1
$\begingroup$

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.

  • 2
    The subsequence 1, 1, 1, 1, 1, etc. converges to 1.2012-10-09

1 Answers 1

1

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

  • 0
    T$h$ank you all. I understand my flaw of reasoning now2012-10-09