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 ∞) k⁴ / (eᵏ⁵)

Accepted Answer

Below 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 positive infinity of m power 2 divided by e to the power of M to the power of 3 converge or diverge. Awesome. So it appears for this particular prom we're asked to take the series that is provided to us by the prom itself and we're asked to. Confirm whether it A converges or B diverges. So now that we know what we're ultimately trying to solve for, we must first recall and use the ratio test because the series, as we should be able to recognize, has positive terms, and the terms involve factorial or exponential like growth. Where comparing successive terms is convenient. So using our ratio test for a series, which will write our series, which is the sum. Of a subscript N. With a subscript, so once again I said as N, but we could say the sum of a subscript M, M as in mom, with a subscript M is greater than 0. So once again, using our ratio test for a series, we have the sum of a subscript M. With a subscript M greater than 0, we can compute that L capital L is equal to the limit, as M approaches positive infinity of a subscript M + 1 divided by a subscript M. And we need to note that if L is less than 1, the series converges absolutely, which I'll call CA, so it will converge absolutely. So therefore converges if L is less than 1, but If L is greater than 1, the series will diverge, so I put a D for diverge, and if L is equal to 1, the test will be inconclusive, so I put an I for inconclusive. So with that in mind, we will need to compute the value of L or a subscript M is equal to M 2 divided by Em of 3. So following a step by step approach, our first step that we need to take is we need to write the term A subscript M is equal to M power of 2. So once again Our value A subscript M will be equal to N 2 divided by E of the power of 3, and we need to note that M is greater than or equal to 1. Fantastic. Moving right along here. So now our second step is we need to form our ratio. So we need to take a subscript M + 1 divided by a subscript M. is equal to parenthesis M + 12 divided by E M + 1 of 3, fantastic. And then we need to divide this expression by m to power 2 divided by. E to the power of M to the power of 3. Which will be equal to parenthesis M + 12 divided by M2, multiplied by E power of M3. Minus parenthesis M + 1 to the power of 3. Fantastic. So, our third step now is we need to expand the cubic difference. So we need to take parentheses M + 13 minus M3, which will be equal to 32 + 3M + 1. So we could go ahead and write that. And to the power of 3, so we're taking. M 3 minus parenthesis M + 13 will be equal to -3 2 + 3M + 1. Fantastic. So now we need to substitute this into our ratio. So we need that will be our 4th step, so we're substituting back into our ratio. So we need to take a subscript M + 1 divided by a subscript M is equal to parentheses 1 + 1 divided by M to the power of 2. Multiplied by E to the power of negative, 32 + 3M + 1. Fantastic. Our 5th step now is we need to take the limit as M approaches positive infinity, and we will observe that parentheses 1 + 1 divided by M to the power 2 approaches 1 as M approaches positive infinity. So we will also notice that E of negative 32 + 3M + 1 will approach 0 as M approaches positive infinity. Because the exponent, which let's make a note of this, because the exponent negative parentheses 32 + 3M + 1 will approach negative infinity. Fantastic. So therefore, capital L will be equal to the limit, as M approaches positive infinity of a subscript M + 1 divided by a subscript M, which will be equal to 1 multiplied by 0, which will be equal to 0. And since L is equal to 0 and 0 is less than 1, the ratio test shows that the series, which is the sum evaluated from M equals 1 to positive infinity of M 2 divided by E to the power of M the power of 3, converges absolutely. Therefore, without a doubt, it converges. So our final answer is it converges. And when we look at our multiple choice answers, our final answer without a doubt will have to be multiple choice answer letter A, which once again is convergence. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.
∑ (from k = 1 to ∞) k⁴ / (eᵏ⁵)

Key Concepts

Convergence of Infinite Series

Comparison Test

Exponential vs. Polynomial Growth

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