1873_solutions

# A n n exp log n n log n log lim n1 bn b n is a

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Unformatted text preview: the divergence of a n from Raabe’s test but 2 there is really no need for such big guns. Since the limit above is negative we know that a n 1  1 an for sufficiently large n and so a n cannot approach 0 as n Ý. b. nn e n n! If we define an  nn e n n! for each n then n1 n1 lim n 1 nÝ a n 1 an  n Ýn lim 1 e n1 n1 ! nn e n n!  1 n Ýn e 1 1 e lim n and so it follows from Raabe’s test that a n diverges. 11. 2n ! n n! 2 4 n 1 2 p where p is a given number We already know that this series diverges when p  1 and from the comparison test it follows that the series also diverges when p  1. From now on we suppose that p  1. We define an  2n ! 4 n n! 2 p for each n and observe that a n 1  an 2n  1 2n  2 p and so 2n  1 p 2n  2 2n  2 p 2n  1 p  n Ýn lim 2n  2 p p To see that the latter limit is 2 , we use mean value theorem to chooce a number c n between 2n  1 and 2n  2 such that lim n 1 nÝ n a n 1 an  n Ýn 1 lim 2n  2 p 2n  1 2n  2 p and observe that 336 p n p1 pc n 2n  2 p n p 2n  1 2n  2 p1 p1 pc n 2n  2 n p n p p 2n  2 2n  2 p1 . p We deduce from Raabe’s test that the given series converges when p  2 and diverges when p  2. We must now consider the case p  2. In this case we shall use the more powerful form of Raabe test: a n1  lim n log n 1 1 2n  1 2 lim n log n 1 1 an n n nÝ nÝ 2n  2 5n  4 4n n  1  n Ý n log n lim 0 2 and we conclude that the series diverges when p  2. n 2 12. 2 n n1 e n n3 2 n 11/12 Ü jn1 2j1 jj 2j 1 2 j! Solution: Start off by supplying the definition n n 2 an  1 2j  1 2j 2 j j j! Ü n n3 j1 2 n n1 e 2 n 11/12 to Scientific Notebook and make the selection that the subscript n is a function argument. You then have  a n 1  an 2 n1 n1 1 e n1 2 n1 n1 3 2 n1 11/12 n Ü 2 2 n n1 e n Ü j1 1 n n3 2 n 11/12 n j1 2j1 2j1 jj 2j 1 2 j! . 2j 1 2 jj j! Do not ask Scientific Notebook to simplify this expression as it stands. First observe that n1 2 n1 n1 3 n1 1 e 2 2 n 3  a n 1  an 2 n1 2n2 1 2 n1 n1 n1 11/12 n 1 ! n 2 n n3 2 n n1 e 2 n 11/12 and then evaluate to obtain a n 1  4 an n1  n1 23 n 12  12n  9 4n 2 n 2n  3 5 12 n2 n 11 e n2 and then supply the definition gn 4 n1  n1 23 n 12 4n 2 n  12n  9 2n  3 n2 to Scientific Notebook. Thus n g n  aa 1 . n Finally you can use the Evaluate button to obtain lim g n  1 nÝ lim n 1 nÝ lim n log n nÝ 1 and so the series diverges. 337 gn 1 n 1 gn 01 5 12 n 11 e n2 13.  1 2  n  1 2  n 1 1 where  and  are given numbers Our understanding of this problem is that  is automatically prohibited from being 0 or any negative integer. For each n we define  1 2  n 1 an   1 2  n 1 and we observe that |  n | n lim lim n 1 aa 1  n Ý n 1 nÝ n |  n |   n   . n a n converges if     1 and diverges if     1. In the  n Ýn 1 lim We deduce from Raabe’s test that event that     1 we have  1 2  n 1 an    1 2  n 1 Exce...
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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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