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 is a sequence which has no convergent subsequence, and assume to the contrary that . Then we have three other possibilities for the infinite limit: (1) , (2) , where L is finite, or (3) does not exist, in that the 's alternate between certain values.
Clearly case (1) is impossible, since for all
Consider case (2): If it means that for some , implies for all . 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 . Call this sequence where for . Clearly, this sequence converges to L also, since all terms of are getting close to L. But this contradicts our original assumption, and so, cannot be the case.
Consider case (3): If the '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 has no convergent subsequence.
Thus, it must be the case that
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.