# Existence of sequence

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!

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

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

converges to x...

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

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?

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) ?

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.

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.