# Vector Analysis & Linear Algebra

I'm given a positive number, a unit vector $u \in \mathbb{R} ^n$ and a sequence of vectors $\{ b_k \} _{ k \geq 1}$ such that $|b_k - ku| \leq d$ for every $k=1,2,...$.

This obviously implies $|b_k| \to \infty$ . But why does this imply $\angle (u,b_k) \to 0$ ? I've tried proving it using some inner-product calculations, but without any success.

In addition, why the given data implies that there must exist $i<j$ such that $|b_i| \leq \frac{\delta}{4} |b_j|$ ?

Thanks a lot !

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You have $|b_k| \leq |b_k -k u| + k=d+k$, and $|b_k| \geq k-|b_k -k u|=k-d$. This gives: $$\lim_{k \to \infty} \frac{\langle u, b_k-ku\rangle}{|b_k|} = 0, \;\; \lim_{k \to \infty} \frac{\langle u, ku\rangle}{|b_k|} = 1,$$ so it follows that $$\lim_{k \to \infty} \frac{\langle u, b_k \rangle}{|b_k|} = \lim_{k \to \infty} \frac{\langle u, b_k -ku\rangle + \langle u, ku\rangle}{|b_k|} = 1.$$
Fix $i$. Since $|b_j| \geq j-d$, you can choose $j$ large enough so that $|b_j| >\frac{4}{\delta} |b_i|$. –  copper.hat May 22 '12 at 16:04
From $|b_k-k\,u|\le d$ you get that for all sufficiently large $k$ $$k-d\le |b_k|\le k+d$$ and $$k-d\le u\cdot b_k\le k+d.$$ Then $$\frac{k-d}{k+d}\le\cos(\angle(u,b_k))\le\frac{k+d}{k-d}.$$ It follows that $\cos(\angle(u,b_k))\to1$ and $\angle(u,b_k)\to0$