Monday, 15 August 2016

real analysis - conditionally convergent but not absolutely convergent series



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+:={nN:an0} and
A:={nN:an<0}, thus A+A=N and A+A=. Then both of the series nA+an and nAan are not absolutely convergent.




nA+an and nAan can't be both absolutely convergent at the same time is straightforward since it follows that nA+Aan=nNan 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) nA+an absolutely convergent, nAan not absolutely convergent;




(2) nA+an not absolutely convergent, nAan 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



mn=1an=mn=1bnmn=1cn,mn=1|an|=mn=1bn+mn=1cn.



If an converges and |an| diverges, then both bn and cn diverge, since



2mn=1bn=mn=1|an|+mn=1an,2mn=1cn=mn=1|an|mn=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:nN,bn0}={an:nA+,an0},{cn:nN,cn0}={an:nA},
and it easily shown that




nA+an=n=1bn=+nAan=n=1cn=


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