Does the following sequence converge or diverge? If it converges what does it converge to?
$\displaystyle a$_{n}$ = $\cos (\frac {6\pi n^{2}}{3n+1})$$
This is my first use of LaTex. It's really cool!
Does the following sequence converge or diverge? If it converges what does it converge to?
$\displaystyle a$_{n}$ = $\cos (\frac {6\pi n^{2}}{3n+1})$$
This is my first use of LaTex. It's really cool!
$\displaystyle \lim_{n \to \infty} a_n=\lim_{n \to \infty} \cos \left(\frac{6 \pi n^2}{3n+1}\right)=\cos \left({\lim_{n \to \infty} \frac{6 \pi n^2}{3n+1}}\right)$
$\displaystyle \lim_{n \to \infty} a_n=\cos \left(6\pi \cdot \lim_{n \to \infty} \frac{ n^2}{3n+1}\right)=\cos \left(\4 \pi \cdot \lim_{n \to \infty} n\right)$
$\displaystyle -1 \leq \cos \left(\4 \pi \cdot \lim_{n \to \infty} n\right) \leq 1$ , therefore sequence diverge .
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.
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 $\displaystyle {\lim _{n \to \infty }}\cos (n)$ 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 $\displaystyle \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 $\displaystyle a_{n}$ converges to -1/2 by observing values of $\displaystyle a_{n}$ as n gets large.
$\displaystyle a_{5} = -.3827$
$\displaystyle a_{6} = -.4017$
$\displaystyle a_{7} = -.4154$
$\displaystyle a_{20} = -.4700$
$\displaystyle a_{100} = -.4940$
$\displaystyle a_{1000} = -.4994$
$\displaystyle a_{10000} = -.4999$
Whatever, if I'm wrong, then explain why it appears that $\displaystyle a_{n}$ 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 $\displaystyle a_{n}$ 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.
By "long division",
$\displaystyle \frac{3n^2}{3n+1} = n - \frac{1}{3} + \frac{1}{3(3n+1)}$
so
$\displaystyle \cos \left( 2 \pi \cdot \frac {3n^2}{3n+1} \right) = \cos \left( 2 n \pi - \frac{2 \pi}{3} + \frac{2 \pi}{3(3n+1)} \right) = \cos \left(- \frac{2 \pi}{3} + \frac{2 \pi}{3(3n+1)} \right)$
$\displaystyle \to \cos \left( - \frac{2 \pi}{3} \right) = - \frac{1}{2} $
as $\displaystyle n \to \infty$