1550251515-Classical_Complex_Analysis__Gonzalez_

188 Chapter4

In the convergence case it is easily seen that

Theorem 4. 7 The series I;~ an converges iff both series of real terms
I;~ Re an and I;~ Im an are convergent.

Proof Suppose that an= bn + icn, and let Sn= I:;~=l ak, Bn = I:;~=l bk,
and Cn = I:;~=l Ck. Clearly,

and by Theorem 4.1(10), if Bn -+ B and CN -+ C, then Sn -+ B + iC.
Conversely, if Sn-+ B + iC, then Bn-+ Band CN-+ C as n-+ oo.
This theorem reduces the question of convergence of a series of complex
terms to the investigation of the convergence of the real series whose terms
are the real parts and the imaginary parts, respectively, of the terms of
the original series.

Example The series I;~( -l)n-l [(1-i)/n] converges, since the real series

and

1

1

1 1

• 
  
  
  
  • --···
    2 3
  
  
  
  

1 1 1

--+---+··· 1 2 3

converge by the alternating series test (Leibniz's rule).

We note that the series of the absolute values of the terms of the given

series, namely,

00

2=-=v2 v2 ( 1+-+-+···^1 1 )

n=l n^2 3

diverges. This shows that the converse of Theorem 4.6 does not hold.

If an = bn + icn and the series I;~ an converges absolutely, then the

series I::~ bn and I;~ en are both absolutely convergent, as follows from

the inequalities lbnl S lanl and icnl S lanl and the comparison test for series

of real nonnegative terms. This observation, combined with Theorem 4.7,

provides an alternative proof of Theorem 4.6. In fact, I;~ lanl convergent

implies I;~ lbnl and I;~ lcnl convergent, which implies I;~ bn and I;~ Cn

convergent, and this in turn implies I;~ an convergent.