Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
up vote 7 down vote accepted

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.

share|cite|improve this answer
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 $|\alpha-x_{n_k}|<2^{-k}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.