pn=an for an>0, =0 for an<0; qn=an for an<0, =0 for an>0

[felvt]

Let ?an be given. For each integer n, let pn=an if an is positive and assign pn=0 if an is negative. Let qn=an if an is negative and qn=0 if an is positive. Show that if ?an converges conditionally, then both ?pn and ?qn diverge.

I am not sure where to start with this. I believe I am supposed to use the fact that ? abs(an) diverges since it is conditional convergance.

 

[Pklarreich]

Re: Proof concerning Infinite Series

OK, the pn's are the positive terms and the qn's are the negative terms.

if ?an converges conditionally, then ? |an| diverges, i.e. -> infinity -- def of cond conv.

To prove both ?pn and ?qn diverge, assume at least one converges, say ?pn.

So ?pn is finite; call it P. From ? |an|,which is infinite, delete the pn terms. What remains must still be infinite, and what remains is simply ? |qn| = infinity. Now these qn are the negative terms, so ? |qn| = ? (- qn), so = ? qn = - infinity.

That means, formally, that given any P, which can be P = ? pn, there exists N such that ?(up to N)|qn| > P.

Therefore the entire sum (no absolute values) must diverge, not converge -- the negative partial sums at some point overwhelm the complete positive sum.

The argument if it is the ? qn that converges is similar.

 

[felvt]

Re: Proof concerning Infinite Series

Right on, thank you much.