Example 4.4.  Show that the series  [Graphics:Images/ComplexSequenceSeriesMod_gr_155.gif]  is divergent.

Solution.  We know that the real series  [Graphics:Images/ComplexSequenceSeriesMod_gr_156.gif]  is divergent.  Hence, Theorem 4.4 implies that the given complex series is divergent.

Explore Solution 4.4.

Enter the formula for the series, and determine if the series converges or diverges.

[Graphics:../Images/ComplexSequenceSeriesMod_gr_157.gif]




[Graphics:../Images/ComplexSequenceSeriesMod_gr_158.gif]

[Graphics:../Images/ComplexSequenceSeriesMod_gr_159.gif]

[Graphics:../Images/ComplexSequenceSeriesMod_gr_160.gif]

 

 

Remark 1.  Since the answer returned for the imaginary part was  [Graphics:../Images/ComplexSequenceSeriesMod_gr_161.gif],  this means that a sum was not found.
It is known that the partial sums of the harmonic series grow slowly without bound.  
For example, adding 10,  100,  1000  and  10000 terms yields:

 

[Graphics:../Images/ComplexSequenceSeriesMod_gr_164.gif]

[Graphics:../Images/ComplexSequenceSeriesMod_gr_165.gif]

[Graphics:../Images/ComplexSequenceSeriesMod_gr_166.gif]

Therefore, the series slowly diverges to [Graphics:../Images/ComplexSequenceSeriesMod_gr_167.gif].  

 

 

Remark 2.  The integral test could also be used.

 

[Graphics:../Images/ComplexSequenceSeriesMod_gr_162.gif]

[Graphics:../Images/ComplexSequenceSeriesMod_gr_163.gif]

 

 

Use Mathematica to construct some of the partial sums of the infinite series.

[Graphics:../Images/ComplexSequenceSeriesMod_gr_168.gif]




[Graphics:../Images/ComplexSequenceSeriesMod_gr_169.gif]

[Graphics:../Images/ComplexSequenceSeriesMod_gr_170.gif]

[Graphics:../Images/ComplexSequenceSeriesMod_gr_171.gif]

 

 

 

 

[Graphics:../Images/ComplexSequenceSeriesMod_gr_172.gif]

[Graphics:../Images/ComplexSequenceSeriesMod_gr_173.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) 2006 John H. Mathews, Russell W. Howell