1873_solutions

# A fd b 1 j1 x j 1 x j for each j we have since the

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Unformatted text preview: log n n 0 Now we define 1 an  n 1 log n log log n /n and for each n. Since we deduce that n n1 8. bn  1 n b n is a divergent p-series and since 1 lim lim a n  n Ý n Ý bn log n log n exp n 1 log log n a n is divergent. n log n Solution: For each n we have n n1 n log n  exp n n1 n log n log . We now observe that lim n log nÝ n n1 nÝ lim log n log n  1 1 n and an easy application of L’Hôpital’s rule shows that n lim n log  1. nÝ n1 We can therefore regard the expression n log n n n1 as being of the order of 1 log n  1 exp n for large values of n and this observation suggests how we can solve the problem. For each n we define n log n n an  n1 and bn  1 n Now exp lim a n  n Ý lim n Ý bn n log n log n n 1 nÝ lim 1 n  n Ý exp log n  n log n log lim n log n log exp log n exp n n 1 n n1 Now lim log n  n log n log nÝ n n1  n Ý log n lim nÝ lim 1  n log 1  n log n n 1 1 log n and, using L’Hôpital’s rule twice, we can see that lim log n  n log n log nÝ 327 n n1 0 n n1 Thus n lim a n  n Ý exp log n  n log n log lim n1 bn b n is a divergent p-series, it follows that a n is divergent.  exp 0  1 nÝ Since n n1 9. n log n 2 Solution: We have already seen that n n1 lim nÝ n log n 1 1 n and so for all sufficiently large n we have n n1 n log n e  n. For all such n we have 2 n log n n log n n n  n1 n1 As long as n is also large enough to make log n  3 we have n n1 1 n2 Since 1 log n 10. n log n 2 log n  e n  n3  12 . n n n log n n n1 is a convergent p-series we deduce that log n  n n log n 2 is convergent. 3 Solution: As n Ý, the expression log n increases much more slowly than n. So we proceed as follows. For each n we define an  1 log n 3 and bn  1 . n Since n lim Ý lim a n  n Ý bn log n 3 b n is a divergent p-series we deduce that a n is divergent. nÝ and since 11. 1 log n n Solution: Since log n  2 whenever n  e 2 , the inequality n holds whenever n is sufficiently large. Since n 1 is convergent. log n 12. 1 log n 1  1n 2 log n 1/2 n is a convergent (geometric) series, we deduce that log n Solution: Since log log n  2 whenever n  exp exp 2 328 we know that the inequality log n log n  exp log n log log n holds whenever n is sufficiently large. Therefore  exp 2 log n  n 2 log n 1  12 log n n for all sufficiently large n and it follows from the fact that 1/n 2 is a convergent p-series that the series log n 1 is convergent. log n 1 log log n 13. log n Hint: Use the fact that log log log n  2 whenever n  exp exp exp 2 Thus, for n sufficiently large we have 1 log log n Since  1 exp log n log log log n  1 exp 2 log n 1 is a convergent p-series it follows that the series n2 1 log n 14. log n . 1 log log n  12 . n log n is convergent. log log n Solution: Since log log n 2 0 log n  log n holds whenever n is sufficiently large. Therefore, if n is lim nÝ we know that the inequality log log n 2 sufficiently large we have log n log log n  exp log log n 2  exp log n  n which...
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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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