# squeeze theorem

• Apr 21st 2009, 08:29 PM
needhelp101
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?
• Apr 21st 2009, 08:38 PM
mr fantastic
Quote:

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 $\cos \frac{\pi}{n}$ and this converges to 1 as n --> +oo.
• Apr 21st 2009, 08:45 PM
needhelp101
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.
• Apr 21st 2009, 11:36 PM
redsoxfan325
Quote:

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 $\lim_{n\to\infty}\cos\left(\frac{n\pi}{n^2}\right) = 1$ because as $n\to\infty, \frac{n\pi}{n^2}\to 0$ and $\cos(0)=1$.

Perhaps you meant $\lim_{n\to\infty}\frac{\cos(n\pi)}{n^2}$? This does equal zero because $\cos(n\pi)\leq 1$ for all $n$ so $\lim_{n\to\infty}\frac{\cos(n\pi)}{n^2} \leq \lim_{n\to\infty}\frac{1}{n^2} = 0$
• Apr 22nd 2009, 12:25 AM
woof
Quote:

Originally Posted by redsoxfan325
Perhaps you meant $\lim_{n\to\infty}\frac{\cos(n\pi)}{n^2}$? This does equal zero because $\cos(n\pi)\leq 1$ for all $n$ so $\lim_{n\to\infty}\frac{\cos(n\pi)}{n^2} \leq \lim_{n\to\infty}\frac{1}{n^2} = 0$

And to add to that, now the "squeeze" part:

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

Now both "ends" converge to 0, squeezing the middle to zero.
• Apr 22nd 2009, 12:32 AM
redsoxfan325
Right, I forgot to finish squeezing in my reply.
• Apr 22nd 2009, 12:42 AM
woof
But you got the important part :)
• Apr 22nd 2009, 03:49 AM
needhelp101
thank u all so much. u all have been very helpful 2 me for this problem.