# Existence of sequence

1. Nov 4, 2011

### autre

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!

2. Nov 4, 2011

### micromass

Take the constant sequence $a_k=x$. Surely, the sequence

$$x,~x,~x~,...$$

converges to x...

3. Nov 4, 2011

### autre

Sorry, forgot to mention that the sequence is in Q.

4. Nov 4, 2011

### micromass

Well, what can you go from?? Can you use that Q is dense in R?? Or must you use the definition of R?

5. Nov 4, 2011

### autre

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. Nov 4, 2011

### micromass

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. Nov 4, 2011

### SteveL27

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.