**Problem 6.**Prove or disprove each of the statements.(a) The sequence \( x_n = (-1)^n \) is convergent.(b) The sequence \( x_n = \frac{(-1)^n}{n} \) is convergent.(c) Let \((x_n)\) be a sequence such that the sequence \(|x_n|\) of absolute values converges. Then \((x_n)\) converges.(d) Let \((x_n)\) and \((y_n)\) be two unbounded sequences. Then the product sequence \((x_n \cdot y_n)\) is unbounded.

Name: **Problem 6.**Prove or disprove each of the statements.(a) The sequen
Uploaded: 2024-03-20T06:38:58.000Z
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Description: **Problem 6.**Prove or disprove each of the statements.(a) The sequence \( x_n = (-1)^n \) is convergent.(b) The sequence \( x_n = \frac{(-1)^n}{n} \) is convergent.(c) Let \((x_n)\) be a sequence such that the sequence \(|x_n|\) of absolute values

Since, you have post the multiple subpart's. I am answering the first three parts. Please post the question again for the remaining parts with specific number. (a) Consider the given sequence. xn=−1n As, in the above sequence, if n is odd then the value sequence is -1. If, the value of n is even then the value of the sequence is 1. Let's take subsequence of the given sequence. a2n and a2n+1 The above sequence are subsequence. If both converge same limit. Then, the series is convergent.The sequence are defined as, a2n=−1n=1 Thus, a2n→1 as n→∞. And, a2n+1=−1n=-1 Thus, a2n+1→-1 as n→∞. As, the limit is different of both sequence. The given sequence is not convergent.(b) Consider the given sequence. xn=-1nn Let's an and bn be two sequence. bn=1n Now, apply the alternating series test. Take the limit and apply. limn→∞bn=limn→∞1n=0 If we increase the denominator then numerator decrease. Thus, the given sequence decreasing sequence. Thus, by the alternating series test, the given sequence is convergent.(c) Consider the given sequence, xn We have to show that if the sequence xn absolutely converge then sequence is xn. Let the sequence is converge and the limit of the sequence is converge to l. xn−l<ε ∀ n≥m And, xn−l≤xn−l<ε For every ε>0 there exist a number m belongs to natural number. such that, xn−l<ε ∀ n≥m Thus, the sequence xn will be convergent.