Sequence upper limit

Finding

{lim^{―}}_{k \to \infty} \cos (k), k \in N

Proof

Let $x_{k}$ be the sequence of arguments of $\cos (k)$ with step $d = 1$ . Consider the ratio $α$ of the step of $x_{k}$ to the period $T = 2 π$ :

α = d / T = 1 / (2 π)

Let us express $\cos (k)$ through $α$ . By the periodicity property, we can discard the integer part of the argument:

\cos (k) = \cos (2 π (k α)) = \cos (2 π k α)

At the same time, $π \notin Q ⟹ 1 / (2 π) \notin Q ⟹ α \notin Q$ . Since $α \notin Q$ , according to Kronecker's theorem, the sequence of fractional parts $k α$ is dense on the interval $[0, 1]$ . This means that the argument $2 π k α$ is dense in $[0, 2 π]$ . Therefore, by density, there exists a subsequence $k_{j}$ such that

2 π k_{j} α \to 0 (1)

By the definition of sequential continuity at $x_{0}$ :

\forall x_{k} : lim_{k \to \infty} x_{k} = x_{0} ⟹ lim_{k \to \infty} f (x_{k}) = f (x_{0})

Let $x_{0} = 0$ . Then, by the continuity of $\cos (x)$ and (1), we have:

lim_{j \to \infty} 2 π k_{j} α = 0 ⟹ lim_{j \to \infty} \cos (2 π k_{j} α) = \cos (0) = 1 (2)

Since $\cos (k) \leq 1 ⟹ {lim^{―}}_{k \to \infty} \cos (k) \leq 1$ , it follows from (2) by definition that

{lim^{―}}_{k \to \infty} \cos (k) = 1.