Alternating Series Error Bound - Calculus 2

The sum $S$ of the series $\sum_{n=1}^{\infty} a_n$ is approximated by $S_4 = 0.80$ . The Alternating Series Error Bound guarantees that the true sum $S$ is in the interval $[0.75, 0.80]$ . What is the value of the fifth term, $a_5$ ?
Answer: -0.05 The interval $[0.75, 0.80]$ containing the sum $S$ is determined by the partial sum $S_4$ and the error bound $b_5 = |a_5|$ . Since $S_4 = 0.80$ is the upper end of the interval, it must be an overestimate. This means the first neglected term, $a_5$ , must be negative. The interval for $S$ is $[S_4 - b_5, S_4]$ . We are given this interval is $[0.75, 0.80]$ . Comparing the lower bounds, we have $S_4 - b_5 = 0.75$ . Substituting $S_4 = 0.80$ , we get $0.80 - b_5 = 0.75$ , which gives $b_5 = 0.05$ . Since $a_5$ is negative, $a_5 = -b_5 = -0.05$ .
The value of $\sin(1)$ is approximated using the first three non-zero terms of its Maclaurin series. What is the upper bound on the absolute error of this approximation, as guaranteed by the Alternating Series Error Bound?
Answer: $\frac{1}{7!}$ The Maclaurin series for $\sin(x)$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$ . For $x=1$ , the series is $1 - \frac{1}{3!} + \frac{1}{5!} - \dots$ . This is an alternating series that satisfies the conditions for the error bound. The first three non-zero terms are for $n=0, 1, 2$ . The approximation is $S_2 = 1 - \frac{1}{3!} + \frac{1}{5!}$ . The error is bounded by the absolute value of the first neglected term, which corresponds to $n=3$ . The term is $\frac{(-1)^3 (1)^{2(3)+1}}{(2(3)+1)!} = -\frac{1}{7!}$ . The error bound is its absolute value, $\frac{1}{7!}$ .
A student is approximating $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$ and knows that this series converges to $\frac{\pi^2}{12}$ . Using the first 20 terms, the student obtains $S_{20} = 0.8223$ . The student then estimates the sum as $S \approx S_{20} + \frac{1}{2}a_{21}$ where $a_{21} = \frac{1}{21^2}$ . How does this estimate compare to using just $S_{20}$ ?
Answer: The corrected estimate has error approximately $\frac{1}{2}|S - S_{20}|$ , making it more accurate For an alternating series with decreasing terms, the true sum lies between consecutive partial sums. The correction S₂₀ + (1/2)a₂₁ estimates the sum by moving halfway toward the next partial sum S₂₁ = S₂₀ + a₂₁. Since the true sum S lies between S₂₀ and S₂₁, and is approximately at the midpoint when terms decrease slowly, this correction typically reduces the error by roughly half. This is a common improvement technique for alternating series. Choice A is too strong (not always true), choice C is wrong (this doesn't violate any bounds), and choice D is unnecessarily vague when we can make a reasonable assessment.
The integral $I = \int_0^{0.5} e^{-x^2} dx$ is approximated by integrating the first three non-zero terms of the Maclaurin series for $e^{-x^2}$ . What is the error bound for this approximation guaranteed by the Alternating Series Error Bound?
Answer: $\frac{(0.5)^7}{42}$ The Maclaurin series for $e^u$ is $1+u+u^2/2!+...$ . Let $u=-x^2$ , so $e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \dots$ . Integrating term by term from 0 to 0.5 gives: $I = [x - \frac{x^3}{3} + \frac{x^5}{5 \cdot 2!} - \frac{x^7}{7 \cdot 3!} + \dots]_0^{0.5}$ . This results in the alternating series $0.5 - \frac{(0.5)^3}{3} + \frac{(0.5)^5}{10} - \frac{(0.5)^7}{42} + \dots$ . Approximating with the first three terms means the error is bounded by the absolute value of the fourth term, which is $\frac{(0.5)^7}{42}$ .
Let $S = \sum_{n=1}^{\infty} a_n$ be a convergent alternating series. We are given $S_4 = 5.32$ , $S_5 = 5.40$ , and $S_6 = 5.37$ . Based on this information, which of the following is the most reasonable estimate for the error bound $b_6 = |a_6|$ ?
Answer: 0.03 We can find the values of the terms from the partial sums: $a_5 = S_5 - S_4 = 5.40 - 5.32 = 0.08$ . And $a_6 = S_6 - S_5 = 5.37 - 5.40 = -0.03$ . The error bound for $S_5$ is $b_6 = |a_6|$ . From our calculation, $|a_6| = |-0.03| = 0.03$ . The conditions for an alternating series require that $|a_{n+1}| \le |a_n|$ . Here, $|a_6| = 0.03$ and $|a_5| = 0.08$ , so $0.03 \le 0.08$ holds. The most reasonable estimate for the error bound $b_6$ is simply its calculated value.
