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Unformatted text preview: lim n !1 6 & 1 n ± 1 = 1 7.4.16 P 1 n =1 1 (1+ln n )) 2 diverges by the Limit Comparison Test using the divergent p & series P 1 =n : lim n !1 a n b n = lim n !1 n (1 + ln n ) 2 = lim n !1 1 2 (1 + ln n ) ± 1 n = lim n !1 n 2 + 2 ln n = lim n !1 1 2 n = 1 : Alternatively, you can show divergence using Direct Comparison: 1 (1 + ln n ) 2 > 1 (2 p n ) 2 = 1 4 n : This gives a BIGGER THAN BIG comparison. 7.5.2 P n 2 e n converges by the Ratio Test: & = lim n !1 ( n + 1) 2 e n +1 ± e n n 2 = lim n !1 ( n + 1) 2 n 2 ± 1 e = 1 e < 1 : 7.5.12 P (ln n ) n n n converges by the Root Test: & = lim n !1 & (ln n ) n n n ± 1 =n = lim n !1 ln n n = 0 < 1 : 7.5.16 P n ln n 2 n convergest by the Ratio Test: & = lim n !1 ( n + 1) ln ( n + 1) 2 n +1 ± 2 n n ln n = lim n !1 n + 1 n ± ln ( n + 1) ln n ± 1 2 = 1 2 < 1 : 2...
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 Spring '09
 CHEUNG
 Calculus, Mathematical Series, lim, Convergence tests

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