Let $θ$ be the angle between the vectors $A =
(1,1,…,1)$ and$ B = (1,2,…,n)$ in $\mathbb{R}^n$. Find the limiting value of θ as n → ∞.
For this question, I want to apply the equation for the angle between two vectors: $θ=arccos\frac{A·B}{\|A\|\|B\|}$. For A·B, I use $A·B=\sum_{k=1}^{n}a_kb_k$. And for the norm of A and B, I simply use the definition to get $\|A\|=\sqrt{n}$ and $\|B\|=\sqrt{\sum_{k=1}^{n}k^2}$. But I'm stuck here. I don't know how to calculate the limit of θ as n→ ∞ by these equations.
Best Answer
Knowing that
$$\begin{cases} \sum_{k=1}^n k &= \frac{n(n+1)}{2}\\ \sum_{k=1}^n k^2 &= \frac{n(n+1)(2n+1)}{6} \end{cases}$$
You get
$$\frac{A_n \cdot B_n}{\Vert A_n \Vert \Vert B_n \Vert} = \frac{n(n+1)/2}{\sqrt n \sqrt{n(n+1)(2n+1)/6}} \to \frac{\sqrt 3}{2}$$ as $n \to \infty$ and therefore
$$\theta = \frac{\pi}{6}$$ as $\arccos$ is continuous.