42–76. Convergence or divergence Use a convergence test of you...

Question

42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.

∑ (from k = 1 to ∞)3 / (2 + eᵏ)

Accepted Answer

Welcome back, everyone. In this problem, we want to figure out if the series 2 divided by 3 to the M + 5 between M equals 1 and infinity converges or diverges. A says it converges, while B says it diverges. Now notice here that we have a positive term series, so the limit comparison test would be convenient. Recall that in the limit comparison test, so let's just make a note of that here. OK. In this test If our limit L is equal to the limit as M up. Approaches infinity of the ratio AM to BM where AM and BM are positive sequences. So if this limit exists, OK. And Or value L is positive, so that means it's between 0 and infinity, OK. Then the series AM and the series for the sequence BM. Either will both converge or both diverge, OK? Both converge or both diverge. So essentially what we need to do is to identify our general term, find our limit L, and if it exists and is positive, that means we can make a conclusion about AM based on what we know about BM or vice versa. So first, let's identify our general term. Now notice here that that would be from our problem in our series 2 divided by 3 to the M + 5. I know if we think about it for a large value of M, the denominator 3 M + 5 will be approximately 3M. So that means our term AM will behave like the approximate value of 2 divided by 3 to the m, OK. So now, in this case, since we know that 2 divided by 3 to the end, the series for this expression is a convergent geometric series where the ratio 1/3 is less than 1, then we suspect convergence. So let's choose a value of BM and see if we can use it in the limit comparison test to see uh if that ratio will uh is is finite and non-zero as well. So no, here. If we choose. BM choosing. BM To be equal to 1 divided by 3 to the m. OK. Then our limit L. Is going to be equal to or limit as M approaches infinity of AM that's 2 divided by 3 to the M + 5 divided by BM, which is 1 divided by 3 to the M. And now if we simplify that, that will be our limit of 2 multiplied by 3 tom divided by 3 tom plus 5. Now if we divide our numerator and denominator by 3 to the m, then we should find that our limit would simplify to 2 divided by 1 + 5 divided by 3 to the m. And know if we apply our limit. Then notice here as m approaches infinity 3 to the m also approaches infinity and our quotient will approach 0 because it's going to get smaller. So really then this is going to be 2 divided by 1 + 0, which means L or limit will be equal to 2. No, notice here that it satisfies the condition. L is is greater than 0 but less than infinity. It's finite and non-zero. So in this case, since our series for BM convert, which equals. The series for 1 divided by 3 to the M, like we said earlier converges, then by the limit comparison test or series for AM will also converge. Therefore, if we look back at our answer choices, that means A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.
∑ (from k = 1 to ∞)3 / (2 + eᵏ)

Key Concepts

Infinite Series and Convergence

Comparison Test for Series

Exponential Growth and Its Impact on Series Terms

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