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Unformatted text preview: gives us
1
log n
Since log log n 1/n is a divergent pseries 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 pseries 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 pseries, 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 pseries
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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 Fall '08
 STAFF
 Math, Calculus

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