1873_solutions

# 2 and 2 n 1 j1 hold whenever n n we see that n j1

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Unformatted text preview: x  e x 2x for every number x then, since f 1  e 2  0 and f x  0 whenever x 1 it follows that f x  0 whenever x 1. (As a matter of fact it is easy to show that f x  0 for every real number x.) For each positive integer k we define a k  exp k 1 k stands for the composition of exp with itself k times. Since a  2 k for each k it where exp k follows that a k Ý as k Ý. Now if x is any positive number and k is a positive integer and Lk x 1 then L k x  log k x  1 and so x  a k . Since the inequality x  a k must hold for sufficiently large k we deduce that L k x  1 for sufficiently large k. the meaning of the expression Ý Ü Lk n k0 is now obvious. b. For each positive integer n, we define an  1 Ý Ü k0 L k n . Using the integral test (or otherwise), show that the series a n is divergent. For every number x 1 we define 1 gx  . Ý Ü k0 L k x The function g is a positive decreasing function on 1, Ý . Now we observe that 331 Þ 1 g x dx  Þ 1 e ee Þe exp 3 1 Þ exp 21 e 1 dx  1 x ee g x dx  Þe g x dx  Þ exp 1 dx  1 x log x exp 3 1 21 1 dx  1 x log x log log x and, in general, exp n1 1 Þ exp for each integer n n1 g x dx  1 0. Since exp n 1 Þ1 g x dx lim Þ 1 g x dx  Ý n for each n we deduce that n nÝ and so it follows from the integral test that the series a n ia divergent. Some Exercises on the Ratio Tests Test the following series for convergence. In the event that the series diverges, determine whether or not its nth term approaches 0 as n Ý. 2n ! 3n ! 1. 3 2 Solution: We define an  2n ! 3n ! 3 2 for each n. For each n we have 2 n1 ! 3 3 n1 ! 2 2n  2 3 2n  1 3 a n 1   an 2n ! 3 3n  3 2 3n  2 2 3n  1 2 3n ! Ý. We deduce from d’Alembert’s test that a n is convergent. as n 2. 3 2 64 729 n2 n! For each n we have 3 n1 2 n 1 ! 3 n2 2 n 1 3 n1 n! and since the latter expression approaches Ý as n 2 3 n is divergent. n! 3. 3 n log n n! Solution: For each n we define 332 Ý, it follows from d’Alembert’s test that an  3 n log n n! . Since e  3 we have the inequality n log n n log n n e n n! n! n! To show that a n is divergent, it is sufficient to show that n n /n! is divergent. But this fact follows at once from d’Alembert’s test and the fact that an  3 n1 n1 n 1 ! nn n! lim nÝ 4. en  n Ý n1 lim n n  e  1. log n n log 2 log 3  log n Solution: For each n we define an  en log n n log 2 log 3  log n Then for each n we have a n 1  an log n  1 n e log n n exp n log log n  1 1 e exp n log log n Now for each n we observe that n log log n  1  1 exp n log log n  1 e n log log n  and by using L’Hôpital’s rule we can deduce that lim n log log n  1 nÝ log log n  1 log log n 1 n n log log n  0. We deduce that n lim aa 1  1  1 e n a n diverges. nÝ and it follows from d’Alembert’s test that 5.  1 2 n! n1 6.  1 2 n! n1 2 p where  is a given number where  and p are given numbers. Solution: For each n we define an   1 2 n! n1 Then for each n we have a n 1  an...
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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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