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n? +1 2. Determine whether the series is convergent or divergent. If it's 2n² +1 n=1 convergent, find its sum. Show All Your Work! Please use the attached series test, to correctly solve the problem. *Also when you find the solution, please plug the answer and into this following sentence frame as a conclusion. Sentence Frame: By the is_divergent or convergent_ test, the series 2 2 +

n? +1 2. Determine whether the series is convergent or divergent. If it's 2n² +1 n=1 convergent, find its sum. Show All Your Work! Please use the attached series test, to correctly solve the problem. *Also when you find the solution, please plug the answer and into this following sentence frame as a conclusion. Sentence Frame: By the is_divergent or convergent_ test, the series 2 2 +

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. Please use the attached series test, to correctly solve the problem.

*Also when you find the solution, please plug the answer into the following sentence frame as a conclusion.

Sentence Frame: By the _____ test, the series _________ is _divergent or convergent____________.

*If you actually solve the question correctly, fill out the above sentence correctly, and show all of your work, I will 100% leave a thumbs up for you. Thank you.

Test Condition Conclusion Note If {b,} is positive, decreasing sequence and lim b, = 0, then Z a, =E (-1)" b, is convergent. b, = |anl Remainder = Error = |S- S,I < bn1 Error found by using the nth partial sum is less than the first omitted term. Try this test when Z a, involves factorials and powers. ALTERNATING SERIES TEST Series Test (E (-1)" b, ) When we determine whether an infinite series is convergent or divergent, we only need to consider the tail (for n2 M). We don't need to worry RATIO TEST If lim n a. M-1 =L and about the head (for 1sns M-1). If needed, we can rewrite the series as Ea, =Ea, +Ea, , and the finite series is always convergent. %3D Use this test to find the interval and radius of convergence of a power series (or Taylor series). 1) then the series is abs conv. 1) L<1, 2) L> 1 (or L = 0), 3) L= 1, 2) then the series is div. 3) then the test is inconclusive. Test Condition Conclusion Note DIVERGENCE ROOT TEST If lim a, |= L and Try this test when Z a, involves n" powers. If lim a, +0 or then the series Za, is divergent. This test does NOT give the convergence. TEST If lim a, = 0, then the series Ea, is may or may not conv. Use this test to find the interval and radius of convergence 1) L<1, 2) L> 1 (or L= 00), 3) L= 1, 1) then the series is abs conv. 2) then the series is div. lim a, = DNE →の of a power series (or Taylor series). INTEGRAL TEST Assume f(n) = an , and Use the integral test if the integral is easy to evaluate. 3) then the test is inconclusive. If, f(x)dx is convergent (divergent), then (for series with positive terms) f(x) is positive, decreasing and continuous on [1, 0). The integral is always an improper integral, so (S(x)dx = lim [" f(x)dx. a, is also convergent (divergent). R JI Special Series/Sequence n=1 Assume 0s a, s b, for n 2 M, 1) If E a, is divergent, 2) If E b, is convergent, COMPARISION Step 1: Guess whether the given series is conv. or div. a conv. to 0if -1<r<1 1) then Z b, is also divergent. 2) then Z a, is also convergent. if |r|<1 TEST Step 2: a) If guessing the given series is convergent, then need to find a bigger convergence series. Geometric series: ar" =Ear"- ={1-r (for series with positive terms) Geometric SEQUENCE: {ar"} ={conv. to a if r =1 div if |r|21 n=0 div. if rs-l or r>1 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. 1 -is convergent (= 1). Harmonic series: >- 1 is divergent . LIMIT Telescoping series: If lim 9 = L and 一→ b in(n+1) To form Z b, form a quotient of dominant terms of numerator and denominator of E an. COMPARISON TEST (for series with positive terms) 1) L>0, 2) L= 00 and Za, conv., 1) then both series conv. or both div. 2) then Zb, conv. 1 conv if p>1 Alternating Harmonic series: 5(-1)" is (conditionally) conv. (= In 2) p-series: 3) L= 0 and Eb, conv., 3) then Za, conv. div if p<1
