correct answer)
$\left|S-S_5\right|\le \frac{3}{5\ln 5}$
$\left|S-S_5\right|\le \frac{3}{7\ln 7}$
$\left|S-S_5\right|\le \frac{3}{6(\ln 5)}$
$\left|S-S_5\right|\le \frac{3}{(6\ln 6)^2}$

Explanation: This problem involves finding the alternating series error bound for a series with a more complex general term. For $S=\sum_{k=2}^{\infty}(-1)^k\frac{3}{k\ln k}$ , when approximating by $S_5$ , we include terms for $k=2,3,4,5$ . The error bound equals the absolute value of the first omitted term, which is at $k=6$ : $|(-1)^6\frac{3}{6\ln 6}| = \frac{3}{6\ln 6}$ . Choice B incorrectly uses $k=5$ (the last included term) rather than $k=6$ (the first excluded term). When a series starts at an index other than 0 or 1, carefully track which term is the first one you're not including in your partial sum.

Question 6

Using $S_6$ for $\sum_{k=1}^{\infty}(-1)^{k}\frac{2}{k+1}$ , what alternating-series error bound applies?

$\le \dfrac{2}{7}$
$\le \dfrac{2}{8}$ (correct answer)
$\le \dfrac{2}{9}$
$\le \dfrac{1}{7}$
$\le \dfrac{1}{2}\cdot\dfrac{2}{8}$

Explanation: The skill being tested here is the alternating series error bound, which provides a way to estimate the error when approximating an infinite alternating series with a partial sum. For an alternating series that satisfies the conditions of the alternating series test—terms alternating in sign, decreasing in absolute value, and approaching zero—the error in using the partial sum S_n is less than the absolute value of the next term, a_{n+1}. This bound works because the remainder after n terms is bracketed between zero and the first omitted term, ensuring the actual sum lies within S_n ± |a_{n+1}|. In this case, for the series ∑ (-1)^k * 2/(k+1) approximated by S_6, the next term is 2/8, so the error is at most 2/8. A tempting distractor like ≤ 2/7 fails because it uses the last included term instead of the next one, which underestimates the bound since the error is actually smaller than the next term, not the previous. A transferable strategy for error bounds in alternating series is to always identify the first omitted term after verifying the series meets the convergence criteria.

Question 7

For $S=\sum_{k=1}^{\infty}(-1)^{k+1}\frac{1}{k^3}$ , what is the maximum error using the first $12$ terms?

$\le \dfrac{1}{12^3}$
$\le \dfrac{1}{13^3}$ (correct answer)
$\le \dfrac{1}{14^3}$
$\le \dfrac{1}{2\cdot 13^3}$
$\le \dfrac{1}{13^3-12^3}$

Explanation: This problem tests the alternating series error bound, which states that the error when truncating an alternating series is at most the absolute value of the first omitted term. Since we're using the first 12 terms of $\sum_{k=1}^{\infty}(-1)^{k+1}\frac{1}{k^3}$ , the first omitted term is the 13th term, which has $k=13$ . The absolute value of this term is $\left|(-1)^{13+1}\frac{1}{13^3}\right| = \frac{1}{13^3}$ . Choice E, $\frac{1}{13^3-12^3}$ , incorrectly attempts to use the difference between consecutive denominators rather than the next term itself. For alternating series satisfying the conditions (decreasing terms approaching zero), always use the absolute value of the first omitted term as your error bound.

Question 8

Approximating $\sum_{k=1}^{\infty}(-1)^{k-1}\frac{3}{k^2+1}$ by $S_8$ , what is the maximum error?

$\le \dfrac{3}{8^2+1}$
$\le \dfrac{3}{9^2+1}$ (correct answer)
$\le \dfrac{3}{10^2+1}$
$\le \displaystyle\sum_{k=9}^{\infty}\dfrac{3}{k^2+1}$
$\le \dfrac{3}{2(9^2+1)}$

Explanation: The skill being tested here is the alternating series error bound, which provides a way to estimate the error when approximating an infinite alternating series with a partial sum. For an alternating series that satisfies the conditions of the alternating series test—terms alternating in sign, decreasing in absolute value, and approaching zero—the error in using the partial sum S_n is less than the absolute value of the next term, a_{n+1}. This bound works because the remainder after n terms is bracketed between zero and the first omitted term, ensuring the actual sum lies within S_n ± |a_{n+1}|. In this case, for the series ∑ (-1)^{k-1} * 3/(k^2+1) approximated by S_8, the next term is 3/(9^2+1), so the error is at most 3/(9^2+1). A tempting distractor like ≤ 3/(8^2+1) fails because it uses the last included term instead of the next one, which underestimates the bound since the error is actually smaller than the next term, not the previous. A transferable strategy for error bounds in alternating series is to always identify the first omitted term after verifying the series meets the convergence criteria.

