real part series and imaginary part series
Theorem 1. Given the series
converge and their sums are and , respectively. The converse is valid as well.
Proof. Let be an arbitrary positive number. Denote the partial sum of (1) by
(). When (1) converges to the sum , then there is a number such that for any integer we have
But a complex number is always absolutely at least equal to the real part (see the inequalities in modulus of complex number), and therefore , similarly as soon as . Hence, and as . This means the convergences
Q.E.D. The converse part is straightforward.
Theorem 2. Notations same as in the preceding theorem. The series
converges if and only if the series
converge absolutely (http://planetmath.org/AbsoluteConvergence).
Proof. Use the inequalities
for using the comparison test.
Theorem 3. If the series converges, then also the series converges and we have
Proof. By theorem 2, the convergence of implies the convergence of and , which, by theorem 1, in turn imply the convergence of . Since for every the triangle inequality guarantees the inequality
then we must have the asserted limit inequality, too.
|Title||real part series and imaginary part series|
|Date of creation||2013-03-22 17:28:08|
|Last modified on||2013-03-22 17:28:08|
|Last modified by||pahio (2872)|