Example 4.4. Show
that the series
is
divergent.
Solution. We know that the real
series
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_158.gif]](../Images/ComplexSequenceSeriesMod_gr_158.gif)
![]()
![[Graphics:../Images/ComplexSequenceSeriesMod_gr_160.gif]](../Images/ComplexSequenceSeriesMod_gr_160.gif)
Remark 1. Since the
answer returned for the imaginary part was
, 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]](../Images/ComplexSequenceSeriesMod_gr_164.gif)
![[Graphics:../Images/ComplexSequenceSeriesMod_gr_165.gif]](../Images/ComplexSequenceSeriesMod_gr_165.gif)
![[Graphics:../Images/ComplexSequenceSeriesMod_gr_166.gif]](../Images/ComplexSequenceSeriesMod_gr_166.gif)
Therefore, the series slowly diverges to
.
Remark 2. The integral test could also be used.
![[Graphics:../Images/ComplexSequenceSeriesMod_gr_162.gif]](../Images/ComplexSequenceSeriesMod_gr_162.gif)
![[Graphics:../Images/ComplexSequenceSeriesMod_gr_163.gif]](../Images/ComplexSequenceSeriesMod_gr_163.gif)
Use Mathematica to construct some of the partial sums of the infinite series.
![[Graphics:../Images/ComplexSequenceSeriesMod_gr_169.gif]](../Images/ComplexSequenceSeriesMod_gr_169.gif)
![]()
![]()
![[Graphics:../Images/ComplexSequenceSeriesMod_gr_172.gif]](../Images/ComplexSequenceSeriesMod_gr_172.gif)
![[Graphics:../Images/ComplexSequenceSeriesMod_gr_173.gif]](../Images/ComplexSequenceSeriesMod_gr_173.gif)