# solHMK7 - Section 3.6#4(a Let M > 0 Let K N be greater than...

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Section 3.6 #4) (a) Let M > 0. Let K N be greater than M 2 . Then for all n K , n > M . Hence, n is properly divergent to + . (b) n < n + 1, so by the Comparison Theorem, n + 1 tends to + . (c) n is the 1-tail of n - 1, so n - 1 tends to + since its tail does. (d) lim(( n/ n + 1) / n + 1) = lim( n/ ( n + 1)) = 1, so by (b) and the Limit Comparison Test, n/ n + 1 tends to + . #5) No. In Section 3.4, subsequences ( x n k ) and ( y n j ) of (sin n ) were constructed such that x n k 1 / 2 and y n j ≤ - 1 / 2. Thus, n k x n k tends to + and n j y n j tends to -∞ . Therefore, ( n sin n ) has two subsequences tending toward diﬀerent inﬁnities. #6) ( x n ) is properly divergent, so ( | x n | ) tends to + . There exists K such that n K implies | x n | > 0. ( x n y n ) converges, and so is bounded. Let M be an upper bound for | x n y n | . Then for n K , 0 ≤ | y n | ≤ M | x n | . By the Squeeze Theorem, ( | y n | ) 0 and hence ( y n ) 0. #7)

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solHMK7 - Section 3.6#4(a Let M > 0 Let K N be greater than...

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