Proving Divergence of (tn + sn)

  • Thread starter Thread starter dmac1215
  • Start date Start date
  • Tags Tags
    Divergence
dmac1215
Messages
1
Reaction score
0
I have a super round about way to prove this, but I'm having trouble finding a succinct proof

Let (tn) be diverge and (sn) converge. Show (tn+sn) diverges

The way I was doing involved considering that tn was unbounded, then showing it (sn+tn) is divergent.

Then I had to consider that tn is bounded and oscillatory, consider convergent subsequences, and show (sn+tn) had no unique limit, and therefore diverges. This part of the proof seemed less clear and I'm not sure if I can assert that because I have convergent subsequences of (tn) with multiple limits that (sn+tn) also has multiple limits.

I figure there has got to be some simple contradiction proof involving some triangle inequality trick that I'm just missing.

Thanks
 
Physics news on Phys.org
Hint: the sum (or difference) of two convergent sequences is convergent.
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top