real analysis - Sequence Convergence in $mathbb{R}^n$

Let {$\mathbf u_k$} be a sequence in $\mathbb{R}^n$ and let $\mathbf u$ be a point in $\mathbb{R}^n$. Suppose that for every $\mathbf v$ in $\mathbb{R}^n$, $$\lim_{k\to \infty} \langle\mathbf u_k, \mathbf v\rangle = \langle\mathbf u, \mathbf v\rangle.$$

Prove that {$\mathbf u_k$} converges to $\mathbf u$.

Using the Triangle Inequality I get

$$\begin{align}\text{dist}(\mathbf u,\mathbf v) &\le \text{dist}\mathbf (u,\mathbf u_k)+ \text{dist}(\mathbf u_k,\mathbf v),\ \quad \forall k \\
\\ \therefore \text{dist}(\mathbf u,\mathbf v) &\le \lim_{k\to \infty}[\text{dist}(\mathbf u,\mathbf u_k)+ \text{dist}(\mathbf u_k,\mathbf v)] \end{align}$$

and
$$\lim_{k\to \infty}\text{dist}(\mathbf u,\mathbf u_k) = 0 $$ because {$\mathbf u_k$} $ \to \mathbf u$.

But I don't know what to do next.

Answer

Since you are working on $\mathbb{R}^n$ with the standard scalar product (I suppose?), you can just insert the $i$-th standard-basis vector $e^i$ for $v$. Then your equation reads

$$
\lim_{k \to \infty} (v_k)^i = v^i
$$
(the upper index $i$ denotes the $i$-th component). This already implies that $u_k$ converges to $u$ since a series of vectors converges if and only if all components converge.