Prove dot product of sequences is equal to dot product of their limits

Question

Assume that $a_j$ and $b_j$ are sequences in $\mathbb{R}^n$ such that $a_j \rightarrow a$ and $b_j \rightarrow b$. Is it true that $a_j \cdot b_j \rightarrow a \cdot b$ How do you know?

Answer 1

Lets $\{a_n\}$ and $\{b_n\}$ your sequences in $\mathbb{R}^n$

One good question to ask is it is convergent but what really means; what means : $$ a_n \to a$$

If we take the norm $$ ||x||=\sqrt{\sum_{i=1}^n x_i^2}$$

So saying : $$ a_n \to a$$ Means :

$$ ||a_n-a|| \to0$$

Which means :

$$ \sqrt{\sum_{i=1}^n (a_{n,i}-a_i)^2} \to_{n \to \infty} 0 $$

Because all terms are positive it imply :

$$ \forall i \in [1,n], a_{n,i} \to_{n \to \infty} a_i $$

That means that in $\mathbb{R}^n$, the convergence of $a_n$ is equivalent to the convergence of all its component

And each $i$-th component of $a_n$ converges to the $i$-th component of $a$.

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So because

$$ (a_n \cdot b_n)_i=a_{n,i}b_{n,i}$$

Your result is explained.