Question 9

Let $S=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{6}{\ln(n+1)}$ . What is the error bound after $9$ terms?

$\le \dfrac{6}{\ln(10)}$
$\le \dfrac{6}{\ln(11)}$ (correct answer)
$\le \dfrac{6}{\ln(12)}$
$\le \dfrac{3}{\ln(11)}$
$\le \left|\dfrac{6}{\ln(11)}-\dfrac{6}{\ln(10)}\right|$

Explanation: This question applies the alternating series error bound to $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{6}{\ln(n+1)}$ after 9 terms. Using terms from $n=1$ to $n=9$ , the first omitted term is at $n=10$ . The absolute value of this term is $\left|(-1)^{10-1}\frac{6}{\ln(10+1)}\right| = \frac{6}{\ln(11)}$ . Choice E, $\left|\frac{6}{\ln(11)}-\frac{6}{\ln(10)}\right|$ , represents a common error of using the difference between consecutive terms instead of the next term itself. For alternating series satisfying the required conditions, the error is always bounded by the absolute value of the first omitted term.

Question 10

A constant is approximated by $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{8}{3n-1}$ ; what is the error bound after $25$ terms?

$\le \dfrac{8}{3\cdot 25-1}$
$\le \dfrac{8}{3\cdot 26-1}$ (correct answer)
$\le \dfrac{8}{3\cdot 27-1}$
$\le \dfrac{4}{3\cdot 26-1}$
$\le \dfrac{8}{(3\cdot 26-1)-(3\cdot 25-1)}$

Explanation: This question tests the alternating series error bound for $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{8}{3n-1}$ after 25 terms. Using terms from $n=1$ to $n=25$ , the first omitted term occurs at $n=26$ . The absolute value of this term is $\left|(-1)^{26-1}\frac{8}{3(26)-1}\right| = \frac{8}{77}$ , which equals $\frac{8}{3\cdot 26-1}$ . Choice E, $\frac{8}{(3\cdot 26-1)-(3\cdot 25-1)}$ , incorrectly uses the difference between consecutive denominators, which would give $\frac{8}{3} = \frac{8}{77-74}$ , a much larger value. Always remember that the alternating series error bound is simply the absolute value of the next term in the series.

Question 11

Approximate $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{\sqrt{n}}$ by $S_{49}$ ; what is the error bound?

$\dfrac{1}{\sqrt{49}}$
$\dfrac{1}{\sqrt{50}}$ (correct answer)
$\dfrac{1}{\sqrt{51}}$
$\dfrac{1}{50}$
$\dfrac{1}{2\sqrt{50}}$

Explanation: This problem involves finding the alternating series error bound for a series with square root terms. For convergent alternating series where terms decrease in absolute value, the error from using n terms is bounded by the (n+1)st term's absolute value. Since we use $S_{49}$ (sum of first 49 terms), the error is bounded by the 50th term. The 50th term is $(-1)^{50-1} \cdot \frac{1}{\sqrt{50}} = -\frac{1}{\sqrt{50}}$ , so the error bound is $\frac{1}{\sqrt{50}}$ . Students might choose $\frac{1}{\sqrt{49}}$ thinking it's the last included term, but error bounds use the first excluded term. Remember: alternating series error bound = $|\text{first omitted term}|$ .

Question 12

If $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{6}{(3n+2)}$ is approximated by $S_{18}$ , what is the error bound?

$\le \frac{6}{3\cdot 18+2}$
$\le \frac{6}{3\cdot 19+2}$ (correct answer)
$\le \frac{6}{3\cdot 17+2}$
$\le \frac{6}{3\cdot 18+3}$
$\le \frac{6}{(3\cdot 19+2)^2}$

Explanation: This problem asks for the alternating series error bound when approximating $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{6}{3n+2}$ by $S_{18}$ . The Alternating Series Error Bound states that the error in truncating a convergent alternating series at the nth term is bounded by the absolute value of the (n+1)th term. Using $S_{18}$ means we include terms through $n=18$ , so the error is bounded by the 19th term: $|a_{19}| = \frac{6}{3(19)+2} = \frac{6}{57+2} = \frac{6}{59} = \frac{6}{3 \cdot 19 + 2}$ . Choice A ( $\le \frac{6}{3 \cdot 18+2}$ ) represents the last included term rather than the first omitted one, a common conceptual error. Remember that the alternating series error bound always uses the first term you don't include in your approximation.

