32–49. Choose your test Use the test of your choice to determi...

Question

32–49. Choose your test Use the test of your choice to determine whether the following series converge absolutely, converge conditionally, or diverge.

∑ (from k = 1 to ∞) (−1)ᵏ k (2ᵏ⁺¹ / (9ᵏ − 1))

Accepted Answer

Hello. In this video, we are going to be determining whether the given series converges or diverges. Now, the series that is given to us is an infinite series starting at 1 of -1 to the power, multiplied by M, multiplied by 4M + 1, divided by 11M minus 1. Before we approach this problem, let's just go ahead and combine this entire sequence together by multiplying all the terms to the numerator of the last quantity. So we can go ahead and rewrite the sequence to be the sequence starting at M is equal to 1 and going to infinity of -1 to the power, multiplied by M, multiply by 4, raise the power of M + 1. And this will all be divided by 11 to the power of M minus 1. Now, in this case here, notice that we have an alternating term, -1 to the M power. Whenever M is odd, this will be negative, and whenever M is even, this will be positive. So this term indicates that we should decide to use the ratio or the root test. Now, in this case here we have a rational expression within the sequence, so we will go ahead and take the approach of the ratio test. So, the ratio test states that we want to go ahead and take the limit as M approaches affinity of a AM plus 1 divided by AM. And so what this means is that we want to go ahead and take the M+1 version of the sequence and divide it by the original. Now, that will go ahead and leave us with a complex fraction, but if we plug in a sub M + 1 into the sequence and multiply by the reciprocal of a sub-M, we can go ahead and take the limit, as M approaches infinity of the quantity. 11 raised to the power of M minus 1, divided by M multiplied by 4 power of M + 1. And this will be multiplied by the Ace of M+1 series, which is going to be M + 1. Multiply by 4, raise the power of M + 1 + 1, which will be M + 2. And this will be divided by 11, raise the power of m + 1 minus 1. Now, in terms of the negative 1 to the nth power, the absolute absolute value. Eliminates that term. So, now what we need to do is we just need to go ahead and simplify. Now, let's go ahead and group up some of these terms together. Starting with the M terms, we have M + 1 in the numerator, and we have M in the denominator. We can go ahead and combine those into their own term, and call that M + 1 divided by M. Then we will go ahead and combine the 11 m minus 1. In the numerator and the 11 m + 1 minus 1 in the denominator. So that will give us the term, 11, raise the power of M minus 1, divided by 11, raise the power of M + 1 minus 1. Then we will multiply the four coefficients together, which will give us 4 power of M + 2, divided by 4 power of M + 1. Now, there's a bit of simplification that we can do here. Looking at the last term, 4 power of M + 2 can be rewritten as 4 the power of M multiplied. By 4 squad, and the same can be done in the denominator here. 4 of M + 1 can be rewritten as 4 power of M multiplied by 4. And so that's going to reduce the last term by canceling out the 4 to the end power and canceling out one of the multiples of 4, and the last term will reduce to just 4. Now, the first term, M + 1 divided by M is already in its simplest form, so there's not much that we can do here. But what about this middle term? Now, the first thing that we can do with the middle term, is we can go ahead and factor out an 11 to the power of M. Doing so is going to leave us with 11 to the power of M multiplied by 1 minus 11 to the negative m. Divided by 11 to the power of M. Multiplied by 11. Minus 11 to the power of negative and this is going to further simplify to leave us with 1 divided by 11. Or apologies, not yet, because we haven't taken the limit yet. But that is going to reduce the sequence to just be 1 minus 11 to the power of negative M divided by 1 minus 11 of negative M + 1. So, this is going to further reduce to be the limit. As M approaches infinity. Of M + 1 divided by M, multiplied by 4, multiplied by 1 divided by 11, multiplied by 1 minus 11 of negative, divided by 1 minus 11 of negative M + 1. And by taking the limit as it goes to infinity, we will be left with 1 multiplied by 4 divided by 11, multiplied by 1, and this is going to give us a value of 4 divided by 11 for the ratio test. Now, because this value is less than 1, that meets the condition and states that since the limit is less than 1, this series is going to converge. And more importantly, because we use the ratio test, this is going to converge absolutely. And so with that being said, the solution to this problem is going to be a. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

32–49. Choose your test Use the test of your choice to determine whether the following series converge absolutely, converge conditionally, or diverge.
∑ (from k = 1 to ∞) (−1)ᵏ k (2ᵏ⁺¹ / (9ᵏ − 1))

Key Concepts

Absolute and Conditional Convergence

Alternating Series Test

Ratio Test

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