Question

Comparison Tests: Problem 7 Previous Problem Problem List Next Problem (1 point) Library/Rochester/setSeries6Comp Tests/ur_sr_6_14.pg For each sequence an find a number k such that nkan has a finite non-zero limit. (This is of use, because by the limit comparison test the series 8 WE n both converge or both diverge.)

A. an = (4 + 4n) 5 k= B. An k= in Ca, = 2 15 12 On 3n 14 K- D.an= (2n +5141 03n 14.) KU

Answer 1

The question pertains to finding a constant k for sequences to ensure the limit of n^k*a_n is finite and non-zero, related to the limit comparison test. For each sequence, this involves normalizing the power of n to achieve a convergent series.

Step-by-step explanation:

The student's question involves finding a value k for each sequence an such that the series nkan converges to a finite non-zero limit. This is related to the limit comparison test which helps to determine the convergence or divergence of series by comparing it to a known convergent series. For the provided sequences, this involves identifying the highest power of n in the numerator and denominator, and choosing k to normalize the power of n so the limit equals a non-zero constant.

Let's tackle each sequence individually:

• For A. an = (4 + 4n), we can see that the sequence is linear, and the highest power of n is 1. Thus, we can choose k = 1 to ensure that the limit of nkan is finite and non-zero.
• For B. an = 2n15, the highest power of n is already obvious. To make the series comparable, one might choose k = -14 to counterbalance the power of n in the sequence. However, without additional context or sequences, it is difficult to determine what constitutes a finite limit for this specific sequence.
• For D. an = (2n + 5)14/3n14, after simplification, the highest power of n in both the numerator and the denominator is 14. Therefore, we can choose k = 0 in this case, which means simply the series itself without any additional n to the power.