Question 13

A value is modeled by $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{4}{3^n}$ ; what is the maximum error using $S_8$ ?

$\dfrac{4}{3^8}$
$\dfrac{4}{3^9}$ (correct answer)
$\dfrac{4}{3^{10}}$
$\dfrac{4}{2\cdot 3^9}$
$\dfrac{4}{9}$

Explanation: This question involves finding alternating series error bounds for geometric-like series. When approximating a convergent alternating series with decreasing terms using n terms, the error is bounded by the $(n+1)$ st term's absolute value. Using $S_8$ means summing the first 8 terms, so the error bound is the 9th term. The 9th term is $(-1)^{9+1} \cdot \frac{4}{3^9} = \frac{4}{3^9}$ . A common error is using $\frac{4}{3^8}$ (the 8th term) instead of the 9th term. Remember that alternating series error bounds always use the first omitted term, not the last included term.

Question 14

A constant is approximated by $\sum_{k=1}^{\infty}(-1)^{k-1}\frac{1}{k(k+1)}$ ; what is the maximum error after $n$ terms?

$\dfrac{1}{n(n+1)}$
$\dfrac{1}{(n+1)(n+2)}$ (correct answer)
$\dfrac{1}{(n+1)^2}$
$\dfrac{1}{n+1}$
$\dfrac{1}{(n+2)^2}$

Explanation: The skill being tested here is the alternating series error bound, which provides an upper limit on the approximation error for convergent alternating series. The series $\sum_{k=1}^{\infty} (-1)^{k-1} \frac{1}{k(k+1)}$ has alternating signs and decreasing magnitudes to zero. After n terms, the error is bounded by $1/((n+1)(n+2))$ , the (n+1)th term's magnitude. This occurs because the tail series alternates with decreasing terms, bracketing the true sum within less than that first tail term. A tempting distractor like $1/(n(n+1))$ might appeal if one uses the nth term, but it incorrectly applies the bound to the included term. Always identify the general term and ensure the error bound is the absolute value of the (n+1)th term for transferable strategy across alternating series.

Question 15

For $S=\sum_{k=1}^{\infty}(-1)^{k+1}\frac{1}{k^3}$ , what is the maximum error when approximating $S$ by $\sum_{k=1}^{n}(-1)^{k+1}\frac{1}{k^3}$ ?

$\dfrac{1}{n^3}$
$\dfrac{1}{(n+1)^3}$ (correct answer)
$\dfrac{1}{(n+1)^2}$
$\dfrac{1}{n+1}$
$\dfrac{1}{2(n+1)^3}$

Explanation: The skill being tested here is the alternating series error bound, which provides an upper limit on the approximation error for convergent alternating series. For the series S = ∑_{k=1}^∞ (-1)^{k+1} (1/k^3), the terms are alternating in sign, positive magnitudes decreasing to zero, satisfying the conditions for the alternating series test. When approximating S by the partial sum up to k=n, the remainder is less than the magnitude of the first omitted term, which is 1/(n+1)^3. This bound arises because the remainder is an alternating series itself starting with that term, and its absolute value is less than the first term due to the decreasing nature. A tempting distractor like 1/n^3 might seem correct if one mistakenly uses the last included term instead of the next one, but that would overestimate the error since the bound is strictly the next term. Always identify the general term and ensure the error bound is the absolute value of the (n+1)th term for transferable strategy across alternating series.

Question 16

Let $S=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{5}{k^4}$ . What is the alternating-series error bound after summing through $k=n$ ?

$\dfrac{5}{(n+1)^4}$ (correct answer)
$\dfrac{5}{n^4}$
$\dfrac{5}{(n+1)^3}$
$\dfrac{5}{n+1}$
$\dfrac{5}{2(n+1)^4}$

Explanation: The skill being tested here is the alternating series error bound, which provides an upper limit on the approximation error for convergent alternating series. In the series S = ∑_{k=1}^∞ (-1)^{k-1} (5/k^4), the signs alternate, and the positive terms decrease monotonically to zero. The partial sum through k=n has an error bounded by the magnitude of the (n+1)th term, which is 5/(n+1)^4. This bound holds as the remainder forms its own alternating series with decreasing terms, ensuring the total error is less than the leading term of the remainder. A tempting distractor such as 5/n^4 might be chosen if confusing the bound with the last term included, but it incorrectly uses a larger value than necessary. Always identify the general term and ensure the error bound is the absolute value of the (n+1)th term for transferable strategy across alternating series.