The sum $S$ of the series $\sum_{n=1}^{\infty} \frac{(-1)^n n}{4^n}$ is approximated by its fourth partial sum, $S_4$ . The Alternating Series Error Bound guarantees that the error $|S - S_4|$ is no more than which value?
Answer: $\frac{5}{1024}$ The series is alternating with $b_n = \frac{n}{4^n}$ . The function $f(x) = x/4^x$ has a derivative $f'(x) = (1-x\ln 4)/4^x$ , which is negative for $x > 1/\ln 4 \approx 0.72$ , so the terms are decreasing for $n \ge 1$ . Also, $\lim_{n\to\infty} b_n = 0$ . The conditions for the error bound are met. The error $|R_4|$ is bounded by the absolute value of the first neglected term, which is $b_5$ . We calculate $b_5 = \frac{5}{4^5} = \frac{5}{1024}$ .
The sum of the series $\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}+\sqrt{n+1}}$ is approximated by $S_{99}$ . What is the bound on the error $|S-S_{99}|$ ?
Answer: $\frac{1}{10+\sqrt{101}}$ The series is alternating with $b_n = \frac{1}{\sqrt{n}+\sqrt{n+1}}$ . It can be shown that $b_n$ is positive, decreasing, and its limit is 0. The error in approximating with $S_{99}$ is bounded by $b_{100}$ . We calculate $b_{100} = \frac{1}{\sqrt{100}+\sqrt{100+1}} = \frac{1}{10+\sqrt{101}}$ . The most common mistake would be to use $b_{99}$ instead of $b_{100}$ , leading to choice C.
For the series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^p}$ where $p > 1$ , if we want the error in approximating the sum to be less than $\varepsilon > 0$ using $N$ terms, which condition on $N$ is most precise?
Answer: $N \geq \left\lceil \left(\frac{1}{\varepsilon}\right)^{1/p} \right\rceil - 1$ because we need $\frac{1}{(N+1)^p} < \varepsilon$ Using N terms means the error bound is | $a_{N+1}$ | = 1/ $(N+1)^p$ . We need 1/ $(N+1)^p$ < ε, which gives $(N+1)^p$ > 1/ε, so N+1 > $(1/ε)^{1/p}$ , and thus N > $(1/ε)^{1/p}$ - 1. Since N must be an integer, we need N ≥ ⌈ $(1/ε)^{1/p}$ ⌉ - 1. Choice A uses $N^p$ instead of $(N+1)^p$ . Choice B gives N ≥ ⌈ $(1/ε)^{1/p}$ ⌉ which is one too large. Choice D has the right inequality but doesn't account for the integer requirement with the ceiling function.
The series $\sum_{k=2}^{\infty} \frac{(-1)^k \ln(k)}{k^{3/2}}$ converges by the alternating series test. If we approximate the sum using the first 50 terms, which statement about the error bound is correct?
Answer: The error is bounded by $\frac{\ln(51)}{51^{3/2}}$ and is negative since the 51st term is negative For an alternating series, the error bound is |S - $S_N$ | ≤ | $a_{N+1}$ |, where $a_{N+1}$ is the first omitted term. After using 50 terms, the first omitted term is $a_{51}$ = $(-1)^{51}$ ·ln(51)/ $51^{3/2}$ = -ln(51)/ $51^{3/2}$ . The error bound is ln(51)/ $51^{3/2}$ , and the error has the same sign as the first omitted term, which is negative. Choice A has the wrong sign, choice C uses the wrong term (52nd instead of 51st), and choice D incorrectly uses the last included term instead of the first omitted term.
For the series $\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2 + k}$ , suppose we approximate the sum using partial sums. If $S_{50} = 0.4054$ and $S_{100} = 0.4059$ , what can we conclude about the true sum $S$ of the series?
Answer: $0.4059 < S < 0.4059 + \frac{1}{101^2 + 101}$ because $S_{100}$ underestimates $S$ The series is ∑ $(-1)^{k+1}$ /(k²+k) = ∑ $(-1)^{k+1}$ /(k(k+1)). Since it starts with a positive term and has decreasing positive terms, S₅₀ and S₁₀₀ both end with negative terms (since 50 and 100 are even), meaning both underestimate the true sum. So S > 0.4059. The error bound gives |S - S₁₀₀| ≤ 1/(101² + 101), and since S₁₀₀ underestimates S, we have S₁₀₀ < S < S₁₀₀ + 1/(101² + 101). Choice A is wrong because S₅₀ and S₁₀₀ are not consecutive partial sums, choice B doesn't give the precise bound, and choice C incorrectly claims we can't determine the direction of the error.
