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examples of convergent questions
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Question:I heard the example that the integral of sin(x)/x from 1 to infinity is convergent, however, it is not absolutely convergent.
How do you show this? I can't find suitable integrands e.g. to establish that the absolute value of the integral is greater than a divergent integral. Of course, the problem is that the primitive function cannot be expressed in closed form.
Answers:Integral [from 1 to M] sin x / x by parts to get cos 1  ( cos M ) / M + Integral[1 to M] cos x / x cos x / x integral converges by comparison with 1/x . To show divergence, take Riemann sum: [n=0 to ] Integral[n to (n+1) ]  sin(x)/x  dx = [n=0 to ] Integral[0 to ] sin(v)/(v+n ) dv [n=0 to ] (1/(n+1) ) Integral[0 to ]  sin(v) dv = 2 / (n+1) which diverges
Answers:Integral [from 1 to M] sin x / x by parts to get cos 1  ( cos M ) / M + Integral[1 to M] cos x / x cos x / x integral converges by comparison with 1/x . To show divergence, take Riemann sum: [n=0 to ] Integral[n to (n+1) ]  sin(x)/x  dx = [n=0 to ] Integral[0 to ] sin(v)/(v+n ) dv [n=0 to ] (1/(n+1) ) Integral[0 to ]  sin(v) dv = 2 / (n+1) which diverges
Question:Provide an example of convergent and divergent evolution, adaptive radiation, and coevolution. Then, respond to the following:
Choose one of the examples you provided and discuss the implications this example may have for future humans.
Please please help me with this one.......
Answers:Convergent evolution: Two different species adapt similarly when they share similar niches. Australia has several examples. Tasmanian tigers are very similar to wolves or dogs. The marsupial lion had many features similar to cats They even had a saber toothed version of a marsupial similar to saber toothed cats. divergent evolution: When species separate into dissimilar niches, they evolve quite differently. The formation of limbs in early amphibians compared to their fish ancestors is an example. Adaptive radiation: When a species spreads in to different niches it forms multiple species. A classic example is the Darwin's finch. The common ancestor split into several species as it occupied the various niches. Coevolution is where two species evolved together. Many species of plants produce flowers that only bats can fertilize. The bats evolved to take advantage of the flowers and the flowers rewarded the bats with nectar and flowering at night. Coevolution of humans and cows: The cows allowed us to exploit land that wasn't suitable to crops. They greatly increase our food supply. Perhaps in the future we will use them in many different ways. The way they convert cellulose to sugar in the guts with microbes could lead to the easy conversion of grass into alcohol to power our cars.
Answers:Convergent evolution: Two different species adapt similarly when they share similar niches. Australia has several examples. Tasmanian tigers are very similar to wolves or dogs. The marsupial lion had many features similar to cats They even had a saber toothed version of a marsupial similar to saber toothed cats. divergent evolution: When species separate into dissimilar niches, they evolve quite differently. The formation of limbs in early amphibians compared to their fish ancestors is an example. Adaptive radiation: When a species spreads in to different niches it forms multiple species. A classic example is the Darwin's finch. The common ancestor split into several species as it occupied the various niches. Coevolution is where two species evolved together. Many species of plants produce flowers that only bats can fertilize. The bats evolved to take advantage of the flowers and the flowers rewarded the bats with nectar and flowering at night. Coevolution of humans and cows: The cows allowed us to exploit land that wasn't suitable to crops. They greatly increase our food supply. Perhaps in the future we will use them in many different ways. The way they convert cellulose to sugar in the guts with microbes could lead to the easy conversion of grass into alcohol to power our cars.
