# why does this converge to 0.

• Jan 11th 2007, 08:59 PM
rcmango
why does this converge to 0.
Hello,

problem is: An = ( cos(n*pi) )/n

I thought this problem would converge to -1 since cos(pi) equals -1

the answer is that it converges to 0.

Is this because of the Monotonic Rule, which states that it will converge to an unknown limit that can be its lower bound or less.

in other words, it was possible that this could have converged to -1, but instead it converged to 0?

I solved this problem by dividing by the greatest n in the denominator.

thanks.
• Jan 11th 2007, 10:38 PM
Jameson
Think about the cosine function. $\displaystyle -1 \le \cos(x) \ge 1$. The magnitude of this function will never be more than 1. What about n in the denominator. It is increasing to infinity linearly and will have an infinite magnitude. So take this oscillating small number will be vastly outpowered, you could say, by the steadily increasing denominator. That's my non-mathematical explanation.
• Jan 11th 2007, 10:53 PM
CaptainBlack
Quote:

Originally Posted by rcmango
Hello,

problem is: An = ( cos(n*pi) )/n

I thought this problem would converge to -1 since cos(pi) equals -1.

This is irrelevant as $\displaystyle n\,\pi$ does not go to $\displaystyle \pi$ as $\displaystyle n \to \infty$.

RonL
• Jan 15th 2007, 10:04 PM
rcmango
Okay, thanks for the feedback, I see how n approaches infinity.

however, I divided by largest n in the denominator for this problem. Was that the wrong way to achieve the solution for this problem?

When should I know that dividing by the largest n in the denominator is the correct way to solve?

thanks again.
• Jan 15th 2007, 11:09 PM
CaptainBlack
Quote:

Originally Posted by rcmango
Okay, thanks for the feedback, I see how n approaches infinity.

however, I divided by largest n in the denominator for this problem. Was that the wrong way to achieve the solution for this problem?

When should I know that dividing by the largest n in the denominator is the correct way to solve?

thanks again.

$\displaystyle 0 \le |\cos(n\,\pi)/n| \le 1/n$

hence as $\displaystyle n$ becomes large $\displaystyle |\cos(n\,\pi)/n|$ is trapped between zero and a number which becomes arbitarily small, and hence tends to zero. But if the absolute value of a sequence tends to zero so does the sequence itself.

RonL