1873_solutions

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 n1 n1 lim n 1 nÝ a n 1 an  n Ýn lim 1 e n1 n1 ! 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 n1  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 n1 e n n3 2 n 11/12 Ü jn1 2j1 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 n3 j1 2 n n1 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 n1 n1 1 e n1 2 n1 n1 3 2 n1 11/12 n Ü 2 2 n n1 e n Ü j1 1 n n3 2 n 11/12 n j1 2j1 2j1 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 n1 2 n1 n1 3 n1 1 e 2 2 n 3  a n 1  an 2 n1 2n2 1 2 n1 n1 n1 11/12 n 1 ! n 2 n n3 2 n n1 e 2 n 11/12 and then evaluate to obtain a n 1  4 an n1  n1 23 n 12  12n  9 4n 2 n 2n  3 5 12 n2 n 11 e n2 and then supply the definition gn 4 n1  n1 23 n 12 4n 2 n  12n  9 2n  3 n2 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 01 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...
View Full Document

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.

Ask a homework question - tutors are online