Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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 $\frac{k+\sqrt{k^2+1}}{k+\sqrt{k^2+1}}$.

share|cite|improve this answer
Why? By Bolzano Weistress Thrm, the original sequence isn't bounded, so it doesn't have a convergent subsequence. The odd terms may be bounded, bu the original one isn't – Hawk Oct 9 '12 at 4:56
You are missing something. Whether it's the accurate statement of B-W, or the definition of convergent, or of subsequence, I'm not sure --- but you are missing something big. – Gerry Myerson Oct 9 '12 at 4:58
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
Thank you all. I understand my flaw of reasoning now – Hawk Oct 9 '12 at 5:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.