Unformatted text preview: so we can split the series into two series.) ∞ X n =1 1 6 ! 1 6 ! n1 + ∞ X n =1 1 3 ! 1 3 ! n1 = 1 / 6 11 / 6 + 1 / 3 11 / 3 = 1 5 + 1 2 = 7 10 4. Determine whether the sequence is convergent or divergent. If it is convergent, ﬁnd the limit. ± 3 + 5 n 2 n + n 2 ² ∞ n =1 lim n →∞ 3 + 5 n 2 n + n 2 = lim n →∞ (3 + 5 n 2 )1 /n 2 ( n + n 2 )1 /n 2 = lim n →∞ 3 /n 2 + 5 1 /n + 1 = 0 + 5 0 + 1 = 5 Have a good Spring Break! Remember not to drink and derive (or integrate)....
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 Spring '08
 Bonner
 Calculus, Limit of a sequence, lim sn, · · ·

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