The Real And Complex Number Systems

Limits And Continuity
Limits of sequence
4.1 Prove each of the following statements about sequences in C.
(a) z n 0if|z| 1; z n diverges if |z| 1.
Proof: For the part: z n 0if|z| 1. Given 0, we want to find that there exists a
positive integer N such that as n N, wehave
|z n 0| .
Note that log|z| 0since|z| 1, hence if we choose a positive integer N log |z|
1,
then as n N, wehave
|z n 0| .
For the part: z n diverges if |z| 1. Assume that z n converges to L, thengiven
1, there exists a positive integer N 1 such that as n N 1 ,wehave
|z n L| 1
|z| n 1 |L|. *
However, note that log|z| 0since|z| 1, if we choose a positive integer
N max log |z|
1 |L| 1, N 1 , then we have
|z| N 1 |L|
which contradicts (*). Hence, z n diverges if |z| 1.
Remark: 1. Given any complex number z C 0, lim n|z| 1/n 1.
2. Keep lim n n! 1/n in mind.
3. In fact, z n is unbounded if |z| 1. ( z n diverges if |z| 1. ) Since given
M 1, and choose a positive integer N log |z|
M 1, then |z| N M.
(b) If z n 0andifc n is bounded, then c n z n 0.
Proof: Since c n is bounded, say its bound M, i.e., |c n| M for all n N. In
addition, since z n 0, given 0, there exists a positive integer N such that as n N,
we have
|z n 0| /M
which implies that as n N, wehave
|c n z n| M|z n| .
That is, lim n c n z n 0.
(c) z n /n! 0 for every complex z.
Proof: Given a complex z, and thus find a positive integer N such that |z| N/2.
Consider (let n N).
z n
z N
z nN
n! N! N 1N 2 n
Hence, z n /n! 0 for every complex z.
zN
N!
1
2
nN
0asn .
Remark: There is another proof by using the fact
n1
a n converges which implies
z
a n 0. Since n
n1
converges by ratio test for every complex z, then we have
n!