## Problems

Let $a_n$ be any sequence with partial limits 2008 and 2009. Suppose that for all nautuarl $n$
\[ |a_{n+1}-a_n|<\frac{1}{2}. \]
Prove that $a_n$ has at least three partial limits.
## Answer

Let $k_n$ and $u_n$ be monotonic increasing sequences of natural numbers, such that
\[ \begin{array}{l} \displaystyle \lim_{n\to\infty} a_{k_n} =2008 \\\\
\lim_{n\to\infty} a_{v_n}=2009. \end{array} \]
Denote
\[ \begin{array}{ll} A = \left\{n \; | \; |a_n-2008|<0.1 \right\} \\
B = \left\{n \; | \; |a_n-2009|<0.1 \right\}
\end{array} \]
Clearly, $\mathbb{N}-A$ is not finite, otherwise, using mathematical induction, one may conclude that up to finite amount of members
$A$ and the set of all natural numbers are equals. Hence, there exists infinite sequence of naturals, $t_n$ in $\mathbb{N}-A$, such that
$t_n-1\in A$ and $t_n\in \mathbb{N}-A$. According to the definition of limits $A$ and $B$ are infinite sets.
Suppose there exist a natural number $n$ such that $t_n\in B$. Then,
\[ \begin{array}{l} |a_{t_n-1}-2008|<0.1 \\
|a_{t_n}-2009|<0.1.
\end{array} \]
According to the data in the question $|a_{t_n}-a_{t_n-1}|<\frac{1}{2}$. Applying the triangle inequality (on the last two inequality)
yields $0.5+0.1>|a_{t_n-1}-2009|$ and since $t_n-1\in A$ we get $0.1>|a_{t_n-1}-2008|$. Once again we can apply the triangle inequality to get
$0.5+0.2>1$, but this is false. Thus, $t_n\not\in B\cap A$, for any natural $n$. Since $t_n-1\in A$, for any natural $n$, we have
$|a_{t_n-1}-2008|<\frac{1}{10}$ which give us $|a_{t_n-1}|<2009$. From the data given in the question we get $|a_{t_n}-a_{t_n-1}|<\frac{1}{2}$ and
this give us $|a_{t_n}|<\frac{1}{2}+|a_{t_n-1}|<\frac{1}{2}+2009<2010$, hence the sequence $a_{t_n}$ is bounded. We now can use
the Bolzano-Weierstrass theorem to conclude that $A_{t_n}$ has a convergent subsequence $q_n$. Non of the members of $a_{t_n}$ is in $A\cap B$, therefor
all the members of $q_n$ is not in $A\cap B$. Thus, the value $\lim_{n\to \infty}q_n$ is not equal to 2008 or to 2009.