# sum of 2 convergent sequences

• Sep 24th 2009, 12:47 PM
spikedpunch
sum of 2 convergent sequences
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.
• Sep 24th 2009, 01:54 PM
spikedpunch
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?
• Sep 24th 2009, 02:15 PM
Plato
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 $\left( {p_n } \right)$ 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 $\left( {\exists (a,b)} \right)\left[ {p + q \in (a,b)} \right]$.
• Sep 24th 2009, 04:48 PM
spikedpunch
Still not clear...
• Sep 24th 2009, 05:15 PM
Plato
Quote:

Originally Posted by spikedpunch
Still not clear...

Well think about how you do this with epsilon-neighborhoods.
Then adapt that to open intervals.
• Sep 25th 2009, 08:02 AM
spikedpunch
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.
Let N=max(N1,N2).

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.
• Sep 25th 2009, 08:42 AM
Plato
You must start with $p+q\in (a,b)$.
So $\left( {\exists \varepsilon > 0} \right)\left[ {\left( {p + q - \varepsilon ,p + q + \varepsilon } \right) \subset (a,b)} \right]$
Now consider two open intervals: $O_p = \left( {p - \frac{\varepsilon }{2},p + \frac{\varepsilon }{2}} \right)\;\& \;O_q = \left( {q - \frac{\varepsilon }{2},q + \frac{\varepsilon }{2}} \right)$
It should be clear that $p\in O_p\;\& \;q\in O_q$.
If $x\in O_p\;\& \;y\in O_q$ can you show that $x+y\in (a,b)?$
• Sep 25th 2009, 10:22 AM
spikedpunch
I don't see how the sequences Pn and Qn fit in to what you are saying.
• Sep 25th 2009, 10:35 AM
Plato
Quote:

Originally Posted by spikedpunch
I don't see how the sequences Pn and Qn fit in to what you are saying.

Almost all of the terms of $\left(p_n\right)$ are in $O_p$ by convergence.
Almost all of the terms of $\left(q_n\right)$ are in $O_q$ by convergence.
Therefore Almost all of the terms of $\left(p_n+q_n\right)$ are in $(a,b)$
Because $p+q\in(a,b)$ that is sufficient to prove that $\left(p_n+q_n\right)\to(p+q)$
• Sep 25th 2009, 10:42 AM
spikedpunch
I can't use the phrase "almost all the terms" in my proof. I can only use the definition given in my initial post.

Again, apologies.