Question 17

For $\sum_{k=1}^{\infty}(-1)^{k+1}\frac{3}{\sqrt{k}}$ , what is the error bound when using the partial sum through $k=n$ ?

$\dfrac{3}{\sqrt{n}}$
$\dfrac{3}{\sqrt{n+1}}$ (correct answer)
$\dfrac{3}{n+1}$
$\dfrac{3}{(n+1)^{3/2}}$
$\dfrac{3}{2\sqrt{n+1}}$

Explanation: The skill being tested here is the alternating series error bound, which provides an upper limit on the approximation error for convergent alternating series. For ∑_{k=1}^∞ (-1)^{k+1} (3/√k), the series alternates with decreasing positive parts approaching zero. Using the partial sum through k=n, the error is less than 3/√(n+1), the size of the next term. The reason is that the remainder alternates and nests within an interval smaller than that term due to the decrease. A tempting distractor like 3/√n fails by mistakenly bounding with the current term, which is larger and not the proper bound. Always identify the general term and ensure the error bound is the absolute value of the (n+1)th term for transferable strategy across alternating series.

Question 18

A sum is defined by $\sum_{k=1}^{\infty}(-1)^{k+1}\frac{1}{k^{1.5}}$ ; what is the error bound after $n$ terms?

$\dfrac{1}{(n+1)^{1.5}}$ (correct answer)
$\dfrac{1}{n^{1.5}}$
$\dfrac{1}{(n+1)^{2.5}}$
$\dfrac{1}{n+1}$
$\dfrac{1}{2(n+1)^{1.5}}$

Explanation: The skill being tested here is the alternating series error bound, which provides an upper limit on the approximation error for convergent alternating series. The sum $\sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k^{1.5}}$ alternates with terms decreasing to zero. After n terms, the error bound is $1/(n+1)^{1.5}$ , the (n+1)th term. This holds because the remainder is alternating and decreasing, ensuring error less than the first tail term. A tempting distractor like $1/n^{1.5}$ might tempt if one bounds with the nth term, but that's not the theorem's application. Always identify the general term and ensure the error bound is the absolute value of the (n+1)th term for transferable strategy across alternating series.

Question 19

For $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n\ln(n+1)}$ , what is the error bound using $S_{20}$ ?

$\le \frac{1}{20\ln(21)}$
$\le \frac{1}{21\ln(22)}$ (correct answer)
$\le \frac{1}{21\ln(21)}$
$\le \frac{1}{20\ln(22)}$
$\le \frac{1}{\ln(22)}$

Explanation: This problem involves finding the alternating series error bound for $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n\ln(n+1)}$ when using $S_{20}$ . The Alternating Series Error Bound guarantees that when truncating a convergent alternating series at the nth term, the error is at most the absolute value of the (n+1)th term. Using $S_{20}$ means including terms through $n=20$ , so the error is bounded by the 21st term: $|a_{21}| = \frac{1}{21\ln(21+1)} = \frac{1}{21\ln(22)}$ . Choice C ( $\le \frac{1}{21\ln(21)}$ ) incorrectly evaluates the logarithm at 21 instead of 22, missing the shift in the original formula. When the general term involves $n+1$ , carefully substitute the correct index value to find your error bound.

Question 20

Using $S_{15}$ to approximate $\sum_{k=1}^{\infty}(-1)^{k+1}\frac{4}{2k-1}$ , what is the maximum error?

$\le \frac{4}{29}$
$\le \frac{4}{31}$ (correct answer)
$\le \frac{4}{30}$
$\le \frac{2}{31}$
$\le \frac{4}{(31)^2}$

Explanation: This question asks for the alternating series error bound when approximating $\sum_{k=1}^{\infty}(-1)^{k+1}\frac{4}{2k-1}$ by $S_{15}$ . The Alternating Series Error Bound theorem states that for convergent alternating series satisfying the necessary conditions, the error using partial sum $S_n$ is bounded by $|a_{n+1}|$ . With $S_{15}$ , we sum through $k=15$ , so the error is bounded by the 16th term: $|a_{16}| = \frac{4}{2(16)-1} = \frac{4}{31}$ . Choice A ( $\le \frac{4}{29}$ ) uses $k=15$ instead of $k=16$ , forgetting that the error bound comes from the first omitted term, not the last included one. Always remember to use the next term after your partial sum cutoff for the alternating series error bound.

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