Limits And Continuity
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چکیده
4.1 Prove each of the following statements about sequences in C. (a) zn 0 if |z| 1; zn diverges if |z| 1. Proof: For the part: zn 0 if |z| 1. Given 0, we want to find that there exists a positive integer N such that as n N, we have |zn 0| . Note that log|z| 0 since |z| 1, hence if we choose a positive integer N log|z| 1, then as n N, we have |zn 0| . For the part: zn diverges if |z| 1. Assume that zn converges to L, then given 1, there exists a positive integer N1 such that as n N1, we have |zn L| 1 |z| 1 |L|. * However, note that log|z| 0 since |z| 1, if we choose a positive integer N max log|z|1 |L| 1,N1 , then we have |z| 1 |L| which contradicts (*). Hence, zn diverges if |z| 1. Remark: 1. Given any complex number z C 0, limn|z| 1. 2. Keep limnn! in mind. 3. In fact, zn is unbounded if |z| 1. ( zn diverges if |z| 1. ) Since given M 1, and choose a positive integer N log|z|M 1, then |z| N M. (b) If zn 0 and if cn is bounded, then cnzn 0. Proof: Since cn is bounded, say its bound M, i.e., |cn | M for all n N. In addition, since zn 0, given 0, there exists a positive integer N such that as n N, we have |zn 0| /M which implies that as n N, we have |cnzn | M|zn | . That is, limn cnzn 0. (c) zn/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). zn n! zN N! znN N 1N 2 n zN N! 1 2 nN 0 as n .
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