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Unformatted text preview: the divergence of
a n from Raabe’s test but
2
there is really no need for such big guns. Since the limit above is negative we know that
a n 1 1
an
for sufficiently large n and so a n cannot approach 0 as n Ý.
b. nn
e n n!
If we define
an nn
e n n! for each n then
n1 n1 lim n 1 nÝ a n 1
an n Ýn
lim 1 e n1 n1 !
nn
e n n! 1 n Ýn e
1 1
e lim
n
and so it follows from Raabe’s test that
a n diverges.
11. 2n !
n n! 2
4 n 1
2 p where p is a given number We already know that this series diverges when p 1 and from the comparison test it follows that
the series also diverges when p 1. From now on we suppose that p 1. We define
an 2n !
4 n n! 2 p for each n and observe that
a n 1
an 2n 1
2n 2 p and so
2n 1 p
2n 2
2n 2 p 2n 1 p
n Ýn
lim
2n 2 p
p
To see that the latter limit is 2 , we use mean value theorem to chooce a number c n between 2n 1
and 2n 2 such that
lim n 1 nÝ n a n 1
an n Ýn 1
lim 2n 2 p 2n 1
2n 2 p and observe that 336 p n p1 pc n
2n 2 p n p 2n 1
2n 2 p1 p1 pc n
2n 2 n p n p p 2n 2
2n 2 p1 . p We deduce from Raabe’s test that the given series converges when p 2 and diverges when
p 2.
We must now consider the case p 2. In this case we shall use the more powerful form of Raabe
test:
a n1 lim n log n 1 1
2n 1 2
lim n log n 1 1
an
n
n
nÝ
nÝ
2n 2
5n 4
4n n 1 n Ý n log n
lim 0 2 and we conclude that the series diverges when p 2.
n 2 12.
2 n n1 e n n3
2
n 11/12 Ü jn1 2j1
jj 2j 1
2 j! Solution: Start off by supplying the definition
n n 2 an 1 2j 1 2j 2
j j j! Ü n n3 j1
2 n n1 e 2 n 11/12
to Scientific Notebook and make the selection that the subscript n is a function argument. You then have
a n 1
an 2 n1 n1 1 e n1
2 n1 n1 3
2 n1 11/12 n Ü 2
2 n n1 e n
Ü j1 1 n n3
2
n 11/12 n
j1 2j1 2j1
jj 2j 1
2 j! . 2j 1
2 jj j! Do not ask Scientific Notebook to simplify this expression as it stands. First observe that
n1
2
n1 n1 3
n1 1 e
2 2 n 3 a n 1
an 2 n1 2n2 1
2 n1 n1 n1 11/12 n 1 ! n 2 n n3
2 n n1 e 2 n 11/12 and then evaluate to obtain
a n 1 4
an n1 n1 23 n
12 12n 9 4n 2 n 2n 3 5
12 n2 n 11 e n2 and then supply the definition
gn 4 n1 n1 23 n
12 4n 2 n 12n 9 2n 3
n2 to Scientific Notebook. Thus
n
g n aa 1 .
n
Finally you can use the Evaluate button to obtain
lim g n 1
nÝ lim n 1 nÝ lim n log n nÝ 1 and so the series diverges. 337 gn
1
n 1
gn 01 5
12 n 11 e n2 13. 1 2 n
1 2 n 1
1 where and are given numbers Our understanding of this problem is that is automatically prohibited from being 0 or any negative
integer.
For each n we define
1 2 n 1
an
1 2 n 1
and we observe that
 n 
n
lim
lim n 1 aa 1 n Ý n 1
nÝ
n
 n 
n .
n
a n converges if 1 and diverges if 1. In the
n Ýn 1
lim We deduce from Raabe’s test that
event that 1 we have
1 2 n 1
an
1 2 n 1
Exce...
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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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