11–27. Alternating Series Test Determine whether the following...

Question

11–27. Alternating Series Test Determine whether the following series converge.

∑ (k = 0 to ∞) (−1)ᵏ / (2k + 1)

Accepted Answer

Welcome back, everyone. Does the series sigma from M equals 2 up to infinity of -1 to the power of M divided by M2 converge? A yes and B no. For this problem, let's go ahead and rewrite our series as sigma from M equals 2 up to infinity of -1 to the power of M multiplied by 1 divided by m2. We can see that this series has a form of alternating series. So we're going to use the alternating CS test. Before we use it, we're going to define our term B subscript M, which in this case is equal to 1 divided by M squad, right? And we want to ensure that it is greater than or equal to 0 for all values of M before we apply the test, right? So, indeed, because we have a square of a number in the denominator, B subscript M is always going to be greater than 0, right? So, this condition is satisfied, and now we want to check to conditions for convergence. First of all, we are going to identify the limit as M approaches infinity of B subscript M, which would be limit as M approaches infinity of 1 divided by M2, and this is going to be equal to 0. Because we're dividing a number by an infinitely large number, right? So our first condition for convergence is satisfied. We want to ensure that this limit is equal to 0, and the second condition is to ensure that B subscript M is a decreasing sequence. So what we're going to do is simply compare two terms. One of them is going to be 1 divided by M2, and then we're going to take B subscript M + 1, right? So we're going to get 1 divided by M + 1 squared. We want to ensure that our term 1 divided by M + 1 squad is lower than 1 divided by m2, and this is the case. We can show that 1 divided by m2 is greater than 1 divided by M + 1 squad. Why is that? We can notice that our second fraction has a higher denominator. We have the same numerators. And a greater denominator essentially means that we're dividing by a larger number. So for any value of M 1 divided by M + 1 squared is lower than 1 divided by m2. So B subscript M is a decreasing sequence. So we have successfully satisfied two conditions for the convergence of a series, according to the alternating series test, meaning the answer is yes, the series converges. Because the limit of B subscript M as M approaches infinity is 0, and our sequence B subscript M is a decreasing sequence. Thank you for watching.

11–27. Alternating Series Test Determine whether the following series converge.
∑ (k = 0 to ∞) (−1)ᵏ / (2k + 1)

Key Concepts

Alternating Series Test

Monotonic Decreasing Sequence

Limit of the Terms Approaching Zero

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