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A sequence $$ does not converge to $L$.Then how can we prove that there exist a monotone subsequence of $$ which also does not converge to $L$.

My try: Since $$ does not converge to $L$,then there exist a subsequence which does not converge to $L$ and every sequence has a monotone subsequence .So the result follows.Thank you.plz check.

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    In "my try", (1) the first part is meaningless; if the sequence doesn't converge, you may as well take the whole thing as the subsequence, and (2) just because a sequence doesn't converge doesn't mean that a monotone subsequence also doesn't converge. – 2017-02-21
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    can you please give me some hint to prove this? – 2017-02-21
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    Unfortunately your argument is not sufficient. The subsequence of $\langle a_n \rangle$ which does not converge to $L$ could just as well be the entire sequence. Consider the case that every montone subsequence converges to $L$. – 2017-02-21
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    @hardmath: plz check:suppose every monotone subsequence converges which leads to $a_n $ converges ,so there exist some monotone subsequence which does not converges since $a_n$ does not converges. – 2017-02-21
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    Prove that for all monotone subsequence of $$ which also are converge to $L$, then $$ converges to $L$. – 2017-02-21
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    I'm the \langle \rangle fairy, here to let you know that $\langle, \rangle$ plays nicer with TeX than <, > does :) – 2017-02-21

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You should indeed use the lemma : every sequence of real numbers has a monotonic subsequence.

Since $(a_n)$ does not converge to $L$, there exists $\epsilon>0$ such that $\forall N\in\mathbb{N},\exists n\ge N;\left|a_n-L\right|\ge\epsilon$.

This implies the existence of some subsequence $(a_{\phi(n)})$ such that $\forall n\in\mathbb{N},\left|a_{\phi(n)}-L\right|\ge\epsilon$.

Now take a monotonic subsequence of $(a_{\phi(n)})$