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?

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- Apr 21st 2009, 07:29 PMneedhelp101squeeze 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, 07:38 PMmr fantastic
- Apr 21st 2009, 07:45 PMneedhelp101
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, 10:36 PMredsoxfan325
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 $\displaystyle \lim_{n\to\infty}\frac{\cos(n\pi)}{n^2} \leq \lim_{n\to\infty}\frac{1}{n^2} = 0$ - Apr 21st 2009, 11:25 PMwoof
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. - Apr 21st 2009, 11:32 PMredsoxfan325
Right, I forgot to finish squeezing in my reply.

- Apr 21st 2009, 11:42 PMwoof
But you got the important part :)

- Apr 22nd 2009, 02:49 AMneedhelp101
thank u all so much. u all have been very helpful 2 me for this problem.