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
k0 is now obvious.
b. For each positive integer n, we define
an 1 Ý
Ü k0 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 k0 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 n1 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 n1 ! 3
3 n1 ! 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 n1 2 n 1 !
3 n2 2 n 1
3
n1 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 n1 n1
n 1 !
nn
n! lim nÝ 4. en n Ý n1
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! n1 6. 1 2
n! n1 2 p where is a given number where and p are given numbers. Solution: For each n we define
an 1 2
n! n1 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.
 Fall '08
 STAFF
 Math, Calculus

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