# Thread: Bounded Sequence

1. ## Bounded Sequence

Suppose (an) is bounded, and that every convergent subsequence of (an) has limit L. Prove that $\lim_{n\rightarrow\infty}a_{n} = L$

2. ## Re: Bounded Sequence

Originally Posted by Kanwar245
Suppose (an) is bounded, and that every convergent subsequence of (an) has limit L. Prove that $\lim_{n\rightarrow\infty}a_{n} = L$
Well any sequence is a subsequence of itself.

3. ## Re: Bounded Sequence

Okay so from Bolzano-Wiererstrass since a_n is bounded, there exists atleast one convergent subsequence. Let that subsequence be a_n itself and we're done?

4. ## Re: Bounded Sequence

Originally Posted by Kanwar245
Okay so from Bolzano-Wiererstrass since a_n is bounded, there exists atleast one convergent subsequence. Let that subsequence be a_n itself and we're done?
Suppose the sequence does not converge to $L$.

What does that mean?

If it does converge to $L$ then if $\varepsilon > 0$ and $(L - \varepsilon ,L + \varepsilon )$ contains almost of the terms. So what is the negation of that?

5. ## Re: Bounded Sequence

Originally Posted by Kanwar245
Okay so from Bolzano-Wiererstrass since a_n is bounded, there exists atleast one convergent subsequence. Let that subsequence be a_n itself and we're done?
No, you can say that there exist at least one convergent subsequence but you [b]cannot[b] assert that it happens to be [itex]\{a_n\}[/itself]. You still need to prove that the sequence is itself convergent. Do, rather, what Plato suggested.

6. ## Re: Bounded Sequence

Originally Posted by HallsofIvy
No, you can say that there exist at least one convergent subsequence but you [b]cannot[b] assert that it happens to be [itex]\{a_n\}[/itself]. You still need to prove that the sequence is itself convergent. Do, rather, what Plato suggested.
okay