# Another Sequence proof too trivial to figure out

1. Dec 6, 2009

### koab1mjr

1. The problem statement, all variables and given/known data
Let Sn be a convergent sequence and the limit of Sn > a. Prove that there exists a number N such that when n > N implies Sn > a

2. Relevant equations

sequence definition of limit

3. The attempt at a solution

This one does not seem to have a point to me. By definition when n > N the sequence is epsilon close to a and they tell us the limit is greater than a so once n>N we are close to the lim of Sn so we must be above a. Am I missing something?

2. Dec 6, 2009

### Dick

Say lim Sn=b. And b>a. Just use the distance from b to a to make a particular choice of an epsilon.

3. Dec 6, 2009

### koab1mjr

I would let epsilon be less than distance between a and b where b = lim Sn. With that selected as my epsilon can I use the fact that Sn being convergen means it can be slower than d(a,b) therefore that N exists?

4. Dec 6, 2009

### Dick

Well, yes. Pick e.g. epsilon=|b-a|/2. Now write out the proof.

5. Dec 7, 2009

### koab1mjr

I think I got it now, my one question is why can't i just say epsilon less than d(a,b). That way when I say Sn is convergent b - epsilon is still bigger than a. You have d(a,b)/2. I thnk for the same reason but i might be overlooking something. Thanks for your help.

6. Dec 7, 2009

### HallsofIvy

Staff Emeritus
No reason why you couldn't. If b= lim Sn> a, and $\epsilon= d(a,b)= b-a$ then there exist N such that if n>N then |Sn-a|< \epsilon= b-a. Then -(b-a)= a- b< b- Sn< b-a so, subtracting b from each part a- 2b< -Sn< -a. Multiplying each part by -1, 2b-a> Sn> a.