Example 4.5. Show
that the series
is divergent.
Solution. Here we set
and
observe that
.
Thus
,
and Theorem 4.5 implies that the series is not
convergent; hence it is divergent.
Explore Solution 4.5.
Enter the formula
for the terms of the sequence, and look at the modulus.
![[Graphics:../Images/ComplexSequenceSeriesMod_gr_180.gif]](../Images/ComplexSequenceSeriesMod_gr_180.gif)
Hence
,
and the series is divergent.
Aside. Mathematica will indicate that there is a problem with convergence.
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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]](../Images/ComplexSequenceSeriesMod_gr_185.gif)
![[Graphics:../Images/ComplexSequenceSeriesMod_gr_186.gif]](../Images/ComplexSequenceSeriesMod_gr_186.gif)
Use Mathematica to compute more of the partial sums of the infinite series.
![[Graphics:../Images/ComplexSequenceSeriesMod_gr_188.gif]](../Images/ComplexSequenceSeriesMod_gr_188.gif)
Use Mathematica to plot some of the partial sums of the infinite series.
![[Graphics:../Images/ComplexSequenceSeriesMod_gr_250.gif]](../Images/ComplexSequenceSeriesMod_gr_250.gif)
![[Graphics:../Images/ComplexSequenceSeriesMod_gr_251.gif]](../Images/ComplexSequenceSeriesMod_gr_251.gif)
We see that the series
is divergent.