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Unformatted text preview: at a n is a decreasing sequence of positive numbers and that for each n we have
n
b n 1 aa 1 ,
n
1 n a n is convergent if and only if the series b n is divergent.
prove that the series
If the series
1 n a n is convergent then 1 n a n 0 as n Ý which implies that a n 0 as n Ý.
Since a n is a decreasing sequence of positive numbers, it follows from the special case of
Dirichlet’s test discussed in Exercise 1 that
1 n a n will converge if a n 0 as n Ý. Now we
b n should
know from Theorem 12.6.11 that the criterion for a n to approach 0 as n Ý is that
diverge. Therefore
1 n a n is convergent if and only if
b n is divergent.
3. Test the following series for convergence and for absolute convergence: 339 1 a. n log n n
If
log x
x fx
for x 0 then for each x 0 we have
1 log x
x2
and we see easily that f is decreasing on the interval e, Ý . Since
log n
lim n 0,
nÝ
1 n log n
the convergence of
follows at once from Dirichlet’s test.
n
fx b. c. sin n/4
n
Of course this series can be handled directly; but it is also a special case of the series
that was discussed in Example 2 of Subsection 12.7.9.
1
2 1 1
3 1 1
n sin nx
n 1 Solution: For each n we define
an 1
2 1
3 1 1 1
n 1 and observe that
an 1 n1 1 1
2 1 1 1
3 1
n and that
1
1 1 1 1 .
n
2
3
1
Since the sequence a n  is decreasing and since it follows from the divergence of
n and
an earlier theorem that a n  0 as n Ý we deduce that a n is convergent. Now to show that
a n fails to converge abolutely we shall use the more powerful form of Raabe’s test:
a n1 
1
lim n log n 1 1
n Ý n log n 1 1
lim
1
n
n
nÝ
n1
a n 
a n  1 n Ý n log n
lim
and so the series
d. 1
2 1 1
3 1
n n1 01 a n  must be divergent.
1 1
n 1 where 0. Hint: The series converges absolutely when 0 1, converges conditionally when 1 and
diverges when 1. Use Raabe’s test when 1 and show that the nth term fails to approach 0 as
n Ý when 1.
The case 0 1:
For each n we define
1
1
an
1
1 1 1
2
3
n
and observe that
1
1 1
1
1
a n  1
2
3
n
If 0 then the series a n  is obviously convergent. Assume now that 0 1. Since 340 lim n 1
nÝ a n1 
a n  1
2 1
n Ýn
lim 1 1
3 1
1 1
2 1
n 1
1
3 1 1 1
n1 1 1
n n
Ý
n1
a n  follows from Raabe’s test.
nÝ
lim the convergence of
The case 1:
For each n we define an 1
2 1 1
3 1 1
n 1 and observe that
1
1 1
1
2
3
it follows from an earlier theorem and the convergence of
a n  n
e. 1 Ý. 1
n
1 that a n  fails to approach 0 as
n 2 log 2 1 3 log 3 1 n log n 1
n! log 2 log 3 log n
For each n we define
2 log 2 1 3 log 3 1 n log n 1
an
n! log 2 log 3 log n
and we observe that
a n1 n 1 log n 1 1
an
n 1 log n 1
and since
n 1 log n 1 1
n
lim
lim n 1 aa 1 n Ý n 1
nÝ
n
n 1 log n 1
n
nÝ
lim
01
n 1 ln n 1
a n is divergent.
we conclude that 4. Determine for what values of x the following series converge and for what values of x the series conv...
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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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