Alternating Series Error Bound - AP Calculus BC

For $S=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{4}{n^5}$ , what is the error bound using the first $6$ terms?
Answer: $\le \dfrac{4}{7^5}$ This question involves the alternating series error bound for $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{4}{n^5}$ using 6 terms. After including terms from $n=1$ to $n=6$ , the first omitted term is at $n=7$ . The absolute value of this term is $\left|(-1)^{7+1}\frac{4}{7^5}\right| = \frac{4}{7^5}$ . Choice D, $\frac{2}{7^5}$ , incorrectly halves the numerator, while choice E, $\frac{4}{7^5-6^5}$ , mistakenly uses the difference of consecutive denominators. For alternating series error bounds, always use the complete first omitted term without modification.
Let $S=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{6}{\ln(n+1)}$ . What is the error bound after $9$ terms?
Answer: $\le \dfrac{6}{\ln(11)}$ 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.
Approximating $\sum_{k=1}^{\infty}(-1)^{k-1}\frac{3}{k^2+1}$ by $S_5$ , what is the error bound?
Answer: $\le \frac{3}{6^2+1}$ This problem tests understanding of the alternating series error bound for rational function terms. The series $\sum_{k=1}^{\infty}(-1)^{k-1}\frac{3}{k^2+1}$ alternates and has decreasing terms approaching zero. When we approximate with $S_5$ , we include terms for $k=1$ through $k=5$ , making the first omitted term at $k=6$ . The alternating series theorem guarantees the error is at most $|(-1)^{6-1}\frac{3}{6^2+1}| = \frac{3}{37}$ . Choice A uses $k=5$ (the last included term) instead of $k=6$ , a common mistake. The key insight is that the error bound always uses the index of the first term you don't include in your approximation.
For $S=\sum_{k=2}^{\infty}(-1)^k\frac{3}{k\ln k}$ , what is the maximum error when using $S_5$ ?
Answer: $\left|S-S_5\right|\le \frac{3}{6\ln 6}$ 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.
For $S=\sum_{k=1}^{\infty}(-1)^k\frac{2k}{3^k}$ , what is the alternating-series error bound after 4 terms?
Answer: $\left|S-S_4\right|\le \frac{10}{3^5}$ This problem tests the alternating series error bound for a series combining polynomial and exponential terms. For $S=\sum_{k=1}^{\infty}(-1)^k\frac{2k}{3^k}$ , when we approximate by $S_4$ , we include terms for $k=1,2,3,4$ . The error bound is the absolute value of the 5th term: $|(-1)^5\frac{2(5)}{3^5}| = \frac{10}{3^5} = \frac{10}{243}$ . Choice A incorrectly uses $k=4$ to get $\frac{8}{3^4}$ , confusing the last included term with the first excluded term. The key to alternating series error bounds is recognizing that the error is bounded by the magnitude of the next term you would add, not the last term you did add.
Let $S=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n(n+1)}$ . What is the maximum error when approximating $S$ by $S_{12}$ ?
Answer: $\left|S-S_{12}\right|\le \frac{1}{13\cdot 14}$ This question applies the alternating series error bound to a series with a product in the denominator. For $S=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n(n+1)}$ , the partial sum $S_{12}$ includes terms through $n=12$ . The error bound equals the absolute value of the 13th term: $|(-1)^{13-1}\frac{1}{13(13+1)}| = \frac{1}{13 \cdot 14}$ . Choice A incorrectly uses $n=12$ to get $\frac{1}{12 \cdot 13}$ , which represents the last term included rather than the first term excluded. The fundamental strategy for alternating series error bounds is to always evaluate your general term at one index beyond your stopping point.
For $\sum_{k=1}^{\infty}(-1)^{k+1}\frac{3}{\sqrt{k}}$ , what is the error bound when using the partial sum through $k=n$ ?
Answer: $\dfrac{3}{\sqrt{n+1}}$ 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.
Using $S_5$ to approximate $\sum_{k=1}^{\infty}(-1)^{k-1}\frac{7}{3^k}$ , what is the error bound?
Answer: $\le \dfrac{7}{3^6}$ 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 ∑ $(-1)^{k-1}$ * 7/ $3^k$ approximated by $S_5$ , the next term is 7/ $3^6$ , so the error is at most 7/ $3^6$ . A tempting distractor like ≤ 7/ $3^5$ 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.
Approximating $\sum_{n=2}^{\infty}(-1)^n\frac{2}{n\ln n}$ by the first 5 terms, what is the error bound?
Answer: $\le \frac{2}{7\ln 7}$ This problem involves the alternating series error bound for a series starting at $n=2$ . The series $\sum_{n=2}^{\infty}(-1)^n\frac{2}{n\ln n}$ alternates with decreasing terms. Using the first 5 terms means we include $n=2, 3, 4, 5, 6$ (since we start at $n=2$ ), so the first omitted term is at $n=7$ . The alternating series error bound gives us $|(-1)^7\frac{2}{7\ln 7}| = \frac{2}{7\ln 7}$ . Choice A would be incorrect as it uses $n=6$ , the last included term. When a series doesn't start at $n=1$ , carefully count which terms you're including to identify the first omitted term correctly.
For $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n(n+1)}$ , what is the maximum error when using $S_{15}$ ?
Answer: $\dfrac{1}{16\cdot 17}$ This problem asks for the alternating series error bound with a product formula in denominators. For convergent alternating series with decreasing terms, using n terms gives an error bounded by the (n+1)st term's absolute value. S₁₅ uses the first 15 terms, so the error bound is the 16th term. The 16th term is $(-1)^{16-1}$ · 1/(16·17) = -1/(16·17), giving error bound 1/(16·17). Students might choose 1/(15·16) thinking it's the 15th term, but that's the last included term, not the first excluded one. Always use the first omitted term for alternating series error bounds.
