proof of Silverman-Toeplitz theorem

First, we shall show that the series n=0amnzn converges. Since the sequence {zn} converges, it must be boundedPlanetmathPlanetmath in absolute valueMathworldPlanetmathPlanetmathPlanetmath — there must exist a constant K>0 such that |zn|K for all n. Hence, |amnzn|K|amn|. Summing this gives


Hence, the series n=0amnzn is absolutely convergent which, in turn, implies that it converges.

Let z denote the limit of the sequence {zn} as n. Then |z|K. We need to show that, for every ϵ>0, there exists an integer M such that


whenever m>M.

Since the sequence {zn} converges, there must exist an integer n1 such that |zn-z|<ϵ4B whenever n>n1.

By condition 3, there must exist constants mo,m1,,mn1 such that

|amn|<ϵ4(n1+1)(K+1)for0nn1andm>mn. (1)

Choose m=max{m0,m1,,mn1}. Then

|n=0n1amnzn|n=0n1|amnzn|<n=0n1|zn|ϵ4(n1+1)(K+1)<ϵ4 (2)

when m>m.

By condition 2, there exists a constant m′′ such that


whenever m>m′′. By (1),


when m>m. Hence, if m>m and m>m′′, we have

|n=n1+1amn-1|<ϵ2(|z|+1). (3)


|n=n1+1amn(zn-z)|n=n1+1|amn||zn-z|<ϵ4Bn=0|amn|ϵ4. (4)

By the triangle inequalityMathworldMathworldPlanetmath and (2), (3), (4) it follows that

Title proof of Silverman-Toeplitz theorem
Canonical name ProofOfSilvermanToeplitzTheorem
Date of creation 2013-03-22 14:51:35
Last modified on 2013-03-22 14:51:35
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 27
Author rspuzio (6075)
Entry type Proof
Classification msc 40B05