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Unformatted text preview: and for sufficiently large n we have 333 n
n1 p n log log n p . n p
n1 p
n1 p n p
n
n1 p
n1 p n p
n1
n
n
n1
n
n1 p
Using the mean value theorem we choose a number u n between n and n 1 such that
n1 p n p
p un p 1.
n1
n
Thus for each n we have
p1
un
n
n
n 1 aa 1 p 1
n
n1
n1
Since
p1
p1
lim n
n Ý n1
lim
1
n Ý n1
n1
we deduce from the sandwich theorem for limits of sequences that
p1
lim u n
1
n Ý n1
and we conclude that
n
lim n 1 aa 1 p 1 .
nÝ
n
We deduce from Raabe’s test that
1 2 n1 p
n!
converges when p 1 1 and diverges when p 1 1.
n1 7. a. a n 1
an n 1 n where is a given number
n!
This problem is an easy application of d’Alembert’s test and is there to lead into parts b and c.
Since
lim nÝ the series
b. n is convergent.
n! n 1
n 1 !
n n1
n nÝ1
lim n 1 n! n n where is a given number
n!
For each n we have
n 1 n 1
n1 !
n n
n!
Since 1 n1 n1
n lim nÝ n n1
n e we see that if 1 then
lim nÝ and 0 n 1 n 1
n1 !
n n
n! n n diverges in this case. If 1 then
n! 334 Ý n n 1 n 1
n1 !
n n
n! lim nÝ n n converges. In the event that 1 we have
n!
n 1 n 1
n1 !
lim
e1
nÝ
n n
n! and and the series
c. nn 0 n n diverges in this case.
n! log n n!
For each n we have
n1 log n1 n1 n
n1 !
n log n
n1
n
n log n
n
n 1 log n1
n!
and we deduce from one of the parts of Exercise 1 of the exercises in indeterminate forms that n1
lim nÝ n1 log n1 n1 !
n n log n
n!
nn from which we deduce that n Ý n1
lim
n n n log n
n 1 log n1 e1 log n diverges.
n!
n n log n which requires Raabe’s
One can actually go a step further and consider the series
e n n!
test (Theorem 12.6.7). This series converges. In fact, we can do even better by replacing log n
n n p which Raabe’s test shows to be convergent when
by a constant p giving us the series
e n n!
p 1 and divergent when p 1 . If p 1 then the stronger form of Raabe’s test (Theorem
2
2
2
12.6.9) shows the series to be divergent. 8. 2n ! n
x where x is a given positive number
4 n n! 2
A simple application of d’Alembert’s test shows that this series converges if x 1 and diverges if
x 1. When x 1 the series reduces to the one we considered in Example 1 of Subsection 12.6.8. 9. n!
x x1 x2 xn 1 where x is a given positive number If we define
an n!
x x1 x2 xn 1 for each n then for each n we have
a n 1
an n1 !
x x1 x2 xn 1
n!
x x1 x2 x n1 1 n1
nx Therefore
a n1 lim n 1 n 1 x 1
an
nx
nÝ
and we deduce from Raabe’s test that
a n is convergent when x 1 and divergent when x 1.
lim n 1 nÝ 335 When x 1 we have a n 1 for all n and, of course, a n diverges. e n n!
nn 10. a. This series is a special case of the series mentioned in the discussion that follows Exercise 7c
above. If we define
n
a n e n!
nn
for each n then
e n1 n1 ! a n 1
an lim n 1 nÝ n Ýn
lim n1 n1
e n n!
nn 1 which can be shown to be 1 . We can deduce...
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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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