1. ## Sequence

Does the following sequence converge or diverge? If it converges what does it converge to?

$a_{n} = \cos (\frac {6\pi n^{2}}{3n+1})$

This is my first use of LaTex. It's really cool!

2. ## Re: Sequence

Originally Posted by browni3141
Does the following sequence converge or diverge? If it converges what does it converge to?

$a_{n} = \cos (\frac {6\pi n^{2}}{3n+1})$

This is my first use of LaTex. It's really cool!
$\lim_{n \to \infty} a_n=\lim_{n \to \infty} \cos \left(\frac{6 \pi n^2}{3n+1}\right)=\cos \left({\lim_{n \to \infty} \frac{6 \pi n^2}{3n+1}}\right)$

$\lim_{n \to \infty} a_n=\cos \left(6\pi \cdot \lim_{n \to \infty} \frac{ n^2}{3n+1}\right)=\cos \left(\4 \pi \cdot \lim_{n \to \infty} n\right)$

$-1 \leq \cos \left(\4 \pi \cdot \lim_{n \to \infty} n\right) \leq 1$ , therefore sequence diverge .

3. ## Re: Sequence

Thank you for the reply, but showing that the limit does not exist does not prove that the sequence diverges. I actually know that it converges and I know what it converges to, but I don't understand how to prove it mathematically.
I constructed this problem intentionally so that it would converge while the limit does not exist, but I made a bad assumption about limits and basically got lucky. It converges to -1/2, but I don't know how to show this.

4. ## Re: Sequence

Originally Posted by browni3141
showing that the limit does not exist does not prove that the sequence diverges.
That is a truly an absurd statement. It has an internal contradiction: a violation of the definition of divergent.

Originally Posted by browni3141
I actually know that it converges and I know what it converges to, but I don't understand how to prove it mathematically.
How can you know that something which is demonstrably false is actually true?
Surely you know that ${\lim _{n \to \infty }}\cos (n)$ does not exist,
If you have doubts about that then you have no business trying this question.

If you do understand that fact, then note that $\left( {\frac{{6\pi {n^2}}}{{3n + 1}}} \right) \to \infty$

5. ## Re: Sequence

Originally Posted by Plato
That is a truly an absurd statement. It has an internal contradiction: a violation of the definition of divergent.

How can you know that something which is demonstrably false is actually true?
Surely you know that ${\lim _{n \to \infty }}\cos (n)$ does not exist,
If you have doubts about that then you have no business trying this question.

If you do understand that fact, then note that $\left( {\frac{{6\pi {n^2}}}{{3n + 1}}} \right) \to \infty$
The problem with what you and princeps have said is that these limits are taken over the real numbers, but n is an integer. The limit over integers may still exist, and it does in this case. Surely you can see that $a_{n}$ converges to -1/2 by observing values of $a_{n}$ as n gets large.
$a_{5} = -.3827$
$a_{6} = -.4017$
$a_{7} = -.4154$
$a_{20} = -.4700$
$a_{100} = -.4940$
$a_{1000} = -.4994$
$a_{10000} = -.4999$

6. ## Re: Sequence

Whatever, if I'm wrong, then explain why it appears that $a_{n}$ converges to -1/2.
I don't see why it's so hard for you to understand what I'm asking. It is clear that $a_{n}$ is approaching -1/2 based on the data provided above, so perhaps you should come back and give the problem more than a second of thought.

7. ## Re: Sequence

By "long division",
$\frac{3n^2}{3n+1} = n - \frac{1}{3} + \frac{1}{3(3n+1)}$
so
$\cos \left( 2 \pi \cdot \frac {3n^2}{3n+1} \right) = \cos \left( 2 n \pi - \frac{2 \pi}{3} + \frac{2 \pi}{3(3n+1)} \right) = \cos \left(- \frac{2 \pi}{3} + \frac{2 \pi}{3(3n+1)} \right)$
$\to \cos \left( - \frac{2 \pi}{3} \right) = - \frac{1}{2}$
as $n \to \infty$

8. ## Re: Sequence

Originally Posted by awkward
By "long division",
$\frac{3n^2}{3n+1} = n - \frac{1}{3} + \frac{1}{3(3n+1)}$
so
$\cos \left( 2 \pi \cdot \frac {3n^2}{3n+1} \right) = \cos \left( 2 n \pi - \frac{2 \pi}{3} + \frac{2 \pi}{3(3n+1)} \right) = \cos \left(- \frac{2 \pi}{3} + \frac{2 \pi}{3(3n+1)} \right)$
$\to \cos \left( - \frac{2 \pi}{3} \right) = - \frac{1}{2}$
as $n \to \infty$
This seems to be exactly what I'm looking for. Thank you for understanding what I was asking.