Transcribed Image Text:Test Condition Conclusion Note If {b,} is positive, decreasing sequence and lim b, = 0, then Z a, =E (-1)" b, is convergent. b, = |anl Remainder = Error = |S- S,I < bn1 Error found by using the nth partial sum is less than the first omitted term. Try this test when Z a, involves factorials and powers. ALTERNATING SERIES TEST Series Test (E (-1)" b, ) When we determine whether an infinite series is convergent or divergent, we only need to consider the tail (for n2 M). We don't need to worry RATIO TEST If lim n a. M-1 =L and about the head (for 1sns M-1). If needed, we can rewrite the series as Ea, =Ea, +Ea, , and the finite series is always convergent. %3D Use this test to find the interval and radius of convergence of a power series (or Taylor series). 1) then the series is abs conv. 1) L<1, 2) L> 1 (or L = 0), 3) L= 1, 2) then the series is div. 3) then the test is inconclusive. Test Condition Conclusion Note DIVERGENCE ROOT TEST If lim a, |= L and Try this test when Z a, involves n" powers. If lim a, +0 or then the series Za, is divergent. This test does NOT give the convergence. TEST If lim a, = 0, then the series Ea, is may or may not conv. Use this test to find the interval and radius of convergence 1) L<1, 2) L> 1 (or L= 00), 3) L= 1, 1) then the series is abs conv. 2) then the series is div. lim a, = DNE →の of a power series (or Taylor series). INTEGRAL TEST Assume f(n) = an , and Use the integral test if the integral is easy to evaluate. 3) then the test is inconclusive. If, f(x)dx is convergent (divergent), then (for series with positive terms) f(x) is positive, decreasing and continuous on [1, 0). The integral is always an improper integral, so (S(x)dx = lim [" f(x)dx. a, is also convergent (divergent). R JI Special Series/Sequence n=1 Assume 0s a, s b, for n 2 M, 1) If E a, is divergent, 2) If E b, is convergent, COMPARISION Step 1: Guess whether the given series is conv. or div. a conv. to 0if -1<r<1 1) then Z b, is also divergent. 2) then Z a, is also convergent. if |r|<1 TEST Step 2: a) If guessing the given series is convergent, then need to find a bigger convergence series. Geometric series: ar" =Ear"- ={1-r (for series with positive terms) Geometric SEQUENCE: {ar"} ={conv. to a if r =1 div if |r|21 n=0 div. if rs-l or r>1 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. 1 -is convergent (= 1). Harmonic series: >- 1 is divergent . LIMIT Telescoping series: If lim 9 = L and 一→ b in(n+1) To form Z b, form a quotient of dominant terms of numerator and denominator of E an. COMPARISON TEST (for series with positive terms) 1) L>0, 2) L= 00 and Za, conv., 1) then both series conv. or both div. 2) then Zb, conv. 1 conv if p>1 Alternating Harmonic series: 5(-1)" is (conditionally) conv. (= In 2) p-series: 3) L= 0 and Eb, conv., 3) then Za, conv. div if p<1
n? +1 2. Determine whether the series Y is convergent or divergent. If it's 2n? +1 n=1 convergent, find its sum. Show All Your Work! Please use the attached series test, to correctly solve the problem. *Also when you find the solution, please plug the answer and into this following sentence frame as a conclusion. 2 n' +1 Σ test, the series 2n? +1 Sentence Frame: By the is_divergent or convergent_ n=1
Transcribed Image Text:n? +1 2. Determine whether the series Y is convergent or divergent. If it's 2n? +1 n=1 convergent, find its sum. Show All Your Work! Please use the attached series test, to correctly solve the problem. *Also when you find the solution, please plug the answer and into this following sentence frame as a conclusion. 2 n' +1 Σ test, the series 2n? +1 Sentence Frame: By the is_divergent or convergent_ n=1
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