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

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- Sep 25th 2007, 09:36 PMcowgirl123limit and convergent subsequence?
If an >= 0 for all n and has no convergent subsequence, show lim n->infinity an=infinity.

- Sep 25th 2007, 10:21 PMJhevon
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 $\displaystyle \{ a_n \}$ is a sequence which has no convergent subsequence, and assume to the contrary that $\displaystyle \lim_{n \to \infty} a_n \ne \infty$. Then we have three other possibilities for the infinite limit: (1) $\displaystyle \lim_{n \to \infty}a_n = - \infty$, (2) $\displaystyle \lim_{n \to \infty}a_n = L$, where L is finite, or (3) $\displaystyle \lim_{n \to \infty}a_n$ does not exist, in that the $\displaystyle a_n$'s alternate between certain values.

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

Consider case (2): If $\displaystyle \lim_{n \to \infty}a_n = L$ it means that for some $\displaystyle N \in \mathbb{N}$, $\displaystyle n > N$ implies $\displaystyle |a_n - L|< \epsilon$ for all $\displaystyle \epsilon > 0$. 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 $\displaystyle \{ a_n \}$. Call this sequence $\displaystyle \{ a_{n_k} \}$ where $\displaystyle n_k = 2k$ for $\displaystyle k \in \mathbb{Z}$. Clearly, this sequence converges to L also, since all terms of $\displaystyle \{ a_n \}$ are getting close to L. But this contradicts our original assumption, and so, cannot be the case.

Consider case (3): If the $\displaystyle a_n$'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 $\displaystyle \{ a_n \}$ has no convergent subsequence.

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

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. - Sep 26th 2007, 02:53 AMPlato
For each positive integer K, there are only finitely many terms of $\displaystyle a_n < K$.

Otherwise, there would be a convergence subsequence.

Hence, the sequence must diverge to infinity. - Sep 26th 2007, 03:24 AMJhevon