Proving Existence of Sequence Converging to x in R

In summary, the goal is to find a sequence {a_k} that converges to some x in the real numbers. One way to do this is by using the definition of the limit of a sequence and setting x as the limit of {a_k}. Another approach is to use the fact that the rational numbers are dense in the real numbers and construct a sequence of rational numbers that converges to x. This can be done by taking successive truncations of x's decimal expansion.
  • #1
autre
117
0
I asked this question before but I totally misunderstood what it was asking. Basically, I need to find that there exists a sequence {a_k} such that it converges to x for some x in R.

Since the real numbers are equivalence classes of convergent Cauchy sequences the result seems fairly obvious, but I was thinking of instead of dealing with equivalence classes I could use the definition of the limit of a sequence and have x be the limit of {a_k}.

I'm having some trouble starting up, though. Do I have to find a sequence that must converge to some arbitrary x to prove its existence?

Thanks!
 
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  • #2
Take the constant sequence [itex]a_k=x[/itex]. Surely, the sequence

[tex]x,~x,~x~,...[/tex]

converges to x...
 
  • #3
Sorry, forgot to mention that the sequence is in Q.
 
  • #4
autre said:
Sorry, forgot to mention that the sequence is in Q.

Well, what can you go from?? Can you use that Q is dense in R?? Or must you use the definition of R?
 
  • #5
does the fact that Q is dense in R tell me that if I take any member of {s_n} it can exist in an interval of, say, (x-1/n,x+1/n) ?
 
  • #6
autre said:
does the fact that Q is dense in R tell me that if I take any member of {s_n} it can exist in an interval of, say, (x-1/n,x+1/n) ?

Q being dense says that for each x in R, there exists a rational in (x-1/n,x+1/n). You can use this to construct a rational sequence.
 
  • #7
autre said:
Sorry, forgot to mention that the sequence is in Q.

Can you just take successive truncations of x's decimal expansion? For example

3, 3.1, 3.14, 3.141, 3.1415, ... is a sequence of rationals converging to pi.
 

1. What does it mean for a sequence to converge to a specific value in the real numbers?

Convergence in the real numbers means that as the terms in the sequence get closer and closer to a specific value, the difference between them and that value becomes smaller and smaller. This is also known as the limit of the sequence.

2. How do you prove the existence of a sequence converging to a specific value in the real numbers?

To prove the existence of a sequence converging to a specific value, we must show that for any given distance away from that value, there exists a term in the sequence that is within that distance. This can be done using the formal definition of a limit or by showing that the terms in the sequence get arbitrarily close to the desired value.

3. What is the formal definition of a limit for a sequence in the real numbers?

The formal definition of a limit for a sequence in the real numbers is that for any given distance ε > 0, there exists a natural number N such that for all n ≥ N, the distance between the nth term of the sequence and the limit value is less than ε.

4. Can you provide an example of a sequence converging to a specific value in the real numbers?

One example of a sequence converging to a specific value in the real numbers is the sequence {1/n} which converges to 0 as n approaches infinity. As n gets larger, the terms in the sequence get closer and closer to 0, and the difference between them and 0 becomes smaller and smaller.

5. Is it possible for a sequence to have more than one limit in the real numbers?

No, a sequence in the real numbers can only have one limit. If a sequence has more than one limit, it is not a convergent sequence. However, a sequence can be divergent and not have a limit.

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