1. ## squeeze theorem

cos(n*pi/n^2)

i have to prove that this sequence is converging to 0. i am unable to see how this would converge to 0. any ideas?

2. Originally Posted by needhelp101
cos(n*pi/n^2)

i have to prove that this sequence is converging to 0. i am unable to see how this would converge to 0. any ideas?
It doesn't converege to zero.

What you have simplifies to $\displaystyle \cos \frac{\pi}{n}$ and this converges to 1 as n --> +oo.

3. yea, i was able to understand that, but according to the question, it is suppose to converge to 0. i was able to get it to converge to 1, but i wasn't sure if there was any other way for it to converge to 0.

4. Originally Posted by needhelp101
cos(n*pi/n^2)

i have to prove that this sequence is converging to 0. i am unable to see how this would converge to 0. any ideas?
As you have it written $\displaystyle \lim_{n\to\infty}\cos\left(\frac{n\pi}{n^2}\right) = 1$ because as $\displaystyle n\to\infty, \frac{n\pi}{n^2}\to 0$ and $\displaystyle \cos(0)=1$.

Perhaps you meant $\displaystyle \lim_{n\to\infty}\frac{\cos(n\pi)}{n^2}$? This does equal zero because $\displaystyle \cos(n\pi)\leq 1$ for all $\displaystyle n$ so $\displaystyle \lim_{n\to\infty}\frac{\cos(n\pi)}{n^2} \leq \lim_{n\to\infty}\frac{1}{n^2} = 0$

5. Originally Posted by redsoxfan325
Perhaps you meant $\displaystyle \lim_{n\to\infty}\frac{\cos(n\pi)}{n^2}$? This does equal zero because $\displaystyle \cos(n\pi)\leq 1$ for all $\displaystyle n$ so $\displaystyle \lim_{n\to\infty}\frac{\cos(n\pi)}{n^2} \leq \lim_{n\to\infty}\frac{1}{n^2} = 0$
And to add to that, now the "squeeze" part:

$\displaystyle -1\leq\cos(n\pi)\leq 1$ for all $\displaystyle n$ so $\displaystyle \lim_{n\to\infty}\frac{-1\ \ }{n^2} \leq\lim_{n\to\infty}\frac{\cos(n\pi)}{n^2} \leq \lim_{n\to\infty}\frac{1}{n^2}$

Now both "ends" converge to 0, squeezing the middle to zero.

6. Right, I forgot to finish squeezing in my reply.

7. But you got the important part

8. thank u all so much. u all have been very helpful 2 me for this problem.