Hmm....I understand that almost all the terms of Pn and Qn will be between a+c and b+d, but how can I show that they converge to p+q?
Defn: Pn converges to point p means that if S is an open interval containing p then there is a positive integer N such that if n is a positive integer and n>N then Pn is in S.
I need to show that if the Pn converges to p and Qn converges to q then Pn+Qn converges to p+q.
I feel like I'm close to getting it, but not quite there yet, so any help would be greatly appreciated.
Note: I can't use the epsilon proof; I must use intervals.
That is just another way of saying ‘converges’. That is, if each open interval that contains p also contains ‘almost all’ [all but a finite collection] of the terms of then the sequence converges to p.
I deleted the post because there was an error. The interval was not general enough.
So here is a correction. Suppose that .
This is what I have so far.
p is in (a,b) and q is in (c,d). let epsilon1= min (p-a,p+b) and epsilon2=min(q-c,q+d).
For some N1, n>N1, -epsilon1<Pn-p<epsilon1 and
for some N2, n>N2, -epsilon2<Qn-q<epsilon2.
Now, I want to show that Pn+Qn converges to (p+q), but I'm not sure if simply combining the inequalities will show this.
Apologies for repeated questions on the same problem, but I'm having a difficult time wrapping my head around this convergence stuff.