MA 124 Fall-08, Exam Answers
- 1)
- 1 a) The sequence converges and the limit as n-> infinity is equal to 1/3.
- 1 b) The series diverges by the n-th term test since the limit as n->
infinity a_n is not equal to zero.
- 2)
- 2 a) The sequence converges and the limit as n-> infinity is equal to 0.
- 2 b) The series converges since it is a geometric series with r = 3/pi
and |3/pi| < 1. The sum of the series is equal to 9/(pi - 3).
- 3)
- 3) a) You can conclude the series \sum a_n diverges by the basic
comparison test.
- 3 b) You cannot conclude anything about the convergence or divergence of
\sum a_n since < divergent series gives no info.
- 4) The series \sum a_n converges absolutely. This can be shown by a)
The Ratio Test OR b) \sum |a_n| converges by Basic Comparison test OR c) \sum
|a_n| converges by the Integral test.
- 5)
- 5 a) The series \sum a_n does NOT converge absolutely. This can by
shown by using Limit Comparison test between \sum |a_n| and \sum 1/n^(1/2) .
- 5 b) The series \sum a_n converges by the Alternating Series Test.
- 5 c) The series \sum a_n does not diverge since it was shown to
converge in b).
- 6)
- 6 a) n = 9 is the smallest value of n that works. This is because
|R_n| less than or equal to b_{n+1} = 1/(n+1)^4, and the smallest n with
1/(n+1)^4 less than or equal to .0001 is n
= 9.
- 6 b) S_9 = -1 + 1/2^4 - 1/3^4 + 1/4^4 - 1/5^4 + 1/6^4 - 1/7^4 + 1/8^4 -
1/9^4.
- 6 c) S_9 is smaller than S since the partial sums of an alternating
series alternate between > S and < S. Since the last tern of S_9 is
negative, we have S_9 < S because adding the next term, which is positive,
would give S_10 > S.
IF YOU HAVE FURTHER QUESTIONS ABOUT HOW TO DO THESE PROBLEMS I WILL BE HAPPY
TO DISCUSS THEM WITH YOU DURING MY OFFICE HOURS OR AT MATH HELP.