Find the angle between the vectors (2,4,6, ..., 2n) and (1,3,5,...,2n-1) as n goes to infinity

Consider the vectors $A = (2,4,6, \ldots, 2n)$ and $B = (1,3,5, \ldots, 2n-1)$ in $\mathbb{R}^n$ . Let $\theta$ denote the angle between $A$ and $B$ . Find the value of $\theta$ as $n \to \infty$ .

If $\theta$ is the angle between $A$ and $B$ then we have

$\cos \theta &= \frac{A \cdot B}{\lVert A \rVert \lVert B \rVert}.$

Computing each of these pieces we have

$\begin{align*} A \cdot B &= \sum_{k=1}^n (2k)(2k-1) \\[9pt] &= \sum_{k=1}^n (4k^2 - 2k) \\[9pt] &= 4 \left( \frac{2n^3 + 3n^2 + n}{6} \right) - 2 \left( \frac{n^2+n}{2} \right) \\[9pt] &= \frac{4n^3 + 3n^2 - n}{3}.\\[9pt] \lVert A \rVert &= \left( \sum_{k=1}^n (2k)^2 \right)^{\frac{1}{2}} \\[9pt] &= \left( 4 \left( \frac{2n^3+3n^2 + n}{6} \right) \right)^{\frac{1}{2}} \\[9pt] &= \left( \frac{4n^3 + 6n^2 + 2n}{3} \right)^{\frac{1}{2}}.\\[9pt] \lVert B \rVert &= \left(\sum_{k=1}^n (2k-1)^2 \right)^{\frac{1}{2}} \\[9pt] &= \left( \sum_{k=1}^n 4k^2 - \sum_{k=1}^n 4k + \sum_{k=1}^n 1 \right)^{\frac{1}{2}} \\[9pt] &= \left( \frac{4n^3+3n^2+2n}{3} \right)^{\frac{1}{2}}. \end{align*}$

Therefore,

$\cos \theta = \frac{A \cdot B}{\lVert A \rVert \lVert B \rVert} = \left( \frac{4n^3+3n^2-n}{3} \right) \left( \frac{9}{16n^3 + o(n^2)} \right)^{\frac{1}{2}}.$

Hence, $\cos \theta \to 1$ as $n \to \infty$ which implies $\theta \to 0$ as $n \to \infty$ .

Stumbling Robot

Find the angle between the vectors (2,4,6, …, 2n) and (1,3,5,…,2n-1) as n goes to infinity

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