squeeze theorem

**needhelp101** · April 21st 2009, 07:29 PM

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?

**mr fantastic** · April 21st 2009, 07:38 PM

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.

**needhelp101** · April 21st 2009, 07:45 PM

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.

**redsoxfan325** · April 21st 2009, 10:36 PM

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 $\lim_{n\to\infty}\frac{\cos(n\pi)}{n^2} \leq \lim_{n\to\infty}\frac{1}{n^2} = 0$

**woof** · April 21st 2009, 11:25 PM

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.

**redsoxfan325** · April 21st 2009, 11:32 PM

Right, I forgot to finish squeezing in my reply.

**woof** · April 21st 2009, 11:42 PM

But you got the important part

**needhelp101** · April 22nd 2009, 02:49 AM

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

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