MA 124 Fall-07, Exam 2 Answers
- 1) a) Diverges
- 1) b) Converges
- 2) a) Converges because geometric series with r = 1/sqrt < 1. The sum
equals sqrt(2)/(sqrt(2)-1)
- 2) b) Diverges by n-th term test
- 3) a) It is bounded since -1 < (-1)^n/(n^2 + 1) < 1
- 3) b) It is not monotonic (i.e. neither increasing nor decreasing)
because the values alternate between positive and negative.
- 3 c) Converges because lim_{n \to \infinity} |a_n| = lim{n \to
\infinity} 1/(n^2 + 1) = 0 hence by the absolute value property lim_{n\to
\infinity} a_n = 0 also.
- 4) a) The series converges absolutely since \sum |a_n| = \sum 1/(n^2
+1) which converges by comparison with the p-series for p=2.
- 4) b) The series converges since converges absolutely implies
converges.
- 4) c) Since the series begins with n = 0 the sum of the first ten terms
is S_9 = \sum_{n=0}^{n=9} and the error |R_9| = 1/(10^2 +1) = 1/101 .
- 5) a) Diverges by the ratio test.
- 5) b) We cannot conclude whether the series diverges or converges. A
series whose limit is one when applying the ratio test may either converge
or diverge.
- 5) c) Converges by the ratio test.
- 6) Converges by the ratio test.
- 7) [-3,3) . The radius of convergence is seen to be 3 by applying the
ratio test. It diverges at x = 3 since one gets \sum 1/n which diverges
since it is the p-series with p=1. It converges at x = -3 since one gets
\sum (-1)^n/n which converges by the alternating series test.
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