The Real And Complex Number Systems

z n /n! 0 for every complex z.
(d) If a n n 2 2 n, then a n 0.
Proof: Since
0 a n n 2 2 n 2
n 2 2 n 1 n for all n N,
and lim n 1/n 0, we have a n 0asn by Sandwich Theorem.
4.2 If a n2 a n1 a n /2 for all n 1, show that a n a 1 2a 2 /3. Hint:
a n2 a n1 1 2 a n a n1 .
Proof: Since a n2 a n1 a n /2 for all n 1, we have b n1 b n /2, where
b n a n1 a n . So, we have
n
b n1 1 b1 0asn . *
2
Consider
which implies that
So we have
n1
a n2 a 2 b k 1
2 n
k2
k1
b n
3a n1
2
b k
a 1 2a 2
2
a n a 1 2a 2 /3 by (*).
1
2
a n1 a 1
.
4.3 If 0 x 1 1andifx n1 1 1 x n for all n 1, prove that x n is a
decreasing sequence with limit 0. Prove also that x n1 /x n 1 2 .
Proof: Claim that 0 x n 1 for all n N. We prove the claim by Mathematical
Induction. As n 1, there is nothing to prove. Suppose that n k holds, i.e., 0 x k 1,
then as n k 1, we have
0 x k1 1 1 x k 1 by induction hypothesis.
So, by Mathematical Induction, we have proved the claim. Use the claim, and then we
have
x
x n1 x n 1 x n 1 x n n x n 1
1 x n 2 1 x n 0since0 x n 1.
So, we know that the sequence x n is a decreasing sequence. Since 0 x n 1 for all
n N, by Completeness of R, (Thatis,a monotonic sequence in R which is bounded is
a convergent sequence.) Hence, we have proved that x n is a convergent sequence,
denoted its limit by x. Note that since
x n1 1 1 x n for all n N,
we have x lim n x n1 lim n 1 1 x n 1 1 x which implies xx 1 0.
Since x n is a decreasing sequence with 0 x n 1 for all n N, we finally have x 0.
For proof of x n1 /x n 1 .Since 2
1 1 x
lim
x0 x 1 2
then we have
*