Question:1 show that the sequences are divergent :
a) ( 1(1)^n+1/n)
b) (sinn /4)
2 Let X= (x_{n}) and Y= (y_{n}) be sequence , and the "shuffled" sequence Z= (z_{n}) be defined by
z1= x1, z2= y1,.....,z_{2n1} = x_{n}, z_{2n}= y_{n}.
show that Z is convergent if and only if both X and Y are convergent and lim X = lim Y
note: n goes to infinity
Answers:1. a) Subsequence of evennumbered terms has value 1/n > 0, subsequence of oddnumbered terms has value 2+1/n > 2, so sequence is divergent. b) Terms with n divisible by 4 are all 0, terms with n1 divisible by 4 are all 1, so sequence is divergent. 2. ==>: Suppose Z is convergent and lim Z = L. Let > 0. Then there is an N > 0 such that for any n > N, z_n  L < . Let M > N/2, then for any m > M, x_m  L = z_{2m1}  L < since 2m1 > N. So X converges to L. Similarly, for any m > M, y_m  L = z_{2m}  L < so Y converges to L. So X and Y both converge and lim X = lim Y. <==: Suppose X and Y converge and lim X = lim Y = L. Let > 0. Then there exist M and N > 0 such that for all m > M, x_m  L < and for all n > N, x_n  L < epsilon; Let K = max{2M, 2N}. Then for any k > K, if k is odd then z_k  L = x_m  L < where k = 2m1 and hence m > M if k is even then z_k  L = y_n  L < where k = 2n and hence n > N So Z is convergent and lim Z = L.
Answers:1. a) Subsequence of evennumbered terms has value 1/n > 0, subsequence of oddnumbered terms has value 2+1/n > 2, so sequence is divergent. b) Terms with n divisible by 4 are all 0, terms with n1 divisible by 4 are all 1, so sequence is divergent. 2. ==>: Suppose Z is convergent and lim Z = L. Let > 0. Then there is an N > 0 such that for any n > N, z_n  L < . Let M > N/2, then for any m > M, x_m  L = z_{2m1}  L < since 2m1 > N. So X converges to L. Similarly, for any m > M, y_m  L = z_{2m}  L < so Y converges to L. So X and Y both converge and lim X = lim Y. <==: Suppose X and Y converge and lim X = lim Y = L. Let > 0. Then there exist M and N > 0 such that for all m > M, x_m  L < and for all n > N, x_n  L < epsilon; Let K = max{2M, 2N}. Then for any k > K, if k is odd then z_k  L = x_m  L < where k = 2m1 and hence m > M if k is even then z_k  L = y_n  L < where k = 2n and hence n > N So Z is convergent and lim Z = L.
Question:Let
(n=1 to infinity) A(n)
Be a absolutely convergent series such that A(n) A(n1)
Prove that,
Lim nA(n) = 0 as n > Infinity.
Answers:This is a subtle result, and usually proved "in the book." See the source for my favorite (and affordable) book on the subject. WLOG, assume the sequence is positive and monotonic nonincreasing. Let be eps > 0 given. Since the series is summable, there's an N for which the tail past N adds up to less than eps/2; i.e. (n=N..infinity) A(n) < eps/2. Now look at terms with index m > 2N. Then eps/2 > (n=N..infinity) A(n) > (n=N..m) A(n) > (mN+1) A(m), because the terms are nonincreasing. But mN+1 > m/2. Therefore m A(m) < eps for all m > 2N; hence since eps was arbitrary, Lim_{n > infinity} n A(n) = 0.
Answers:This is a subtle result, and usually proved "in the book." See the source for my favorite (and affordable) book on the subject. WLOG, assume the sequence is positive and monotonic nonincreasing. Let be eps > 0 given. Since the series is summable, there's an N for which the tail past N adds up to less than eps/2; i.e. (n=N..infinity) A(n) < eps/2. Now look at terms with index m > 2N. Then eps/2 > (n=N..infinity) A(n) > (n=N..m) A(n) > (mN+1) A(m), because the terms are nonincreasing. But mN+1 > m/2. Therefore m A(m) < eps for all m > 2N; hence since eps was arbitrary, Lim_{n > infinity} n A(n) = 0.
From Youtube
Absolute Convergence, Conditional Convergence and Divergence :Absolute Convergence, Conditional Convergence and Divergence for series. In this video, I give the basic result and do 3 examples!
Sequences  Examples showing convergence or divergence :Sequences  Examples showing convergence or divergence. For more free math videos, check out PatrickJMT.com