1873_solutions

1873_solutions

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Unformatted text preview: at a n is a decreasing sequence of positive numbers and that for each n we have n b n  1 aa 1 , n 1 n a n is convergent if and only if the series b n is divergent. prove that the series If the series 1 n a n is convergent then 1 n a n 0 as n Ý which implies that a n 0 as n Ý. Since a n is a decreasing sequence of positive numbers, it follows from the special case of Dirichlet’s test discussed in Exercise 1 that 1 n a n will converge if a n 0 as n Ý. Now we b n should know from Theorem 12.6.11 that the criterion for a n to approach 0 as n Ý is that diverge. Therefore 1 n a n is convergent if and only if b n is divergent. 3. Test the following series for convergence and for absolute convergence: 339 1 a. n log n n If log x x fx  for x  0 then for each x  0 we have 1 log x x2 and we see easily that f is decreasing on the interval e, Ý . Since log n lim n  0, nÝ 1 n log n the convergence of follows at once from Dirichlet’s test. n fx b. c. sin n/4 n Of course this series can be handled directly; but it is also a special case of the series that was discussed in Example 2 of Subsection 12.7.9. 1 2 1 1 3 1 1 n sin nx n 1 Solution: For each n we define an  1 2 1 3 1 1 1 n 1 and observe that an  1 n1 1 1 2 1 1 1 3 1 n and that 1 1 1 1 1 . n 2 3 1 Since the sequence |a n | is decreasing and since it follows from the divergence of n and an earlier theorem that |a n | 0 as n Ý we deduce that a n is convergent. Now to show that a n fails to converge abolutely we shall use the more powerful form of Raabe’s test: |a n1 | 1 lim n log n 1 1  n Ý n log n 1 1 lim 1 n n nÝ n1 |a n | |a n |  1  n Ý n log n lim and so the series d. 1 2 1 1 3 1 n n1 01 |a n | must be divergent. 1 1 n 1 where   0. Hint: The series converges absolutely when 0   1, converges conditionally when   1 and diverges when   1. Use Raabe’s test when   1 and show that the nth term fails to approach 0 as n Ý when   1. The case 0   1: For each n we define 1 1 an  1 1  1 1 2 3 n and observe that 1 1 1 1 1 |a n |  1 2 3 n If   0 then the series |a n | is obviously convergent. Assume now that 0    1. Since 340 lim n 1 nÝ |a n1 | |a n | 1 2 1  n Ýn lim 1 1 3 1 1 1 2 1 n 1 1 3 1 1 1 n1 1  1 n n Ý n1  |a n | follows from Raabe’s test. nÝ lim the convergence of The case   1: For each n we define an  1 2 1 1 3 1 1 n 1 and observe that 1 1 1 1 2 3 it follows from an earlier theorem and the convergence of |a n |  n e. 1 Ý. 1 n 1 that |a n | fails to approach 0 as n 2 log 2 1 3 log 3 1  n log n 1 n! log 2 log 3  log n For each n we define 2 log 2 1 3 log 3 1  n log n 1 an  n! log 2 log 3  log n and we observe that a n1  n  1 log n  1 1 an n  1 log n  1 and since n  1 log n  1 1 n lim lim n 1 aa 1  n Ý n 1 nÝ n n  1 log n  1 n nÝ lim 01 n  1 ln n  1 a n is divergent. we conclude that 4. Determine for what values of x the following series converge and for what values of x the series conv...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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