Test IV, March 22, 2016 6. Suppose that 2 n a n converges. Prove or disprove that a n converges. Proof. Method I: Let a n x n be a power series, which converges at x = 2 by the assumption. Applying Theorem 18, we know that a n x n converges absolutely for all x ( 2, 2); that is, a n x n converges for all x ( 2, 2). Choose x = 1 and obtain that a n converges. Method II: Since 2 n a n converges, lim n 2n a n = 0 by the n-th term test (Theorem 7). By the definition of limit, choosing ε = 1, there exists N N such that That is, Hence, a n n=n 2 n a n 0 < 1 for all n N. a n < 2 n for all n N. 2 n <, which means that a n converges. n=n ŒÞ åà /ÝÝ (1) ã Ratio Test á Á lim b n+1 n b n D3 vœá  y 1 ` = b n [e = b n [e. Q b n [eqp 1J lim < 1.»A {b n } = á Î lim n b n+1 b n n b n+1 b n { 1 2, 1 3 2, 1 2 3, 1 3 4, 1 2 5, 1 3 6,, 1 2 n, 1 3 n+1, 1 2 n+2, 1 3 n+3, }. D3. lim n b n = 1 2 2n 1 + 1 [e 3 2n b 2 n n+1 lim = 0 if n is odd = n 3n+1 b n 3 n. lim n 2 n+1 = if n is even 1
I5Ý!.3J 9 Þ` K W. 2 n a n [e X ã Ratio Test ÿ lim 2 n+1 a n+1 2 n a n < 1 n (hýàý Ratio test) = lim 2 a n+1 < 1 n a n = lim a n+1 1 < n a n 2 = a n converges by Ratio Test.!ø2 Root Test ôb8!ýµ (2) Limit Comparison Test (Theorem 11) 1 b ƒ' b n c n KÎÑó lim n n c n = 0 v u b n c n ÎÑó` î ŒWñ»A {b n } = {c n } = { 1, 0, { 1, 1 2, c n [e J1J b n [e 1 3 0.6, 0, 1 5 0.6, 0,, 0, 1 (2n 1) 0.6, 0, 1 } (2n + 1) 0.6, 0, 1, 1, 3 4 ã Theorem 14 á c n [e ê. b 2n+1 lim = lim n c 2n+1 n 1 5,, ( 1)n+1 n, b 2n 0 lim = lim = 0, n c 2n n c 2n (2n + 1) 0.5 = lim (2n + 1) 0.6 n }. 1 = 0, (2n + 1) 0.1 b ãœí 543 Ï 131 Þ ÿá lim n n c n = 0. Î b n =. b I5!.3J Ï 6 Þ` ýà Limit Comparison Test W lim n a n ±1 2 n = lim a n n 2 n = 0 v 2 n a n [e X ã Limit Comparison Test ÿ a n [e 2