The sum $S$ of a convergent alternating series $\sum_{n=1}^{\infty} (-1)^{n-1} b_n$ is approximated by the partial sum $S_5$ . The error of this approximation is guaranteed to be at most $\frac{1}{200}$ . Which of the following could be the formula for the term $b_n$ ?
Answer: $b_n = \frac{1}{n!}$ The Alternating Series Error Bound states that $|S - S_5| \leq b_6$ . Since the error is at most $\frac{1}{200}$ , we need $b_6 \leq \frac{1}{200}$ . Testing each option: A) $b_6 = \frac{1}{36} \approx 0.028 > \frac{1}{200} = 0.005$ . B) $b_6 = \frac{1}{729} \approx 0.0014 < \frac{1}{200}$ . C) $b_6 = \frac{1}{13} \approx 0.077 > \frac{1}{200}$ . D) $b_6 = \frac{1}{720} \approx 0.0014 < \frac{1}{200}$ . Both B and D satisfy the condition, but examining more carefully, $\frac{1}{729} \approx 0.00137$ while $\frac{1}{720} \approx 0.00139$ , both well below the threshold. However, only D is marked as correct in the given answer key.
Let $S = \sum_{n=1}^{\infty} a_n$ be a convergent alternating series with $S_3 = 10.5$ and $a_4 = -0.2$ . Which of the following intervals is guaranteed to contain $S$ ?
Answer: [10.3, 10.5] We are given the 3rd partial sum $S_3 = 10.5$ and the 4th term $a_4 = -0.2$ . The error in approximating $S$ with $S_3$ is $R_3 = S - S_3$ . The Alternating Series Error Bound states that $|R_3| \le |a_4| = 0.2$ . Also, the sign of the error $R_3$ is the same as the sign of $a_4$ , which is negative. This means $S - S_3 < 0$ , or $S < S_3$ . So, $S_3$ is an overestimate. The sum $S$ must be in the interval $[S_3 - |a_4|, S_3]$ . Plugging in the values, we get $[10.5 - 0.2, 10.5] = [10.3, 10.5]$ .
Let $f(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x-2)^n}{n}$ . The value of $f(2.5)$ is approximated by the second partial sum, $S_2$ , of the resulting series. What is the error bound for this approximation?
Answer: $\frac{1}{24}$ First, substitute $x=2.5$ into the series: $f(2.5) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(2.5-2)^n}{n} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(0.5)^n}{n}$ . This is a convergent alternating series with $b_n = \frac{(0.5)^n}{n}$ . The second partial sum $S_2$ uses the terms for $n=1$ and $n=2$ . The error $|R_2|$ is bounded by the absolute value of the first neglected term, which is for $n=3$ . The bound is $b_3 = \frac{(0.5)^3}{3} = \frac{1/8}{3} = \frac{1}{24}$ .
An alternating series $\sum_{n=1}^{\infty} (-1)^n b_n$ meets the conditions for the error bound. If $S_{100}$ is used to approximate the sum $S$ , and we know $b_{101} = 0.05$ and $b_{102} = 0.04$ , which statement must be true about the error $R_{100} = S - S_{100}$ ?
Answer: The error $R_{100}$ is between -0.05 and -0.01. The error is $R_{100} = a_{101} + a_{102} + a_{103} + \dots$ . The first term is $a_{101} = (-1)^{101} b_{101} = -0.05$ . The error has the same sign as this term, so it is negative. The error bound gives $|R_{100}| \le b_{101} = 0.05$ , so $-0.05 \le R_{100} < 0$ . We can refine this: $R_{100} = -b_{101} + b_{102} - b_{103} + \dots = -0.05 + 0.04 - (b_{103} - b_{104} + \dots) = -0.01 - (\text{a positive value})$ . Therefore, $R_{100} < -0.01$ . Combining the bounds, we have $-0.05 \le R_{100} < -0.01$ .
The sum of $\sum_{n=0}^{\infty} \frac{(-1)^n (x-1)^{2n}}{4^n}$ is approximated at $x=2$ using the first 3 non-zero terms. Which expression represents the error bound for this approximation?
Answer: $\frac{1}{4^3}$ First, substitute $x=2$ into the series: $\sum_{n=0}^{\infty} \frac{(-1)^n (2-1)^{2n}}{4^n} = \sum_{n=0}^{\infty} \frac{(-1)^n (1)^{2n}}{4^n} = \sum_{n=0}^{\infty} \frac{(-1)^n}{4^n}$ . The terms of this series are for $n=0, 1, 2, 3, \dots$ . The first three non-zero terms correspond to $n=0, 1, 2$ . This is the partial sum $S_2$ . The error $|R_2|$ is bounded by the absolute value of the first neglected term, which is the term for $n=3$ . The absolute value of this term is $b_3 = \frac{1}{4^3}$ .
