proof of Silverman-Toeplitz theorem
Hence, the series is absolutely convergent which, in turn, implies that it converges.
Let denote the limit of the sequence as . Then . We need to show that, for every , there exists an integer such that
Since the sequence converges, there must exist an integer such that whenever .
By condition 3, there must exist constants such that
By condition 2, there exists a constant such that
whenever . By (1),
when . Hence, if and , we have
By the triangle inequality and (2), (3), (4) it follows that
|Title||proof of Silverman-Toeplitz theorem|
|Date of creation||2013-03-22 14:51:35|
|Last modified on||2013-03-22 14:51:35|
|Last modified by||rspuzio (6075)|