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Unformatted text preview: ct that
2 1
j! 0e
j0 e
21 ! we deduce that
e 11 1
2
6 e
which gives us e 3.
d. Prove that if n is a positive integer we have n 1
j! 0e
j0 3
n1 ! and deduce that
n lim
nÝ j0 1 e.
j! The inequality
n 0e
j0 1
j! 3
n1 ! 1
j! e
n1 ! follows at once from the inequality
n 0e
j0 and the fact that e 3. The fact that 259 n lim e
nÝ 1
j! j0 0 now follows at once from the sandwich theorem.
e. Prove that the number e is irrational. Solution: To obtain a contradiction, suppose that e is rational. Choose two positive integers p
p
and q such that e q . Since n lim
nÝ 1 e,
j! j0 we have
n lim
nÝ q n jq1 1
j! nÝ
lim 1
j! j0 q j0 1
j! e
j0 1
j! and therefore, multiplying by q!, we obtain
n lim
nÝ
Since both of the numbers q
q!
j0 j! jq1
p
and q q!
j! p
q q! q
j0 q!
j! q! are integers, we deduce that the number
n lim
nÝ q!
j! jq1 must also be an integer. Now, on the one hand, the number
n q!
j! lim nÝ jq1 must be positive because
n lim
nÝ
However, whenever j jq1 q 1 q!
j! q!
1 0.
q1
j!
jq1 q 1 we have
q!
1
j!
q1 q2 j 1
q1 jq and therefore
n lim
nÝ jq1 q!
j! nq n nÝ
lim jq1 1
q1 jq 1 1
nÝ
lim
q1 1 nÝ
lim
1
q 1
1
q 1 j1 1
q1 j nq 1
q 1 We have therefore reached the impossible conclusion that the number
n lim
nÝ jq1 q!
j! is an integer that lies between 0 and 1 and so we have reached the desired contradiction.
9. Prove that if x n is a sequence of positive numbers and if 260 n
lim xx 1
n nÝ then
lim n x n . nÝ Deduce that
lim n e.
n! nÝn Solution: We look first at the case 0. In order to show that
lim n x n nÝ we shall show that no number other than can be a partial limit of the sequence n x n . Suppose that u is
any number that is unequal to and choose a number between 0 and 1 such that the interval
,
does not contain the number u. We now use the facts that
and
lim x n1
n Ý xn
to choose a positive integer N such that the inequality
n
xx 1
n
holds whenever n N.
Now whenever n N we have
xn xN x N1
xN x N2
x N1 xx n1
n and so
x N
xn xN nN nN which we can express as
n xN
N N n xn n N xN
N .
Since
lim
n Ýn xN
N N n Ýn
lim N xN
1
N x n can lie outside of the interval
,
n xn .
and we conclude that u is not a partial limit of
we deduce that no partial limit of the sequence n We now consider the case 0. In order to show that
lim n x n 0
nÝ
and, once again, we shall demonstrate that no number unequal to can be a partial limit of the sequence
n x n . Since it is already clear that a negative number can’t be a partial limit of n x n , we can confine
our attention to positive numbers. Suppose that u 0 and choose a number between 0 and 1 such that
u. We now use the fact that
lim x n1 0
n Ý xn
to choose an integer N such that the inequality 261 x n 1
xn
holds whenever n N. Now whenever n N we have
xn xN x N1
xN x N2
x N1 xx n1
n xN...
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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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