PlanetMath: real part series and imaginary part series

(more info)

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real part series and imaginary part series

(Theorem)

Theorem 1. Given the series

$\displaystyle z_1+z_2+z_3+\ldots$

(1)

with the real parts of its terms $\Re{z_n} = a_n$ and the imaginary parts of its terms $\Im{z_n} = b_n$ ( $n = 1,\,2,\,3,\,\ldots$ ). If the series (1) converges and its sum is

, where

and

are real, then also the series

$\displaystyle a_1+a_2+a_3+\ldots\;\;\;$ and $\displaystyle \;\;\;b_1+b_2+b_3+\ldots$

converge and their sums are

and

, respectively. The converse is valid as well.

Proof. Let $\varepsilon$ be an arbitrary positive number. Denote the partial sum of (1) by

$\displaystyle S_n = z_1+\ldots+z_n = (a_1+ib_1)+\ldots+(a_n+ib_n) = (a_1+\ldots+a_n)+i(b_1+\ldots+b_n) := A_n+iB_n$

( $n = 1,\,2,\,3,\,\ldots$ ). When (1) converges to the sum

, then there is a number $n_\varepsilon$ such that for any integer $n > n_\varepsilon$ we have

$\displaystyle \vert(A_n-A)+i(B_n-B)\vert = \vert(A_n+iB_n)-(A+iB)\vert < \varepsilon.$

But a complex number is always absolutely at least equal to the real part (see the inequalities in modulus of complex number), and therefore $\vert A_n-A\vert \leqq \vert(A_n-A)+i(B_n-B)\vert < \varepsilon$ , similarly $\vert B_n-B\vert \leqq \vert(A_n-A)+i(B_n-B)\vert < \varepsilon$ as soon as $n > n_\varepsilon$ . Hence, $A_n \to A$ and $B_n \to B$ as $n \to \infty$ . This means the convergences

$\displaystyle a_1+a_2+a_3+\ldots = A\;\;\;$ and $\displaystyle \;\;\;b_1+b_2+b_3+\ldots = B,$

Q.E.D. The converse part is straightforward.

Theorem 2. Notations same as in the preceding theorem. The series

$\displaystyle \vert z_1\vert+\vert z_2\vert+\vert z_3\vert+\ldots$

converges if and only if the series

$\displaystyle a_1+a_2+a_3+\ldots\;\;\;$ and $\displaystyle \;\;\;b_1+b_2+b_3+\ldots$

converge absolutely.

Proof. Use the inequalities

$\displaystyle 0 \leqq \vert a_n\vert \leqq \vert z_n\vert,\quad 0 \leqq \vert b_n\vert \leqq \vert z_n\vert$

and

$\displaystyle 0 \leqq \vert z_n\vert \leqq \vert a_n\vert+\vert b_n\vert$

for using the comparison test.

Theorem 3. If the series $\displaystyle\sum_{n=1}^\infty\vert z_n\vert$ converges, then also the series $\displaystyle\sum_{n=1}^\infty z_n$ converges and we have

$\displaystyle \left\vert\sum_{n=1}^\infty z_n\right\vert \leqq \sum_{n=1}^\infty\vert z_n\vert.$

Proof. By theorem 2, the convergence of $\sum\vert z_n\vert$ implies the convergence of $\sum a_n$ and $\sum b_n$ , which, by theorem 1, in turn imply the convergence of $\sum z_n$ . Since for every the triangle inequality guarantees the inequality