Determining Convergence or Divergence
Which of the serie...

Question

Determining Convergence or Divergence

Which of the series in Exercises 17–56 converge, and which diverge? Use any method, and give reasons for your answers.

∑ (from n=1 to ∞) (1 − n) / n2ⁿ

Accepted Answer

Hello there, today we are going to 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. Test the series for convergence. The sum evaluated from n equals 1 to infinity of 1 minus 2 n divided by 3 to the power of n multiplied by n. OK, so it appears for this particular problem we're asked to look at our series that is provided to us. We're asked to determine if this series will either A converge or B diverges. So ultimately we are asked to just take our series and test it for convergence and determine whether or not it will converge or diverge. So now that we know what we're ultimately trying to solve for, our first step that we need to take before we start solving is we need to split the series by linearity. So we need to take the sum evaluated from N equals 1 to infinity of 1 minus 2N. Divided by 3 to the power of n multiplied by n. Fantastic. And this will be equal to the sum evaluated from n equals 1 to infinity of 1 divided by 3 to the power of n multiplied by n minus the sum evaluated from n equals 1 to infinity of 2 n divided by 3 to the power of n multiplied by n. Which will be equal to the sum evaluated from n equals 1 to infinity of 1 divided by 3 to the power of n multiplied by n minus 2 multiplied by the sum evaluated from n equals 1 to infinity of 1 divided by 3 to the power of n. Fantastic. So looking at the first series, we need to recall and use the direct comparison test, and as we should recall, the direct comparison test states that if 0 is less than or equal to n and less than or equal to bn for all n, and the sum of bn converges, then the sum of ann will also converge. And conversely, if 0 is less than or equal to AN and AN is less than or equal to bn for all n, the sum of A N diverges, then the sum of bn diverges. OK. So now we need to let AN be equal to 1 divided by 3 to the power of n multiplied by N. And we need to note for n is greater than or equal to 1, that will mean that we will have that 3 to the power of n multiplied by n will be will be greater than or equal to 3 to the power of n. So that will mean, as a result, we can therefore write that 0 will be less than or equal to 1 divided by n multiplied by 3 to the power of n, which will be less than or equal to 1 divided by 3 multiplied by the, so it's gonna be 1 divided by 3 to the power of n. And since, once again, since we have the sum evaluate from n equals 1 to infinity of 1 divided by 3 to the power of n is equal to the sum evaluated from n equals 1 to infinity of parentheses of 1 divided by 3 to the power of n, and we need to be able to recall and recognize that this is a convergent. Geometric series such as the absolute value of r is going to be equal to the absolute value of 1 divided by 3, which will be less than 1. And we can use the direct comparison test, and when we use the direct comparison test, it implies that the sum evaluated from n equals 1 to infinity of 1 divided by 3 to the power of n multiplied by n converges, which for I'll just call C for short, so we'll know that C represents convergence. So when we use the direct comparison test, it will imply that as a result, our sum will converge as a result. So, moving on to our second step now, focusing our attention on the second series now, which is -2 multiplied by the sum evaluated from N equals 1 to infinity of 1 divided by 3 to the power of n. And this will be a constant multiple of the geometric series, the sum. Evaluated from. N equals 1 to infinity of 1 divided by 3 to the power of n, which will be equal to the sum evaluated from n equals 1 to infinity of parentheses of 1 divided by 3 to the power of n. Therefore, as a result, the second series will also Converge as a result, and since both of these series, so let's make a note, since both of these series converges or converge, it follows that the given series provided to us by the prom itself will also converge. So that's our final answer is gonna be a converges. So looking at our multiple choice answers once again. We can determine that our final answer without a doubt for this particular problem will be once again A converges, and that's it. Thank you so much for watching, hopefully that helped, and I can't wait to see you in the next video. Bye.

Determining Convergence or DivergenceWhich of the series in Exercises 17–56 converge, and which diverge? Use any method, and give reasons for your answers.∑ (from n=1 to ∞) (1 − n) / n2ⁿ

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Determining Convergence or Divergence
Which of the series in Exercises 17–56 converge, and which diverge? Use any method, and give reasons for your answers.
∑ (from n=1 to ∞) (1 − n) / n2ⁿ