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Is there any typical method to prove a sequence has no convergent subsequence? B-W theorem tells us that if a real sequence is bounded, then it has a convergent subsequence. For example, if a series of sequence square is one, any counterexample this sequence has no convergent subsequence?

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  • $\begingroup$ $\frac{6}{\pi^2}\sum^\infty_{k=1}\frac{1}{k}$ $\endgroup$ – Surb Dec 4 '14 at 23:29
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Here is a characterization:

Let $(x_n)$ be such that for every $C>0$, there exists $N$ such that $C<|x_n|$ for every $n >N$. Let $(x_{n_k})$ a subsequence of $(x_n)$ then for every $C >0$ there is some $M$ such that $|x_{n_k}|>C$ for every $k >M$ and so $(x_{n_k})$ is not convergent.

Now let $(x_n)$ be a sequence that admit no convergent subsequence. Let $C >0$ and suppose by absurd that for every $N >0$ there exists $k > N$ such that $|x_{k}| < C$. In particular there is some $k_1>0$ such that $|x_{k_1}| < C$. There is now $k_2 >k_1$ such that $|x_{k_2}| < C$. Continuing this process we get a sequence of index $k_1<k_2<k_3< \ldots$ such that $|x_{k_i}| < C$ for every $i \in \mathbb N$. But then there is a bounded subsquence which admit a converging subsequence (by the theorem of Bolzano-Weierstrass, a contradiction.

It follows that: $(x_n)$ admit no converging subsequence if and only if for every $C >0$ there is some $N>0$ such that $|x_k|>C$ for every $k > N$.

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