Example:
The series
converges conditionally.
Note that we can not simply apply the alternating series test,
because this series is not alternating. We will rather apply the
more sophisticated Abel's test.
To apply Abel's test, we first need to identify two sequences.
Let
and
While it is clear that { b n } is decreasing
and convergent to zero, it is not clear at all whether the
sequence of partial sums of the a n's is bounded
or not. We will make an unfair detour here, using some basic
formulas from complex analysis:
Recall from Complex Analysis that
- cos(t) + i sin(t) = e it
for all real numbers t
where i is the basic complex number given as the square
root of -1. Taking this formula for granted, we have:
( cos(1) + i sin(1) ) +
( cos(2) + i sin(2) ) + ... +
( cos(n+1) + i sin(n+1) ) =
e i + e 2i + ... + e (n+1)i =
e i
( 1 + ( e i ) +
( e i ) 2 + ... +
( e i ) n ) =

Therefore, taking absolute values on both sides we can estimate
the last expression using the triangle inequality and the fact
that | e it | = 1 to obtain:
for all n
Finally, using the fact that
| a |
| a + i b |
for all real numbers a, b, we have
Putting the last two expressions together, we see that the
sequence of partial sums
- | cos(1) + cos(2) + ... + cos(n) |
is bounded for all n. Therefore, both conditions of
Abel's test are verified, and hence the original series
converges.
Of course, this is an 'unfair' proof, since it uses complex
analysis and we have not even defined properly what a complex
number is. Therefore, another proof of this example is
necessary that uses only real analysis. Such a proof is left as
an exercise, as well as the proof that the series does not
converge absolutely.
To Theory |
Glossary |
Map
(bgw)