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8. Determine whether the series (-1)" is absolutely convergent, conditionally 7n-5 n=1 convergent, or divergent Please use the attached series tests, to correctly solve the problem. Please indicate what series test you are using. Also, show all your work. Thanks.

8. Determine whether the series (-1)" is absolutely convergent, conditionally 7n-5 n=1 convergent, or divergent Please use the attached series tests, to correctly solve the problem. Please indicate what series test you are using. Also, show all your work. Thanks.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Hello, Please answer the following attached Calculus question correctly and show all your work completely without skipping any steps. Please also use the attached "series tests", to correctly solve the problem. Thanks.

8. Determine whether the series (-1)" is absolutely convergent, conditionally 7n-5 n=1 convergent, or divergent Please use the attached series tests, to correctly solve the problem. Please indicate what series test you are using. Also, show all your work. Thanks.
Transcribed Image Text:8. Determine whether the series (-1)" is absolutely convergent, conditionally 7n-5 n=1 convergent, or divergent Please use the attached series tests, to correctly solve the problem. Please indicate what series test you are using. Also, show all your work. Thanks.
Series Test When we determine whether an infinite series is convergent or divergent, we only need to consider the tail (for n 2 M). We don't need to worry about the head (for 1 ≤ n ≤ M-1). If needed, we can rewrite the series as Σa, = a + Σa,,, and the finite series is always convergent. M-1 00 *=1 #-M Condition Conclusion Note Test DIVERGENCE TEST If lim a,, #0 or then the series Σa, is divergent. This test does NOT give the convergence. 11-00 If lim a,, = 0, then the series Ea, is may or may not conv. lima = DNE 21-00 Assume f(n) = an, and INTEGRAL TEST (for series with positive terms) If f(x)dx is convergent (divergent), then Use the integral test if the integral is easy to evaluate. The integral is always an improper integral, so f(x) is positive, decreasing and continuous on [1, ∞). 00 a,, is also convergent (divergent). f(x) dx = limf(x) dx. n=1 Assume 0 ≤ a,, ≤ b,, for n ≥ M, COMPARISION TEST Step 1: Guess whether the given series is conv. or div. Step 2: 1) If I an, is divergent, 1) then 2 b, is also divergent. 2) If I b,, is convergent, 2) then 2 a,, is also convergent. a) (for series with positive terms) If guessing the given series is convergent, then need to find a bigger convergence series. b) If guessing the given series is divergent, then need to find a smaller divergence series. Use this as the last resort unless you can see the answer right away or it's a required test. LIMIT If lim = L and COMPARISON TEST n→→ b To form z b₁, form a quotient of dominant terms of numerator and denominator of Σ an- (for series with 1) L > 0, 1) then both series conv. or both div. positive terms) 2) 2) then Eb, conv. 3) 3) then Ea, conv. Land Σa, conv., L = 0 and Σb,, conv., 004-11 Conclusion then 2 a, Σ (-1)" b₁, is convergent. = 1) then the series is abs conv. 2) then the series is div. 3) then the test is inconclusive. 1) then the series is abs conv. 2) then the series is div. 3) then the test is inconclusive. Test Condition ALTERNATING SERIES TEST (Σ (-1)-¹ b₁) If (b) is positive, decreasing sequence and limb = 0, RATIO TEST = L and ant If lim Man 1) L < 1, 2) L> 1 (or L= ∞), 3) L = 1, ROOT TEST If lima = L and 1) L <1, 2) L> 1 (or L 00), 3) L = 1, Geometric series: Σar" =Σar"; n=0 R=1 Telescoping series: is convergent (= 1). n(n+1) Alternating Harmonic series: (-1)-¹ n n=1 Note b= |an| Remainder Error = |S-Snl <bn+1 Error found by using the nth partial sum is less than the first omitted term. Try this test when I a,, involves factorials and powers. Use this test to find the interval and radius of convergence of a power series (or Taylor series). Try this test when Σ an, involves nth powers. Use this test to find the interval and radius of convergence of a power series (or Taylor series). conv. to 0if -1<r<1 Geometric SEQUENCE: {ar"}= conv. to a if r=1 div. ifr<-lorr>1 Harmonic series: is divergent. a if|rkl ={1-r div ifr21 Special Series/Sequence R=In 1 conv if p>1 div if p≤1 is (conditionally) conv. (= In 2) p-series: Nel n N=1
Transcribed Image Text:Series Test When we determine whether an infinite series is convergent or divergent, we only need to consider the tail (for n 2 M). We don't need to worry about the head (for 1 ≤ n ≤ M-1). If needed, we can rewrite the series as Σa, = a + Σa,,, and the finite series is always convergent. M-1 00 *=1 #-M Condition Conclusion Note Test DIVERGENCE TEST If lim a,, #0 or then the series Σa, is divergent. This test does NOT give the convergence. 11-00 If lim a,, = 0, then the series Ea, is may or may not conv. lima = DNE 21-00 Assume f(n) = an, and INTEGRAL TEST (for series with positive terms) If f(x)dx is convergent (divergent), then Use the integral test if the integral is easy to evaluate. The integral is always an improper integral, so f(x) is positive, decreasing and continuous on [1, ∞). 00 a,, is also convergent (divergent). f(x) dx = limf(x) dx. n=1 Assume 0 ≤ a,, ≤ b,, for n ≥ M, COMPARISION TEST Step 1: Guess whether the given series is conv. or div. Step 2: 1) If I an, is divergent, 1) then 2 b, is also divergent. 2) If I b,, is convergent, 2) then 2 a,, is also convergent. a) (for series with positive terms) If guessing the given series is convergent, then need to find a bigger convergence series. b) If guessing the given series is divergent, then need to find a smaller divergence series. Use this as the last resort unless you can see the answer right away or it's a required test. LIMIT If lim = L and COMPARISON TEST n→→ b To form z b₁, form a quotient of dominant terms of numerator and denominator of Σ an- (for series with 1) L > 0, 1) then both series conv. or both div. positive terms) 2) 2) then Eb, conv. 3) 3) then Ea, conv. Land Σa, conv., L = 0 and Σb,, conv., 004-11 Conclusion then 2 a, Σ (-1)" b₁, is convergent. = 1) then the series is abs conv. 2) then the series is div. 3) then the test is inconclusive. 1) then the series is abs conv. 2) then the series is div. 3) then the test is inconclusive. Test Condition ALTERNATING SERIES TEST (Σ (-1)-¹ b₁) If (b) is positive, decreasing sequence and limb = 0, RATIO TEST = L and ant If lim Man 1) L < 1, 2) L> 1 (or L= ∞), 3) L = 1, ROOT TEST If lima = L and 1) L <1, 2) L> 1 (or L 00), 3) L = 1, Geometric series: Σar" =Σar"; n=0 R=1 Telescoping series: is convergent (= 1). n(n+1) Alternating Harmonic series: (-1)-¹ n n=1 Note b= |an| Remainder Error = |S-Snl <bn+1 Error found by using the nth partial sum is less than the first omitted term. Try this test when I a,, involves factorials and powers. Use this test to find the interval and radius of convergence of a power series (or Taylor series). Try this test when Σ an, involves nth powers. Use this test to find the interval and radius of convergence of a power series (or Taylor series). conv. to 0if -1<r<1 Geometric SEQUENCE: {ar"}= conv. to a if r=1 div. ifr<-lorr>1 Harmonic series: is divergent. a if|rkl ={1-r div ifr21 Special Series/Sequence R=In 1 conv if p>1 div if p≤1 is (conditionally) conv. (= In 2) p-series: Nel n N=1
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