Unformatted text preview: pt for the trivial case 0, the series  n  is divergent.
14. 1 2 n
1 2 n 2 1
1
n where and are given numbers Solution: For each n we define
an 1 2 n
1 2 n 2 1
1 and we observe that
lim n 1
nÝ
We deduce from Raabe’s test that
Assume now that n Ýn 1
lim 1
2 . . In this case we apply the more powerful form of Raabe test. Since 1
2 lim n log n
nÝ n 2
2 .
n 2
a n is convergent when 1 and is divergent when
2 a n 1
an 1
n 1 a n 1
an n Ý n log n
lim 1
n 1 1
2 n
1
2 2 n n 2 2 Ý
we deduce that 1 2 n
1 2 n a n is convergent in this case. Therefore convergent when 1
2 and divergent when 1
2 1
1 2 is . 15. Cauchy’s root test says that if a n is a sequence of nonnegative numbers and if n a n
the series a n converges if 1 and diverges if 1. as n Ý then a. Prove Cauchy’s root test.
Suppose that n a n
as n Ý. In the event that 1 we have n a n 1 for all sufficiently
large n which tells us that a n 1 for sufficiently large n. Therefore, in the case 1 the series
a n is divergent.
Now suppose that 1. Choose a number p such that p 1. For all n sufficiently large we
have n a n p and for all such n we have a n p n . Since the series
p n is a convergent
a n is convergent.
geometric series the comparison test guarantees that
b. Review an earlier exercise and then prove that if Cauchy’s root test can be used to test a given series for 338 convergence then so can d’Alembert’s ratio test.
The desired assertion follows at once from that previous exercise which is where all the hard
work lies.
16. Prove the following more powerful root test:
If a n 0 for all n and if
n
1 a n 1/n
p
log n
as n Ý, then the series a n converges if p 1 and diverges if p 1.
This form of the root test is one of the results that are developed in the special document on ratio and root
tests and that can be reached by clicking on the icon . Solution: We suppose that p 1 and that
n
1 a n 1/n
p
log n
as n Ý. Choose a number q such that 1 q p. We know that the inequality
n
1 a n 1/n q
log n
must hold for all sufficiently large n. For all such n we have
q log n n
an 1
n
From the fact that
lim n q 1 nÝ n q log n
n 1 and from the fact that 1/n q is a convergent pseries we deduce that the series
convergent; and it follows from the comparison test that a n is convergent. 1 q log n
n n is Now try the case p 1. Exercises on Conditionally Convergent Series
1. A common test for convergence that one encounters in an elementary calculus course is the alternating series
test, sometimes known as the Leibniz test which says that if a n is a decreasing sequence of positive
numbers and if a n 0 as n Ý then the series
1 n a n is convergent. Prove that the alternating series
test follows at once from Dirichlet’s test.
Since
n 1 j 1 j1 for every n, the Leibniz test follows at once from Dirichlet’s test.
2. Given th...
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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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