Using $S_6$ for $\sum_{k=1}^{\infty}(-1)^{k}\frac{2}{k+1}$ , what alternating-series error bound applies?
Answer: $\le \dfrac{2}{8}$ 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.
For $\sum_{k=1}^{\infty}(-1)^{k+1}\frac{1}{k^3}$ , what is the maximum error when approximating the sum by $S_{12}$ ?
Answer: $\le \frac{1}{13^3}$ This problem requires finding the alternating series error bound for the series $\sum_{k=1}^{\infty}(-1)^{k+1}\frac{1}{k^3}$ when approximated by $S_{12}$ . The Alternating Series Error Bound theorem states that for a convergent alternating series satisfying the conditions (terms decrease in absolute value and approach zero), the error when using partial sum $S_n$ is bounded by the absolute value of the next term, $|a_{n+1}|$ . Since we're using $S_{12}$ , the error is bounded by the 13th term: $|a_{13}| = \frac{1}{13^3}$ . Choice C ( $\le \frac{1}{12^4}$ ) might tempt students who confuse the index or exponent, but the correct bound uses the next term's index with the same exponent as in the series. When applying the alternating series error bound, always identify the first omitted term and use its absolute value as your bound.
For $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{4}{n(n+1)}$ , what is the maximum error using $S_{10}$ ?
Answer: $\le \frac{4}{11\cdot 12}$ This question involves applying the alternating series error bound to a series with factorial-like denominators. The series $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{4}{n(n+1)}$ alternates with decreasing terms that approach zero. Using $S_{10}$ means we sum from $n=1$ to $n=10$ , so the first omitted term occurs at $n=11$ . By the alternating series theorem, the error is bounded by $|(-1)^{11+1}\frac{4}{11(11+1)}| = \frac{4}{11 \cdot 12}$ . Choice A incorrectly uses $n=10$ (the last included term) rather than $n=11$ . Remember that the error bound is determined by the magnitude of the next term in the series, not the last term you included.
Using $S_{10}$ for $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{5}{n^4}$ , what is the alternating-series error bound?
Answer: $\dfrac{5}{11^4}$ This problem tests alternating series error bounds with fourth power denominators. For convergent alternating series where terms decrease, the error from a partial sum $S_n$ is bounded by the absolute value of term (n+1). Using S₁₀ means we sum the first 10 terms, so the error bound is the 11th term's absolute value. The 11th term is $(-1)^{11-1}$ · 5/11⁴ = 5/11⁴. Students often mistakenly use 5/10⁴, thinking the error relates to the last included term rather than the first excluded one. For alternating series, always remember: error bound = |first term not in your partial sum|.
Approximate $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{\sqrt{n}}$ by $S_{49}$ ; what is the error bound?
Answer: $\dfrac{1}{\sqrt{50}}$ 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 = $|\text{first omitted term}|$ .
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?
Answer: $\dfrac{1}{(n+1)(n+2)}$ 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.
For $\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n!}$ , what is the maximum error when using the first $5$ terms?
Answer: $\le \dfrac{1}{6!}$ This problem involves finding the alternating series error bound for $\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n!}$ using 5 terms. After including terms from $n=1$ to $n=5$ , the first omitted term is at $n=6$ . The absolute value of this term is $\left|(-1)^{6}\frac{1}{6!}\right| = \frac{1}{6!}$ . Choice E, $\frac{1}{6!-5!}$ , incorrectly suggests using the difference between consecutive factorial terms, which would give a much larger error bound than the actual bound. The alternating series error bound theorem guarantees that the error is at most the absolute value of the first term not included in your approximation.
Approximating $\sum_{k=1}^{\infty}(-1)^{k}\frac{2}{\ln(k+1)}$ by $S_{20}$ , what is the error bound?
Answer: $\le \dfrac{2}{\ln(22)}$ 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/ln(k+1) approximated by $S_{20}$ , the next term is 2/ln(22), so the error is at most 2/ln(22). A tempting distractor like ≤ 2/ln(21) 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.
Using $S_{12}$ for $\sum_{k=1}^{\infty}(-1)^{k+1}\frac{4}{k^4}$ , what is the alternating-series error bound?
Answer: $\le \dfrac{4}{13^4}$ 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 ∑ $(-1)^{k+1}$ * 4/ $k^4$ approximated by $S_{12}$ , the next term is 4/ $13^4$ , so the error is at most 4/ $13^4$ . A tempting distractor like ≤ 4/ $12^4$ 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.
Let $S=\sum_{n=0}^{\infty}(-1)^n\frac{2}{2n+1}$ . What is the alternating-series error bound using $S_4$ ?
Answer: $\left|S-S_4\right|\le \frac{2}{11}$ This question requires applying the alternating series error bound to find the maximum error when approximating the series by its 4th partial sum. For $S=\sum_{n=0}^{\infty}(-1)^n\frac{2}{2n+1}$ , the partial sum $S_4$ includes terms for $n=0,1,2,3,4$ (that's 5 terms total since we start at $n=0$ ). The error bound is the absolute value of the next term, which occurs at $n=5$ : $|(-1)^5\frac{2}{2(5)+1}| = \frac{2}{11}$ . Choice B incorrectly calculates using $n=3$ instead of $n=5$ , forgetting that $S_4$ means we include the term where $n=4$ . Remember that for series starting at $n=0$ , $S_k$ includes $k+1$ terms, so the error bound uses $n=k+1$ .

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