Example 4.5.  Show that the series [Graphics:Images/ComplexSequenceSeriesMod_gr_174.gif] is divergent.  

Solution.  Here we set  [Graphics:Images/ComplexSequenceSeriesMod_gr_175.gif]  and observe that  

            [Graphics:Images/ComplexSequenceSeriesMod_gr_176.gif].  

Thus [Graphics:Images/ComplexSequenceSeriesMod_gr_177.gif], and Theorem 4.5 implies that the series is not convergent;  hence it is divergent.

Explore Solution 4.5.

Enter the formula [Graphics:../Images/ComplexSequenceSeriesMod_gr_178.gif] for the terms of the sequence, and look at the modulus.

[Graphics:../Images/ComplexSequenceSeriesMod_gr_179.gif]




[Graphics:../Images/ComplexSequenceSeriesMod_gr_180.gif]

 

 

Hence [Graphics:../Images/ComplexSequenceSeriesMod_gr_181.gif], and the series is divergent.

Aside.  Mathematica will indicate that there is a problem with convergence.

[Graphics:../Images/ComplexSequenceSeriesMod_gr_182.gif]


[Graphics:../Images/ComplexSequenceSeriesMod_gr_183.gif]

[Graphics:../Images/ComplexSequenceSeriesMod_gr_184.gif]

 

 

Which is comforting to see that Mathematica knows the answer too!

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

[Graphics:../Images/ComplexSequenceSeriesMod_gr_185.gif]

 

 

 

 

 

[Graphics:../Images/ComplexSequenceSeriesMod_gr_186.gif]

 

 

Use Mathematica to compute more of the partial sums of the infinite series.

[Graphics:../Images/ComplexSequenceSeriesMod_gr_187.gif]



[Graphics:../Images/ComplexSequenceSeriesMod_gr_188.gif]

 

 

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

[Graphics:../Images/ComplexSequenceSeriesMod_gr_249.gif]





[Graphics:../Images/ComplexSequenceSeriesMod_gr_250.gif]

[Graphics:../Images/ComplexSequenceSeriesMod_gr_251.gif]

 

We see that the  series  [Graphics:Images/ComplexSequenceSeriesMod_gr_252.gif] is divergent.   

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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