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 = 1 to ∞) [(-1)ⁿ⁺¹ (1 / n^(3/2))]

Accepted Answer

Determine whether the series, the sum from n equals 1 to infinity of -1 to the n + 1 multiplied by 1 divided by the square root of n converges or diverges. And we have two possible answers being converges or diverges. Now, to solve this, we'll make use of the alternating series test. Now, how this works is that we suppose we have a series of A sub N. Which is of the form AN equals -1 to the NBN or AN equals -1 raised to the N + 1BN. Where b_n is greater than it equals 0 for all n. If we have a limit as it approaches affinity of BN equaling 0, and BN being a decreasing sequence, our series AN is convergent. So let's first identify if this qualifies for the alternating series test. We just need to check if it follows the form, and it does, because we have -1 raised to the n + 1. So we can then say B_N, it's just 1 divided by the square root of N. Now let's check the conditions. We need our limit to equal 0, and this to be decreasing overall. Now we have the limit As it approaches infinity of 1 divided by the square root of n. Now, do we know if this approaches 0? But we notice here that as we have a square root, this always grows to infinity, which means we have one divided by infinity. This limit is going to be zero, so that checks out. Now we need to see if our terms are non-increasing. To do this, we can check. If BN + 1 is less than equalsbn. For all in So, BN + 1 is just 1 divided by the square root of n + 1. Now is 1 divided by the square root of n plus 1 less equals 1 divided by the square root of n. Well, We know the square root of n + 1 is greater than the square root of n. But when it's in the reciprocal form, like it is here, it becomes less thin. Which means this sequence is indeed decreasing. Now, because we said it's decreasing and the limit equals 0, we can confirm. That this series converges. That 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 = 1 to ∞) [(-1)ⁿ⁺¹ (1 / n^(3/2))]

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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 = 1 to ∞) [(-1)ⁿ⁺¹ (1 / n^(3/2))]