11–86. Applying convergence tests Determine whether the follow...

Question

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.

∑ (from k = 1 to ∞) (−7)ᵏ / k!

Accepted Answer

Welcome back, everyone. Does the series sigma from M equals 1 to infinity of negative tweets to the power of M divided by M factorial converge. A says yes and B says no. For this problem, let's recognize that our A subscript M term is -3 to the power of M divided by M factorial. And because we have factorials involved, to test for convergence, we can use the ratio test, right? What we want to do is simply identify the limit L. Where L is equal to limitsm approaches infinity of the ratio. A subscript M + 1 divided by a subscript M and we once take the absolute value of that ratio. Now we can get several values of L if L is less than 1, the series converges. If it is greater than 1, the series diverges. If L is exactly 1, the test is inconclusive. So let's go ahead and evaluate a. Subscript M + 1 divided by a subscript M. Starting with a subscript M + 1, we're going to replace M with M+1, which gives us -3 to the power of M + 1. Divided by M + 1 factorial. And we're dividing by a subscript M. We're basically multiplying by the reciprocal. Which is M factorial. Divided by 3 is the power of M. What we want to do is basically simplify so that now negative tweets, the power of M + 1, according to the properties of exponents can be written as negative tweets the power of 1, multiplied by negative tweets the power of M, right? And now considering our factorial, we can expand it as M + 1. Multiplied by A factorial. And then we can notice that we can cancel out M factorial, we can cancel out negative tweets to the power of M because those are on opposite sides of our fractions. And we basically get -3 divided by M + 1. And seeking the absolute value of this term, we can exclude the negative sign because M is greater than 0, right? So we're taking a limit as M approaches infinity. Of 3 divided by M + 1. Now if M approaches infinity. We're going to have our number in the numerator, which is 3, divided by infinitely infinity plus 1, or basically infinity. And whenever we divide 3 by an infinitely large number, we get a limit of 0. Now, we have our limit L, which is 0, and because it is less than 1, according to the series ratio test, the series converges. Well, once we have our answer and the correct option is a. Yes, the series converges. Thank you for watching.

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.
∑ (from k = 1 to ∞) (−7)ᵏ / k!

Key Concepts

Convergence of Infinite Series

Ratio Test

Factorials and Their Growth

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