Determining Convergence or Divergence
In Exercises 1–14,...

Question

Determining Convergence or Divergence

In Exercises 1–14, determine whether the alternating series converges or diverges. Some of the series do not satisfy the conditions of the Alternating Series Test.

∑ (from n = 2 to ∞) [(-1)ⁿ⁺¹ (1 / ln n)]

Accepted Answer

Determine whether the series, the sum as n equals 4 to infinity of -1 to the n multiplied by 1 divided by natural log n + 1 converges or diverges. Now we have two possible answers being converges or divergence. To answer this, we'll make use of the alternating series test. This test says suppose we have a series AN, where AN follows the form of -1 to NBN or -1 N + 1BN. Where bn is greater than equals 0 for all n. Then, if our limit as it approaches infinity of BN equals 0, and BN is a decreasing sequence, our series AN is convergent. So, based on the alternating series tests, let's identify BN. We notice this does follow the form because we have a -1 raised to the end. We can then say b_n is 1 divided by natural log of n + 1. Now, we just need to check our two conditions. We need to see if this is decreasing or non-increasing. To do this, we can shed the sea. If BN + 1 is equals to bin. So we have 1 divided by natural log. Of n + 2. Or BN + 1. And our original 1 divided by natural log n plus 1. Now, we need to check, is this true? Now, we know that natural log is strictly increasing on its own. And so therefore, natural log n + 2 would be greater than natural log of n + 1. But because we have the reciprocal. This switches around. Because natural log N + 2 is larger, normally, it will be smaller when we have the reciprocal. So this is indeed less than our original. We can confirm that this check works. That will check our other condition if the limit equals 0. We have the limit. As n approaches infinity of 1 divided by natural log n plus 1. Now if we were to use infinity. We end up getting a value that goes to infinity because natural log approaches infinity. We have 1 divided by affinity, which means this fraction approaches 0. In other words, our limit is 0. That confirms our second condition for alternating series, which means this does indeed converge. This means the answer to our problem is answer A. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

Determining Convergence or DivergenceIn Exercises 1–14, determine whether the alternating series converges or diverges. Some of the series do not satisfy the conditions of the Alternating Series Test.∑ (from n = 2 to ∞) [(-1)ⁿ⁺¹ (1 / ln n)]

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Determining Convergence or Divergence
In Exercises 1–14, determine whether the alternating series converges or diverges. Some of the series do not satisfy the conditions of the Alternating Series Test.
∑ (from n = 2 to ∞) [(-1)ⁿ⁺¹ (1 / ln n)]