Animalain or Zarkon. I have another proofs question for you.

Question

anonymous · Answer

attachment

anonymous · Answer

Yes.....?

anonymous · Answer

I posted the png.

anonymous · Answer

Found it.  Maybe by contradiction?

anonymous · Answer

Shall i have to use the BOLZANO-WEIERSTRASS THEOREM.

anonymous · Answer

How about take c(n) subseq of a(n) such that every positive element of a(n) is an element of c(n).  d(n) similarly picks up all the negative elements of a(n).  Summation of c(n) minus summation d(n) is bounded by hypothesis, so a(n) is absolutely convergent.

anonymous · Answer

Whaddaya think?

anonymous · Answer

when you say the summation of c(n) minus d(n) is bounded by hypothesis. what do you mean by hypothesis.

anonymous · Answer

The given fact that all subseq of a(n) converge when summed.

anonymous · Answer

A finite number plus a finite number is a finite number....

anonymous · Answer

Thanks. Ill take this into account. If you want to add more, feel free but i have to go study for an artifical organs test. Yay.

anonymous · Answer

Hope I was helpful.

anonymous · Answer

Key thing that I glossed over before is that summation of c(n) minus summation of d(n) equals summation of absolute value of a(n).  What we have is the positive elements minus the negative elements.