1. ## subsequence

Prove that if $a_n$ is a sequence such that every subsequence of $a_n$ has a further subsequence converging to $0$, then $a_n \to 0$.

Let $\sigma: \mathbb{N} \to \mathbb{N}$ be a strictly increasing function. Suppose that $b_{n} = a_{\sigma(n)}$ is a subsequence of $a_n$. Then $c_{n} = b_{\sigma(n)} \to 0$ is a subsequence of $b_n$. If $\varepsilon > 0$, then there is an $N \in \mathbb{N}$ such that $n \geq N$ implies that $c_{n} \in (-\varepsilon, \varepsilon)$.

From here, I have to show that $a_n \in (-\varepsilon, \varepsilon)$.

Now what? $c_n \leq b_n \leq a_n$. Could I use the Squeeze theorem?

2. actually use the following theorem twice: $a_n \to L \Longleftrightarrow$ every subsequence of $a_n$ converges to $L$?

Is this correct?

3. possibly yes..

EDIT: but if i read and interpret it correct, it doesn't tell you that every subsequence of the subsequence of a_n also converges to 0.

4. Originally Posted by particlejohn
"every subsequence of $a_n$ converges to $L$"
you do not have this condition. we are told "every subsequence of has a further subsequence converging to " it is a different statement

5. yes, but this is a general theorem about subsequences. A subsequence is also a sequence right? Regardless, is my approach correct?

6. if you are going to use that, you have to show that every subsequence of the subsequence of a_n also converges to 0.

7. Originally Posted by kalagota
possibly yes..

EDIT: but if i read and interpret it correct, it doesn't tell you that every subsequence of the subsequence of a_n also converges to 0.

Oh okay. It just has $1$ subsequence converging to $0$?

8. Let $\{a_n\}$ be a sequence such that for any subsequence of $\{a_n\}$ we can find a subsequence which is convergent to zero. Assume that $a_n$ does not converge to zero, then it means for some $\epsilon > 0$ there are some $a_{n_1},a_{n_2}, ...$ such that $|a_{n_k}|\geq \epsilon$. The sequence $\{ a_{n_k}\}$ is a subsequence of $\{a_n\}$ but it does not have any subsequence which is convergent to zero since all its terms are $\geq \epsilon$ in absolute value. This is a contradiction. Thus, $\{ a_n\}$ must be convergent to zero.

This is Mine 11th Post!!!