I'm stuck on the following exercise:
Let ∑∞n=0an be a series of real numbers which is
conditionally convergent, but not absolutely convergent.
Define the
sets A+:={n∈N:an≥0} and
A−:={n∈N:an<0}, thus A+⋃A−=N and A+⋂A−=∅. Then both of the series ∑n∈A+an and ∑n∈A−an are not absolutely convergent.
∑n∈A+an and ∑n∈A−an can't be both absolutely convergent at the same time is straightforward since it follows that ∑n∈A+⋃A−an=∑n∈Nan is absolutely convergent, a contradiction.
What I haven't been able to do to is exclude the possibility that one of the two converges and the other diverges, i.e. that the remaining two cases:
(1) ∑n∈A+an absolutely convergent, ∑n∈A−an not absolutely convergent;
(2) ∑n∈A+an not absolutely convergent, ∑n∈A−an absolutely convergent;
it leads to a contradiction.
So, I would appreciate any hints about how to carry out this part of the proof.
Best regards,
Lorenzo.
Answer
Let bn=(|an|+an)/2 and cn=(|an|−an)/2.
Then the partial sums satisfy
m∑n=1an=m∑n=1bn−m∑n=1cn,m∑n=1|an|=m∑n=1bn+m∑n=1cn.
If ∑an converges and ∑|an| diverges, then both ∑bn and ∑cn diverge, since
2m∑n=1bn=m∑n=1|an|+m∑n=1an,2m∑n=1cn=m∑n=1|an|−m∑n=1an,
and the sum or difference of a divergent and convergent series is divergent.
Furthermore, we have divergence to +∞ in each case, as the partial sums of |an| form a non-negative, non-decreasing sequence.
Note that
{bn:n∈N,bn≠0}={an:n∈A+,an≠0},{cn:n∈N,cn≠0}={−an:n∈A−},
and it easily shown that
∑n∈A+an=∞∑n=1bn=+∞∑n∈A−an=−∞∑n=1cn=−∞
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