Find the angle between the vectors (1,1, …, 1) and (1,2,…,n) as n goes to infinity

by RoRi

April 28, 2016

Consider the vectors $A = (1,1, \ldots, 1)$ and $B = (1,2, \ldots, n)$ 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

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

Thus,

$\lim_{n \to \infty} \cos \theta = \lim_{n \to \infty} \frac{\sqrt{3}}{\sqrt{2}} \left( \frac{n^2 + n}{\sqrt{2n^4 + 3n^3 + n^2}} \right) = \frac{\sqrt{3}}{2}.$

Therefore, $\theta \to \frac{\pi}{6}$ as $n \to \infty$ .

Stumbling Robot

Find the angle between the vectors (1,1, …, 1) and (1,2,…,n) as n goes to infinity

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