# Does this actually have a subsequence?

Basically I am asking if this sequence is "bounded" or not.

Consider the sequence $(n + \cos(n\pi)\sqrt{n^2 + 1})$. Does it have a subsequence that is convergent?

I think not because I tested $n = 2k$ and $n = 2k+1$. The first case $n = 2k$ tells me the sequence is unbounded, but the $n = 2k+1$ tells me that the sequence is bounded i.e. $1 - \sqrt{2} \leq b_{2k+1} \leq 0$.

I am thinking that it still doesn't have a convergent subsequence because the even terms tells me the whole sequence is unbounded.

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You are asking two very different questions: is it bounded, does it have a convergent subsequence. 1, 2, 1, 3, 1, 4, 1, 5, etc., isn't bounded but it has a convergent subsequence. – Gerry Myerson Oct 9 '12 at 4:51
That example you gave. What's the limit for the subsequential limit? I don't see how that one converges – Hawk Oct 9 '12 at 4:54
The subsequence 1, 1, 1, 1, 1, etc. converges to 1. – Gerry Myerson Oct 9 '12 at 4:57

As you observed, the sequence is not bounded. But there is a subsequence that converges to $0$, just pick $n$ odd, and note that for large $k$, $k-\sqrt{k^2+1}$ is close to $0$. This can be seen in various ways, for example by multiplying by $\frac{k+\sqrt{k^2+1}}{k+\sqrt{k^2+1}}$.
B-W says if sequence is bounded, it has convergent subsequence. It does not assert the converse, that if unbounded there is no convergent subsequence. For example, consider $1,0,2,0,3,0,4,0,\dots$. Unbounded, easy convergent subsequence. If you want something less trivial, replace the $0$'s by $1/2,1/3,1/4,\dots$. – André Nicolas Oct 9 '12 at 5:01