# Bounded Sequence and Subsequential Limit

Given $\{a_n\}_{n=1}^\infty$ bounded sequence such that $\lim_{n\to \infty} (a_{n+1}-a_n)=0$. Prove that each point in $[\liminf a_n, \limsup a_n]$ is subsequential limit of the sequence $a_n$.

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What have you tried? –  Mark Bennet May 2 '12 at 20:03

Let $\alpha\in[\liminf a_n,\limsup a_n]$. You wish to find a subsequence of $(a_n)$ that converges to $\alpha$.

I think it should be clear how to proceed if an outline of finding the first term of the required subsequence is given:

Let $\epsilon_1>0$ and select $N_1$ so large that $|a_{n+1}-a_n|<\epsilon_1$ whenever $n\ge N_1$. Now choose $k_1>N_1$ and $l_1>k_1$ so that $a_{k_1}$ is within $\epsilon_1$ of the $\sup$ and $a_{l_1}$ is within $\epsilon_1$ of the $\inf$. Then there is an index $n_1$ between $k_1$ and $l_1$ so that $a_{n_1}$ is within $\epsilon_1$ of $\alpha$.

Informally, there is an index $k_1$ where $a_{k_1}$ is close to the $\sup$, and a latter index $l_1$ where $a_{l_1}$ is close to the $\inf$. Since successive terms of $(a_n)$ are close to each other, in the journey from the $\sup$ to the $\inf$, $a_j$ passes close by to any number between the $\sup$ and the $\inf$ (in particular, close by to $\alpha$).

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what do you mean by: "...so that $a_{k_1}$ is with $\epsilon_1$ of the sup"? –  Anonymous May 2 '12 at 20:27
@Anonymous Sorry for the vaugeness; I mean $|a_{k_1} -\limsup_n a_n|<\epsilon_1$. –  David Mitra May 2 '12 at 20:29
why can I deduce that $|a_{k_1} -\limsup_n a_n|<\epsilon_1?$ –  Anonymous May 2 '12 at 20:36
@Anonymous If $\beta=\limsup_n a_n$, then there is a subsequence of $(a_n)$ that converges to $\beta$. Thus, there is a term of $(a_n)$, say $a_{k_1}$ such that $|a_{k_1}-\beta|<\epsilon_1$. Of course, infinitely many terms of $(a_n)$ satisfy this, but we need to fix a particular term of $(a_n)$ for the argument. –  David Mitra May 2 '12 at 20:39
ok, great, now why can I deduce that $|a_{n_1}-\alpha|<\epsilon_1$? –  Anonymous May 2 '12 at 20:43