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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
n1 8. bn 1
n
b n is a divergent pseries 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
n1 n log n exp n
n1 n log n log . We now observe that
lim n log nÝ n
n1 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
n1
We can therefore regard the expression
n log n
n
n1
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
n1
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
n1 Now
lim log n n log n log nÝ n
n1 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
n1 0 n
n1 Thus
n
lim a n n Ý exp log n n log n log
lim
n1
bn
b n is a divergent pseries, it follows that a n is divergent. exp 0 1 nÝ Since n
n1 9. n log n 2 Solution: We have already seen that
n
n1 lim nÝ n log n 1 1
n and so for all sufficiently large n we have
n
n1 n log n e
n. For all such n we have
2 n log n
n log n
n
n
n1
n1
As long as n is also large enough to make log n 3 we have n
n1
1
n2 Since 1
log n 10. n log n 2 log n e
n n3 12 .
n
n
n log n
n
n1 is a convergent pseries 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 pseries 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 pseries 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 pseries 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.
 Fall '08
 STAFF
 Math, Calculus

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