# Real Analysis

1. Apr 23, 2014

### dmac1215

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

2. Apr 24, 2014

### jbunniii

Hint: the sum (or difference) of two convergent sequences is convergent.

Last edited: Apr 24, 2014