77–87. Absolute or conditional convergence
Determine whe...

Question

77–87. Absolute or conditional convergence

Determine whether the following series converge absolutely, converge conditionally, or diverge.

∑ (from k = 1 to ∞)(−2)ᵏ⁺¹ / k²

Accepted Answer

We'll come back every once another video where the series sigma from N equals 1 up to infinity of negative tweets the power of n divided by n cubed determine whether the series converges absolutely, conditionally, or diverges. A says converges absolutely. B converges conditionally, and C diverges. So for this problem, let's begin by identifying the nth term. That's A N equals -3 to the power of N divided by n cubed, which can also be written as -1 to the power of N multiplied by 3 to the power of N and divided by n cubed. What we're going to do is simply apply the root test. It says that we want to evaluate the limit as an approaches infinity of a N. And we're going to take the absolute value of A N and raise it to the power of 1 divided by N, right? So, this is the limit that we want to evaluate and specifically, we're going to take the limit as N approaches infinity. What we can do is simplify the absolute value of -1 to the power of N multiplied by 3 to the power of N divided by N cubed. Specifically, -1 to the power of N oscillates between -1 and 1. So when we take the absolute value, we can just omit that term. And because N starts with 1, N cubed is always greater than 0, and 3 to the power of N is also greater than 0 because it's an exponential function. So we can just write 3 to the power of N divided by n cubed. And now what we're going to do is simply raise it to the power of 1 divided by N, right? So we have 3 to the power of N divided by N cubed, raise to the power of 1 divided by N. Which is equal to 3 to the power of n multiplied by 1 divided by n gives us 3 and 3 is divided by n raises the power of 3 divided by n so we're applying the properties of exponents, meaning we want to evaluate the limit of 3 divided by N to the power of 3 divided by N. So, in this context, we want to focus on the properties of limits. We are going to take the limit as N approaches infinity of 3, which is 3, and divided by the limit as N approaches infinity of N to the power of 3 divided by N. Let's suppose that L is equal to the limit. As n approaches infinity of and the power of 3 divided by n. And essentially, if we substitute and approaches infinity for every and we're going to get infinity raise to the power of 0, right? And because that's the case, we're going to try to rearrange this expression and. Apply Lapito's rule. Specifically, we're going to take LN of L which gives us limit as N approaches infinity of 3 divided by N multiplied by LN of N using the properties of logarithms. And now what we can do is simply factor out 3, so we got. We multiplied by the limit as n approaches infinity of a n of N divided by n. And now the indeterminate form is infinity divided by infinity, meaning we can apply Lapito's rule and take the derivative of each part of this fraction. So we get 3 multiplied by the limit as N approaches infinity of 1 divided by N divided by 1, right? So the derivative of LN of N is 1 divided by N and the derivative of N is 1. And L as N approaches infinity, one divided by an infinitely large number is going to give us a limit of 0, so the answer is 0. Meaning the limit l itself is equal to e to the power of 0, which is 1. And I substituting this value into the limit of the root test, we get 3 divided by 1, which is 3. And according to the root test, if that limit is greater than 1, the series diverges, meaning the correct answer to this problem is C. Thank you for watching.

77–87. Absolute or conditional convergence
Determine whether the following series converge absolutely, converge conditionally, or diverge.
∑ (from k = 1 to ∞)(−2)ᵏ⁺¹ / k²

Key Concepts

Absolute Convergence

Conditional Convergence

Alternating Series Test

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