MATH 101 Quiz: MATH101 Quiz 6 2016 Winter with Solutions
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(cid:88) n=1 ( 1)n 1 + 4n3: 1 mark the series. (cid:88) ( 1)n 1 + 4n3: 1 mark the series. 2 + 3n3 either: [ca] converges absolutely; [cc] converges con- ditionally; [d] diverges; or [n] none of the above. 2 + 3n3 does not exist, so the test for divergence shows that the series diverges. (note that [n] can"t possibly be the correct answer, since every series either converges absolutely, converges conditionally, or diverges. ) (cid:88) ( 1)n 1 n2 n=1: 2 marks how many terms of the series within an error that is 10 6? are needed in order to evaluate its value. Therefore we need (n + 1)2 106, which is equivalent to n + 1 1000 or n 999. 1 n n=1 which is a divergent harmonic series. = 3|x 2| lim n n2 (n + 1)2 n2 (n + 1)2 = 3|x 2| 1.