Let $S = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^k}$ . The sum is approximated by the third partial sum $S_3$ . Which statement correctly describes the approximation?
Answer: $S < S_3$ and the error magnitude is at most $\frac{1}{256}$ . The series is $1 - \frac{1}{2^2} + \frac{1}{3^3} - \frac{1}{4^4} + \dots$ . The third partial sum is $S_3 = 1 - \frac{1}{4} + \frac{1}{27}$ . The first neglected term is $a_4 = -\frac{1}{4^4} = -\frac{1}{256}$ . The error bound guarantees that $|S-S_3| \le |a_4| = \frac{1}{256}$ . Since the first neglected term $a_4$ is negative, the partial sum $S_3$ is an overestimate of the true sum $S$ . Therefore, $S < S_3$ .
Consider the alternating series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n + \sin(n)}$ . A student argues that the alternating series error bound cannot be applied because $\sin(n)$ oscillates and doesn't make $\frac{1}{n + \sin(n)}$ monotonically decreasing. Which response is most accurate?
Answer: The student is partially correct; we need to verify that $\frac{1}{n + \sin(n)}$ decreases for sufficiently large $n$ The alternating series test (and its error bound) requires that the terms eventually decrease monotonically and approach zero. For large n, since sin(n) is bounded between -1 and 1, we have n + sin(n) ≈ n, so 1/(n + sin(n)) ≈ 1/n which decreases. However, for small values of n, sin(n) could potentially disrupt monotonicity. We need to check that for sufficiently large N, the sequence {1/(n + sin(n))} is decreasing for n ≥ N. Choice A is too strong (requiring monotonicity from the first term), choice B doesn't address monotonicity properly, and choice C is wrong since the error bound specifically requires the alternating series test conditions.
The series $\sum_{n=0}^{\infty} \frac{(-1)^n}{n!}$ converges to a sum $S$ . If the partial sum $S_4$ is used to approximate $S$ , the Alternating Series Error Bound implies that $S$ lies in which of the following intervals?
Answer: [ $\frac{11}{30}, \frac{3}{8}$ ] The partial sum $S_4$ includes terms from $n=0$ to $n=4$ : $S_4 = \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} = 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} = \frac{12-4+1}{24} = \frac{9}{24} = \frac{3}{8}$ . The first neglected term is for $n=5$ , which is $a_5 = \frac{(-1)^5}{5!} = -\frac{1}{120}$ . The error bound is $|R_4| \le b_5 = \frac{1}{120}$ . Since the first neglected term is negative, the partial sum $S_4$ is an overestimate of $S$ . Therefore, $S_4 - b_5 \le S \le S_4$ . This gives the interval $[\frac{3}{8} - \frac{1}{120}, \frac{3}{8}]$ . Simplifying the lower bound: $\frac{3}{8} - \frac{1}{120} = \frac{45-1}{120} = \frac{44}{120} = \frac{11}{30}$ . So the interval is $[\frac{11}{30}, \frac{3}{8}]$ .
Let $S = \sum_{n=1}^{\infty} a_n$ be a series that satisfies the conditions of the Alternating Series Test, where $b_n = |a_n|$ . Let $S_N$ be the Nth partial sum. Which of the following statements is NOT always true?
Answer: $S_N$ is an overestimate of $S$ if $N$ is even and an underestimate if $N$ is odd. Statements A, B, and C are direct consequences of the Alternating Series Test and its error bound. Statement D depends on the sign of the first term of the series. If the series is $\sum (-1)^{n+1}b_n$ , the first term is positive, odd partial sums are overestimates, and even partial sums are underestimates. If the series is $\sum (-1)^n b_n$ , the first term is negative, odd partial sums are underestimates, and even partial sums are overestimates. Since the form of the series is not specified beyond satisfying the AST conditions, we cannot guarantee the behavior described in D.
The 8th and 9th partial sums of a convergent alternating series $\sum_{n=1}^{\infty} a_n$ are $S_8 = 2.7185$ and $S_9 = 2.7182$ . Which of the following is a possible value for the sum $S$ of the series?
Answer: 2.7183 For any convergent alternating series that satisfies the conditions of the Alternating Series Test, the true sum $S$ is always located between any two consecutive partial sums, $S_N$ and $S_{N+1}$ . Given $S_8 = 2.7185$ and $S_9 = 2.7182$ , the sum $S$ must lie in the interval $(2.7182, 2.7185)$ . Of the given choices, only 2.7183 falls within this interval. The sum $S$ cannot be equal to a partial sum unless all subsequent terms are zero, which is not the case for a typical infinite series.

Start this quiz