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Homework Statement
Prove that the lim sup cos(n)=1
Homework Equations
given any [itex]α\in\mathbb{R}-\mathbb{Q} , \exists P_n,Q_n\in\mathbb{N}.\ni.\frac{P_n}{Q_n} \rightarrow {a} [/itex] and by definition, [itex] \frac{1}{Q_n} \rightarrow 0[/itex]
Also, [itex] cos^{2}(x)=\frac{1+cos(2x)}{2}[/itex]
My attempt
I first consider [itex]lim\:sup\:cos^{2}(n) [/itex]
We know that [itex]\exists P_n,Q_n\in\mathbb{N}.\ni.\frac{P_n}{Q_n} \rightarrow {\pi}[/itex] which means [itex]\forall \epsilon>0 \: \exists{N} \: .\ni. \forall{n}>N, \: |\frac{P_n}{Q_n}-\pi|<\epsilon[/itex]
pick [itex]N.\ni.\:\epsilon<\frac{1}{Q_{n}^2}[/itex]
So it follows that [itex]|\frac{P_n}{Q_n}-\pi|<\frac{1}{Q_{n}^2}[/itex]
and therefore [itex]|P_n-Q_n\pi|<\frac{1}{Q_n} [/itex], am I allowed to make assume by continuity that [itex]|cos^{2}(P_n)-cos^{2}(Q_n\pi)|=|cos^{2}(P_n)-1|<ε [/itex] and since [itex] P_k = a_{n_{k}} [/itex] where [itex]a_n=n[/itex]?