# $\{x_n\}$ be a bounded sequence of real number we need to show

Let $\{x_n\}$ be a bounded sequence of real numbers. We need to show that there exist a real number $\alpha$ and positive integers $n_1,n_2,\dots$ such that $n_1<n_2<\dots$ and $\sum_{k}|x_{n_k}-\alpha|<\infty$, please hint!

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Yes. ${x_n}$ might even all be negative... – awllower Feb 10 '13 at 13:31
@awllower. Are you sure about the last comment? If $x_{n}=-1$ for all $n$‚ then $|x_{n}-\alpha|\geq 1$ for every $\alpha>0$ and $n$, so the sum cannot converge. – T. Eskin Feb 10 '13 at 13:48
@ThomasE. I think he was responding to and adding to my comment. – David Mitra Feb 10 '13 at 13:53
@all edited..... – Un Chien Andalou Feb 10 '13 at 14:04
@City: For example, "Showing that a bounded sequence has a subsequence that converges in a strong way" would give more information about what's happening. Or just "Showing that a bounded sequence has a subsequence with a particular property" (if you haven't noticed that the particular property is a strong way of converging) would still be better than the original. You don't need to waste space in the title to introduce which variable letters you're using ($\{x_n\}$) or to specify that the elements are real numbers -- such detail belongs in the question itself. – Henning Makholm Feb 10 '13 at 16:49

You can use Bolzano Weirstrass theorem to find a monotonically increasing/decreasing sunsequence of the original sequence and as the sequence is bounded so this subsequence must converge.

Now call this subsequence which is convergent as $\{x_{n_{k}}\}$ and call $y_k=|x_{n_{k}}-\alpha|$

As this is convergent and decreasing $\exists \{k_i|i\in N\ , k_i<k_{i+1}\}$ such that $y_{k_{i}}<1/2^i$(Terms from any converging series will do)(Using the convergence of $y_k \to 0$).

We are done ,as $1/2^i$ converges so $\sum y_{k_{i}}$ also converges.

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But we are not requiring the convergence of the subsequence, instead that of the series of the subsequence. Per chance this is the same? – awllower Feb 10 '13 at 13:27
Thanks for the clarification. – awllower Feb 10 '13 at 14:06
You are welcome @awllower – Abhra Abir Kundu Feb 10 '13 at 14:23

By Bolzano-Weierstrass there exists a limit point $\alpha$. Then for each $k$ select $n_k>n_{k-1}$ with $|\alpha-x_{n_k}|<2^{-k}$.

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