77–87. Absolute or conditional convergence
Determine whe...

Question

77–87. Absolute or conditional convergence

Determine whether the following series converge absolutely, converge conditionally, or diverge.

∑ (from k = 1 to ∞)(−1)ᵏ⁺¹(k² + 4) / (2k² + 1)

Accepted Answer

Welcome back, everyone. In this problem, we want to evaluate whether the series of -1 to the M + 1 multiplied by 4 M2 + 9 divided by 3M2 + 2 from M equals 1 to infinity converges absolutely, converges conditionally, or diverges. A says it converges absolutely. B that it converges conditionally and C says that it diverges. Now, to figure out what happens to our series, let's first try to find the limit as M approaches infinity. So we can start by considering the absolute value of our terms. No, our absolute value of -1 to the M + 1 multiplied by 4 m2 + 9, all divided by 3 m2 + 2. Is going to be equal to 4 m2 + 9 divided by 3 m2 + 2. Know that we have the absolute value, then our limit that we can call L. Which is the limit as M approaches infinity of 4 m2 + 9 divided by 3M2 + 2 is going to be equal to, if we rewrite it, the limit. Of M2 multiplied by 4 + 9 divided by M2 divided by M2 multiplied by 3 + 2 divided by M2. OK? So we're rewriting it to help us find our limit. We factored out M2. Now in this case, the M2d terms cancel and if we apply our limit, this is going to be 4 plus a number that's approaching 0, a very small number which is gonna be equal to 4 in our numerator, and in our denominator 3 plus a very small number or a number that approaches 0, which is going to be 3 + 0 or 3. So as M approaches infinity or a term approaches for 3. Now that we know what our term approaches, then we can evaluate what happens to the series using the divergence test and the alternating series test. Now, by the divergence test, OK, or series divergence if the limit is not 0, OK. And in this case, since. L our limit is not equal to 0, OK, then that means. The series will diverge. But no, it doesn't stop there. There's a reason why we'll also need the alternating series test because we also need to test if the original series does not converge absolutely. Now, how can the alternating series test help us? What does it say? Well, by the alternating series test, OK. What it basically tells us is that for a series. OK, of -1 to the power of N + 1 multiplied by AN, a term, where or a term A N is positive, OK. Then the series converges. If it satisfies the following two conditions, if first. Or see our term and and decreases, OK. That is A plus 1 the consecutive term is less than or equal to a n, the original term for all n, at least eventually. And second, if the limit as n approaches infinity of our term a n equals zero. Now how does this apply to the term we already have? Well, in this case. We know that our term, if we compare our series here, our general term for the alternating series to the series we already have, we can tell that AN would be equal to 4 m2 + 9 divided by 3 m2 + 2. And we already know. That our limit, the limit as m approaches infinity of AN we've already found that it is going to be equal to 4/3 and 4/3 is not equal to 0. So this, this tells us that condition 2. So thus condition 2 is not satisfied, OK? And that means the alternating series test fails. Therefore, the series diverges. C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

77–87. Absolute or conditional convergence
Determine whether the following series converge absolutely, converge conditionally, or diverge.
∑ (from k = 1 to ∞)(−1)ᵏ⁺¹(k² + 4) / (2k² + 1)

Key Concepts

Absolute Convergence

Conditional Convergence

Alternating Series Test

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