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

Question

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

∑ (k = 1 to ∞) (−1)ᵏ (k¹¹ + 2k⁵ + 1) / [4k(k¹⁰ + 1)]

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Does the series, the sum evaluated from M equals 1 to infinity of parentheses 1 to the power of M multiplied by m 6 plus or m2 + 1 divided by 2 multiplied by M multiplied by parenthesis m of 5 + 3 converge. Awesome. So it appears for this particular prompt, we're asked to examine this particular series and we're asked to determine yes or no does it converge. So will it be either A, yes, the series converges, or B, no, the series diverges. So now that we know what we're ultimately trying to solve for, our first step is we need to recall and use the alternating series test. So we need to suppose that we have a series for the sum of a subscript N. And either a subscript N is equal to parentheses 1 to the power of N multiplied by B subscript N. Or a subscript N is equal to -1 to the power of N + 1 multiplied by B subscript N. Where this B subscript N is greater than or equal to 0 for all ends. Fantastic. Then if we can say that the limit as N approaches positive infinity, so the limit as N approaches positive infinity of B subscript N is equal to 0. And we can say that curly brackets of B subscript N, so the curly brackets of B subscript N is a decreasing sequence, which I'll call DS for short. Then the series, once again, this series. The sum of a subscript N will be convergent. Fantastic. So, with that in mind, our next step now is we need to identify the value B subscript M. So the series alternates, as we should be able to recognize and recall. So the series will alternate due to -1 to the power of M. So we have this -1 to the power of M, so we can let B subscript M, which in in an effort to save time, I'll just say BM. So BM is equal to M 6 + 4 multiplied by m2 + 1 divided by 2 multiplied by M multiplied by parenthesis of the power of 5 + 3. Awesome, we are on a roll. And do not forget your parenthesis. And we need to note that for any value of M is greater than or equal to 1, both the numerator and the denominator will give a positive value. So that will mean that so. We'll have a positive divided by a positive value will be greater than or equal to 0. Fantastic. So, with that in mind, moving right along here, so therefore, BM will be equal to M to the power of 6 plus, so we'll have M 6+ or multiplied by M2 or 42. Plus 1 divided by 2 multiplied by M multiplied by parenthesis of M to the power of 5 plus 3 will be greater than or equal to 1. For all M is greater than or equal to 1. Fantastic. So Our next step now, step 3, is we need to check the limit condition. So we need to take the limit as M approaches positive infinity of B of M. Fantastic. So with that in mind. We need to note that we'll have the limit as M approaches positive infinity of M 6 + 4 multiplied by M 2 + 1 divided by 2 multiplied by M multiplied by parenthesis of M 5 + 3. So now, our next step is we need to expand the denominator, so we need to take 2 multiplied by M, multiplied by parenthesis M to the power of 5 + 3, which will be equal to 2 multiplied by M 6 + 6M. So therefore, the expression will become M 6 + 42 + 1 divided by 26 plus 6M, fantastic. So we now need to divide the numerator and the denominator by the highest power of m in the denominator, which is m the power of 6, so we need to take m the power of 6 divided by m 6 plus 4 m2 divided by m 6 plus 1 divided by m 6 all divided by 2 m 6 divided by m the power. Of 6. Plus 6 M divided by 6, which will be equal to 1 + 4 divided by m the power of 4. Plus 1 divided by M to the power of 6, divided by 2 plus 6 divided by M to the power of 5. Fantastic. So now, our next step is we need to take the limit as M approaches positive infinity. So we need to take the limit as M approaches positive infinity of M 6 + 4 M 2 + 1 divided by 2 multiplied by M multiplied by parenthesis M 5 + 3. Which will be equal to the limit, as M approaches positive infinity of 1 + 4 divided by m of 4 plus 1 divided by M 6 divided by 2 plus 6 divided by m of 5. Which will be equal to 1 + 0 + 0 divided by 2 + 0, which will give us an answer of 12 or 1 divided by 2. So our 4th and final step is we need to make our final conclusions. And since the limit as M approaches positive infinity of B subscript M cannot equal 0, the series diverges even though it alternates. So even though the series alternates, it diverges. So our final answer will be multiple choice answer letter B. Which once again is, no, the series diverges, and that is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

11–27. Alternating Series Test Determine whether the following series converge.
∑ (k = 1 to ∞) (−1)ᵏ (k¹¹ + 2k⁵ + 1) / [4k(k¹⁰ + 1)]

Key Concepts

Alternating Series Test

Behavior of the General Term

Comparison and Simplification of Polynomial Expressions

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