1873_solutions

The key to this exercise is the fact that although n

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Unformatted text preview: and for sufficiently large n we have 333 n n1 p n log log n p . n p n1 p n1 p n  p n n1 p n1 p n  p n1 n n n1 n n1 p Using the mean value theorem we choose a number u n between n  and n  1 such that n1 p n  p  p un p 1. n1 n Thus for each n we have p1 un n n n 1 aa 1  p   1 n n1 n1 Since p1 p1 lim n   n Ý n1 lim 1 n Ý n1 n1 we deduce from the sandwich theorem for limits of sequences that p1 lim u n 1 n Ý n1 and we conclude that n lim n 1 aa 1  p 1   . nÝ n We deduce from Raabe’s test that   1  2   n1 p n! converges when p 1    1 and diverges when p 1    1. n1 7. a. a n 1 an n 1 n  where  is a given number n! This problem is an easy application of d’Alembert’s test and is there to lead into parts b and c. Since lim nÝ the series b. n  is convergent. n! n 1  n 1 ! n n1 n nÝ1 lim n 1 n! n n where  is a given number n! For each n we have n  1  n 1 n1 ! n n n! Since 1  n1 n1 n lim nÝ n  n1 n  e we see that if   1 then lim nÝ and 0 n  1  n 1 n1 ! n n n! n n diverges in this case. If   1 then n! 334 Ý n  n  1  n 1 n1 ! n n n! lim nÝ n n converges. In the event that   1 we have n! n  1  n 1 n1 ! lim e1 nÝ n n n! and and the series c. nn 0 n n diverges in this case. n! log n n! For each n we have n1 log n1 n1 n n1 ! n log n  n1 n n log n n n  1 log n1 n! and we deduce from one of the parts of Exercise 1 of the exercises in indeterminate forms that n1 lim nÝ n1 log n1 n1 ! n n log n n! nn from which we deduce that  n Ý n1 lim n n n log n n  1 log n1 e1 log n diverges. n! n n log n which requires Raabe’s One can actually go a step further and consider the series e n n! test (Theorem 12.6.7). This series converges. In fact, we can do even better by replacing log n n n p which Raabe’s test shows to be convergent when by a constant p giving us the series e n n! p  1 and divergent when p  1 . If p  1 then the stronger form of Raabe’s test (Theorem 2 2 2 12.6.9) shows the series to be divergent. 8. 2n ! n x where x is a given positive number 4 n n! 2 A simple application of d’Alembert’s test shows that this series converges if x  1 and diverges if x  1. When x  1 the series reduces to the one we considered in Example 1 of Subsection 12.6.8. 9. n! x x1 x2  xn 1 where x is a given positive number If we define an  n! x x1 x2  xn 1 for each n then for each n we have a n 1  an n1 ! x x1 x2  xn 1 n! x x1 x2  x n1 1  n1 nx Therefore a n1  lim n 1 n  1  x 1 an nx nÝ and we deduce from Raabe’s test that a n is convergent when x  1 and divergent when x  1. lim n 1 nÝ 335 When x  1 we have a n  1 for all n and, of course, a n diverges. e n n! nn 10. a. This series is a special case of the series mentioned in the discussion that follows Exercise 7c above. If we define n a n  e n! nn for each n then e n1 n1 ! a n 1 an lim n 1 nÝ  n Ýn lim n1 n1 e n n! nn 1 which can be shown to be 1 . We can deduce...
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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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