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absolutely.
a. b. 2n
n
The given series must be divergent if 3x 2  1 because, in this case, the expression
3x 2 n
3x 2  n
fails to approach 0 as n Ý. In the event that 3x 2  1, the series
n
n
converges by a simple application of d’Alembert’s test. We are left with the case 3x 2  1
1n
which is
which occurs when x 1 or x 1. When x 1 the given series becomes
n
3
3
1
conditionally convergent and when x 1 it becomes
which is divergent.
n
Thus the series converges absolutely when 1 x 1, converges conditionally when x 1
3
3
and diverges otherwise.
3x log x n
n
This series converges absolutely when log x  1, converges conditionally when log x 1 and
diverges otherwise. In other words, the series converges absolutely when 1 x e,
e
converges conditionally when x 1 and diverges otherwise.
e 341 c. 1 nxn
log n x
For each n 2 we define
an 1 nxn
log n x and observe that
x
log n
x .
log n 1
a n converges absolutely when x  1. If x  1 then, since a n fails to approach 0
Therefore
as n Ý, the series
a n diverges. When x 1 the series is
log n which diverges. When
1n
x 1 the series is
which is conditionally convergent.
log n
n
lim aa 1
nÝ
n d. n Ýx 
lim 3n ! x n
2n ! n!
For each n we define
3n ! x n
2n ! n! an
and observe that
a n 1
an 3n 3 3n 2 3n 1
2n 2 2n 1 n 1 x  and so
n
lim aa 1
n nÝ 27x 
.
4 3n ! x n
converges absolutely when x 
2n ! n!
a n fails to approach 0 as n Ý, the series
a n diverges.
4
Now suppose x 27 . The series
a n becomes
3n ! 4 n
2n ! n! 27 n
and in this case
3n 3 3n 2 3n 1 4
n
lim
lim n 1 aa 1 n Ý n 1
nÝ
n
2n 2 2n 1 n 1 27
Therefore the series 4
27 . If x  4
27 then, since 1 1
2 Thus it follows from the criterion for the nth term to approach 0 that a n decreases to 0 as
a n is divergent.
n Ý. In spite of this fact, the series
4
Finally, suppose that x 27 . The series
a n becomes
3n ! 4 n
1n
2n ! n! 27 n
3n ! x n
which converges by Dirichlet’s test. Therefore the series
converges absolutely
2n ! n!
4
4
when x  27 , converges conditionally when x 27 and diverges for other values of x.
e. nnxn
n!
For each n we define
nn
an n x
n! and observe that
a n 1
an
and so 342 x  1 1
n
e n n
lim aa 1 ex .
n
n n x n converges absolutely when x 
Therefore the series
n!
to approach 0 as n Ý, the series
a n diverges.
nÝ Now suppose that x 1
e . The series a n becomes 1
e . If x  1
e then, since a n fails n n . Since
e n n! n1 n1 lim n 1 nÝ a n 1
an n Ýn
lim 1 e n1 n1 ! nn
e n n! 1
2 we know from Raabe’s test that
a n is divergent but we also know from the criterion for the
nth term to approach 0 that a n decreases to 0 as n Ý.
n
Finally suppose that x 1 . The series
1 n n which converges by
a n becomes
e
n n!
e
Dirichlet’s test.
nxn
n
Therefore the series
converges if x  1 , converges conditionally if x 1 and
e
e
n!
diverges for other values of x.
5. Find the values of x...
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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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