Proof that lim sup cos(n) is 1

[tt2348]

Homework Statement

Prove that the lim sup cos(n)=1

Homework Equations

given any α\in\mathbb{R}-\mathbb{Q} , \exists P_n,Q_n\in\mathbb{N}.\ni.\frac{P_n}{Q_n} \rightarrow {a} and by definition, \frac{1}{Q_n} \rightarrow 0

Also, cos^{2}(x)=\frac{1+cos(2x)}{2}

My attempt

I first consider lim\:sup\:cos^{2}(n)
We know that \exists P_n,Q_n\in\mathbb{N}.\ni.\frac{P_n}{Q_n} \rightarrow {\pi} which means \forall \epsilon>0 \: \exists{N} \: .\ni. \forall{n}>N, \: |\frac{P_n}{Q_n}-\pi|<\epsilon
pick N.\ni.\:\epsilon<\frac{1}{Q_{n}^2}
So it follows that |\frac{P_n}{Q_n}-\pi|<\frac{1}{Q_{n}^2}
and therefore |P_n-Q_n\pi|<\frac{1}{Q_n}, am I allowed to make assume by continuity that |cos^{2}(P_n)-cos^{2}(Q_n\pi)|=|cos^{2}(P_n)-1|<ε and since P_k = a_{n_{k}} where a_n=n?

 

[HallsofIvy]

Why would you do all of that? You know that cos(x)\le 1 for all x and that cos(0)= 1. Therefore the maximum value of cos(x) is 1 and so is "lim sup".

(Strictly speaking "lim sup" is defined for sets, not functions, but I assume that by "lim sup cos(x)" you mean "lim sup {y| y= cos(x) for some x}".)

 

[tt2348]

It's not for a function... but the sequence cos(n)
where n=1,2,3... and so on
obviously I know that cosine is max at 1, but i have to prove that the lim sup is 1 over natural numbers

 

[tt2348]

to clarify, I need to show that
\stackrel{limit}{N \rightarrow{\infty}}\:sup\left\{cos(n):n\in\mathbb{N} \wedge n>N\right\}=1

 

[tt2348]

also... having a maximum value of 1 at 0 is not the same as having the lim sup being 1...
max value of 1/n is 1 (n=1)... but the lim sup {1/n:n is a natural number and n>N}=0

 

[HallsofIvy]

Yes, now that I see that it is the limit of hte sequence, not the function, you are correct.

 

[tt2348]

but as far as my proof, is it a correct assumption?