The Real And Complex Number Systems

complete.
Supplement
1. Show that the sequence
lim
n
2n!!
2n 1!! 0.
Proof: Let a and b be positive integers satisfying a b 1. Then we have
a!b a!b! a b! ab!. *
So, if we let fn 2n!, then we have, by (*)
2n!!
2n 1!! fn!
fn2n 1! 1
2n 1 0.
Hence, we know that lim n
2. Show that
2n!!
2n1!!
0.
a n 1 1 n 1 2 n 1 n n e 1/2 as n ,
where x means Gauss Symbol.
Proof: Since
x 1 2 x2 log1 x x, for all x 1, 1
we have
k n
k1
k
n 1 2
k
n
k n
2
log an log 1 n
k
k1
k n
k1
Consider i 2 n i 1 2 , then by Sandwish Theorem, we know that
lim n
log a n 1/2
which implies that a n e 1/2 as n .
3. Show that n! 1/n n for all n N. ( n! 1/n as n .)
Proof: We prove it by a special method following Gauss’ method. Consider
n! 1 k n
n n k 1 1
and thus let fk : kn k 1, it is easy to show that fk f1 n for all
k 1, 2, . . . , n. So, we have prove that
n! 2 n n
which implies that
n! 1/n n .
Remark: There are many and many method to show n! 1/n as n . We do not
give a detail proofs about it. But We method it as follows as references.
(a) By A. P. G. P. , we have
k1
n 1
k
n 1
n!
1/n
k
n