1873_solutions

j1 xn x x1 1 differentiating we obtain n jx j 1 1 xn

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Unformatted text preview: gives us 1 log n Since log log n 1/n is a divergent p-series we conclude that  1. n log log n 1 log n is divergent. log n 15. 1 log log log n For n sufficiently large we have log log log log n  2 and for all such n we have 1 log log log n Since log n  1 exp log n log log log log n  1 is a convergent p-series it follows that the series n2 1 exp 2 log n 1 log log n log log n 16.  12 . n log n is convergent. 1 log log n Since log log n  log n for all n sufficiently large we know from Exe5rcise 14 that if n is sufficiently large then 1 log log n and so 1 log log n log log n log log n is divergent. 329  1 log n log log n 1 n 1 log log n 17. log log n log log n Solution: We observe first that if n is sufficiently large then log log n log log n log log n  log log n log log n 2  exp log log n 2 log log log n . Since log log log n log log n lim nÝ as n 0 Ý we know that log log log n  log log n whenever n is sufficiently large and, therefore, that the inequality log log n log log n log log n  exp log log n 3 must hold whenever n is sufficiently large. Now since log log n 3 0 lim nÝ log n we know that log log n 3  log n whenever n is sufficiently large and, therefore, that the inequality log log n log log n log log n  exp 3 log log n  exp log n  n must hold whenever n is sufficiently large. Therefore 1 log log n whenever n is sufficiently large and, since 1 log log n 18. 1 log log n log log n log log n log log n 1 n 1/n is a divergent p-series, we conclude that the series log log n is divergent. log log n log log n Hint: Whenever n  exp exp e we have log log n  exp log log n log log log n log log n  exp log log n  log n. Now use the conclusion you obtained in Exercise 13. 19. sin x n  where x and  are given positive numbers. Solution: Since lim sin t  1 t t0 we have lim nÝ Comparing with the p-series  1. sin x n 1 n  nÝ lim 1/x  we conclude that 330 sin x n  x n sin x n x  x.  converges if   1 and diverges if 20. Prove that if a n is a sequence of positive numbers and a n converges then so does the series a 2 . n a n is convergent. Since a n 0 We assume that a n is a sequence of positive numbers and that as n Ý we have a n  1 for n sufficiently large and for all such n we have a 2  a n . Since a n is n convergent we deduce from the comparison test that a 2 is convergent. n 21. In this exercise we encounter a series that diverges very slowly. We begin by defining log x if x Lx  e if x  e 1 . The graph of this function is illustrated in the following figure: 2.5 2 1.5 1 0.5 -20 0 -10 10 20 If k is any positive number we shall write L k for the composition of the function L with itself k times. Thus if x is a given number then L 3 x  L L L x . We also define L 0 x  x for every x. a. Given any number x, explain why we must have L k x  1 for all sufficiently large values of k. For a given positive integer n, give a simple meaning to the “infinite product” Ý Ü Lk n  L 0 n L 1 n L 2 n . k0 If we define f...
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