-2
$\begingroup$

Prove that the sequence $\dfrac{n^2+n \cos(n^2+1)}{2+n^2}$ has a convergent subsequence.

I know need to use the Bolzano–Weierstrass theorem , every bounded sequense has a convergent subsequense, but i dont know to prove it is a bounded sequence.

  • 1
    Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Determine", etc.) to be rude when asking for help; please consider rewriting your post. – 2017-02-27
  • 3
    Why not prove directly that it converges (to the limit $1$)? – 2017-02-27
  • 0
    As @Did has said, if we prove a sequence is convergent, every subsequence of that sequence is convergent. – 2017-02-27

2 Answers 2

1

Let $ I = \dfrac{n^2+n \cos(n^2+1)}{2+n^2}$. Then

$$ \dfrac{n^2 - n}{2+n^2} \leq I \leq \dfrac{n^2 + n}{2+n^2} $$

Now,

$$ \dfrac{n^2 + n}{2+n^2} \leq \frac{n^{2}}{n^{2} + 2} + \frac{n}{n^{2}+2} \leq 1 + \frac{1}{n} \leq 2.$$

Similarly, calculate for the other term.

EDIT: On a related note, one notes that

$$ \lim_{n} I = \lim_{n} \frac{n^{2}}{n^{2} + 2} +\lim_{n} \frac{n\cos(n^2+1)}{n^{2}+2} = 1 + 0 = 1.$$

Since the sequence converges, it is bounded.

  • 0
    so the other term is , (n^2-n)/(n^2+2) = n^2/(n^2+2) - n/(n^2+2) > 1 +1/n >1 – 2017-02-27
  • 0
    @Karry how did you do that computation? – 2017-02-27
  • 0
    so there are two way to prove it is bounded, first find the range . second is prove it is converges then it implies it is bounded. and the range is [ 0 , 2] ? – 2017-02-27
  • 0
    @Karry You can just prove that all terms are positive since $n^{2} > n$. – 2017-02-27
  • 0
    @Karry yes, usually if a sequence converges, you are asked to find the limit or show the convergence. The question seems atypical in that it asks you to find a convergent subsequence of a convergent sequence. – 2017-02-27
  • 0
    I do not know the exact range. I suspect it might be monotone and then range would be [0,1] – 2017-02-27
-2

yes, because it is bounded ($-1\le\cos(...)\le 1$ and you can easily check that all terms are between $-1$ and $2$) and it's known that every bounded sequence of reals has a convergent subsequence: https://en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem

  • 0
    ya ,so how to prove it ? i dont know how to justfy it is a bounded . – 2017-02-27
  • 0
    I edited the answer – 2017-02-27
  • 0
    "all terms are between −1 and 2" Why send the OP towards the possibility that some terms are negative while none is? – 2017-02-27