limit and convergent subsequence?

**cowgirl123** · September 25th 2007, 09:36 PM

If an >= 0 for all n and has no convergent subsequence, show lim n->infinity an=infinity.

**Jhevon** · September 25th 2007, 10:21 PM

Originally Posted by cowgirl123

If an >= 0 for all n and has no convergent subsequence, show lim n->infinity an=infinity.

Here's what i am thinking. I don't know if it is 100% correct, but it will at least give you something to think about.

Assume $\{ a_n \}$ is a sequence which has no convergent subsequence, and assume to the contrary that $\lim_{n \to \infty} a_n \ne \infty$ . Then we have three other possibilities for the infinite limit: (1) $\lim_{n \to \infty}a_n = - \infty$ , (2) $\lim_{n \to \infty}a_n = L$ , where L is finite, or (3) $\lim_{n \to \infty}a_n$ does not exist, in that the $a_n$ 's alternate between certain values.

Clearly case (1) is impossible, since $a_n \ge 0$ for all $n$

Consider case (2): If $\lim_{n \to \infty}a_n = L$ it means that for some $N \in \mathbb{N}$ , $n > N$ implies $|a_n - L|< \epsilon$ for all $\epsilon > 0$ . In other words, the terms of the sequence all get arbitrarily close to L after a certain point in the sequence. Thus we can construct a subsequence, say, of the even terms of $\{ a_n \}$ . Call this sequence $\{ a_{n_k} \}$ where $n_k = 2k$ for $k \in \mathbb{Z}$ . Clearly, this sequence converges to L also, since all terms of $\{ a_n \}$ are getting close to L. But this contradicts our original assumption, and so, cannot be the case.

Consider case (3): If the $a_n$ 's are alternating, again we can construct a convergent subsequence. This subsequence can consist of any single term that occurs in the sequence repeatedly. Which again, contradicts our assumption that $\{ a_n \}$ has no convergent subsequence.

Thus, it must be the case that $\lim_{n \to \infty}a_n = \infty$

QED

Now what bothers me is that the argument for case (3) is not as "airtight" as i would like. i feel like i'm missing something. also, i'm not sure if i am neglecting to consider some other case.

**Plato** · September 26th 2007, 02:53 AM

For each positive integer K, there are only finitely many terms of $a_n < K$ .
Otherwise, there would be a convergence subsequence.
Hence, the sequence must diverge to infinity.

**Jhevon** · September 26th 2007, 03:24 AM

Originally Posted by Plato

For each positive integer K, there are only finitely many terms of $a_n < K$ .
Otherwise, there would be a convergence subsequence.
Hence, the sequence must diverge to infinity.

well, that was easy

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