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Problem: Given ${a_n}$ and that 2008 and 2009 are partial limits of ${a_n}$. Also, for all $n$, $|a_{n+1} - a_n| \le 1/2$. Prove that ${a_n}$ has at least 3 partial limits.

I attempted to prove it using the definition of partial limits of a sequence and assuming that there are only two partial limits and show some contradiction, but got nowhere.

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Is a "partial limit" the limit of a subsequence? –  Arturo Magidin Jan 26 '11 at 17:45
    
Yes, partial limit is a limit of a subsequence. –  Ma.H Jan 26 '11 at 18:02
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3 Answers 3

up vote 4 down vote accepted

Take a subsequence converging to 2008, and one converging to 2009. Argue that you may assume that they have no terms in common, and in fact, the terms in the first subsequence are "very close" to 2008, and those in the second are "very close" to 2009 (if $b_n\to L$ then for $n$ large, $|b_n-L|<1/8$, and we can restrict ourselves to large values of $n$ in both subsequences).

Argue that you may assume the indices of the sequences alternate. This means that if $a_{n_1}, a_{n_2},\dots $ is the sequence converging to 2008 and $a_{m_1},a_{m_2},\dots$ is the sequence converging to 2009, then we may assume that $n_1<m_1<n_2<m_2<n_3<\dots$ (for this, all you need is to "thin" out your subsequences).

Ok. Now argue that for all $i$, between $n_i$ and $m_i$ there is some $k_i$ such that both $a_{k_i}-a_{n_i}$ and $a_{m_i}-a_{k_i}$ are $\ge1/4$. It is here that you use that $|a_{n+1}-a_n|\le1/2$ for all $n$, noting that 2008 and 2009 are 1 unit apart, and $|a_{n_i}-2008|\le 1/8$ and $|a_{m_i}-2009|\le 1/8$.

Now, the sequence $(a_{k_i}\mid i=1,2,\dots)$ is completely contained in the interval ${}[2008.25,2008.75]$, so it must have a convergent subsequence, and this subsequence must converge somewhere other than 2008 and 2009.


On the other hand, note that you cannot conclude that there are more than 3 "partial limits", by considering the sequence $2008, 2008.5, 2009, 2008.5, 2008, 2008.5,\dots$

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Their must be a monotonic sequence of naturals $t_n$ such that $a_{t_n}$ is neither converge to 2008 nor to 2009, thus $a_{t_n}$ has to have a partial limit that differ from 2008 or 2009. But this partial limit is also a partial limit of $a_n$. See: http://magmath.com/problems/math/partial-limits-sequence

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2008? 2009? Sounds like a Putnam Competition question. (They often include the year.)

Anyway, because of the partial limits, $a_n$ is frequently $<2008\frac14$ and frequently $>2008\frac34$, so must frequently be in $\left[2008\frac14,2008\frac34\right]$ (by the inequality given). Frequently in an interval implies a limit in the interval's closure.

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