Section 3.1 Sequences of Vectors
Given vectors \(\vect x_n=(x_{1n},\dots,x_{Nn})\) and \(\vect x=(x_1,\dots,x_N)\) in \(\mathbb R^N\) we say that \(\vect x_n\) converges to \(\vect x\) as \(n\) goes to infinity if all components \(x_{in}\) of \(\vect x_n\) converge to the corresponding component \(x_i\) of \(\vect x\text{.}\) In symbols
\begin{equation}
\vect x_n\xrightarrow{n\to\infty}\vect x
\iff
x_{in}\xrightarrow{n\to\infty}x_i\text{ for all }i=1,\dots,N\text{.}\tag{3.1}
\end{equation}
We say that \(\vect x\) is the limit of the sequence \((\vect x_n)\text{,}\) and write
\begin{equation*}
\lim_{n\to\infty}\vect x_n=\vect x\text{.}
\end{equation*}
As in case of sequences of numbers if the limit exists. If a sequence of real numbers \((t_n)\) converges to \(t\) then we know that \(|t_n-t|\xrightarrow{n\to\infty} 0\) and vice versa. A similar fact is true for sequences of vectors as the following proposition shows.
Proposition 3.1. Convergence of sequences of vectors.
Let \((\vect x_n)_{n\in\mathbb N}\) be a sequence of vectors in \(\mathbb R^N\text{.}\) Then
\begin{equation*}
\lim_{n\to\infty}\vect x_n=\vect x\text{ if and only if }
\lim_{n\to\infty}\|\vect x_n-\vect x\|=0\text{.}
\end{equation*}
Proof.
By definition of the norm we have
\begin{equation}
\|\vect x_n-\vect x\|
=\Bigl(\sum_{i=1}^N|x_{in}-x_i|^2\Bigr)^{1/2}\text{.}\tag{3.2}
\end{equation}
If
\(\vect x_n\xrightarrow{n\to\infty}\vect x\) then by
(3.1) we know that
\(x_{in}\xrightarrow{n\to\infty}x_i\) for every
\(i=1,\dots,N\text{,}\) and thus every term on the right hand side of
(3.2) goes to zero. This shows that
\(\|\vect x_n-\vect x\|\xrightarrow{n\to\infty}0\) if
\(\vect x_n\xrightarrow{n\to\infty}\vect x\text{.}\)
Now assume that
\(\|\vect x_n-\vect x\|\xrightarrow{n\to\infty}0\text{.}\) Observe that every term in the sum on the right hand side of
(3.2) is non-negative. Hence
\begin{equation*}
0\leq|x_{in}-x_i| \leq\Bigl(\sum_{i=1}^N|x_{in}-x_i|^2\Bigr)^{1/2} =\|\vect x_n-\vect x\|
\end{equation*}
for every \(i=1,\dots,N\text{.}\) By assumption \(\|\vect x_n-\vect x\|\xrightarrow{n\to\infty}0\text{,}\) and thus by the “squeezing lemma” \(|x_{in}-x_i|\xrightarrow{n\to\infty}0\) for all \(i=1,\dots,N\text{.}\)
This shows that thus \(x_{in}\xrightarrow{n\to\infty}x_i\) for all \(i=1,\dots,N\) as required.
As for sequences in \(\mathbb R\) we can give an \(\varepsilon\)-characterisation of convergence. It can also be used as a definition of convergence of a sequence. The proof is similar to the proof of the above proposition and left as an exercise.
Proposition 3.2.
A sequence \((\vect x_n)\) in \(\mathbb R^N\) converges to \(\vect x_0\in\mathbb R^N\) if and only if for every \(\varepsilon\gt 0\) there exists \(n_0\in\mathbb N\) such that \(\|\vect x_n-\vect x_0\|\lt \varepsilon\) for all \(n\gt n_0\text{.}\)
Limits of vectors have the same properties as limits of sequences of real numbers.
Proposition 3.3.
Suppose that \((\vect x_n)\text{,}\) \((\vect y_n)\) are sequences in \(\mathbb R^N\) with limits \(\vect x\) and \(\vect y\text{,}\) respectively. Moreover, assume that \((\alpha_n)\) is a sequence in \(\mathbb R\) with limit \(\alpha\text{.}\) Then
\(\displaystyle \lim_{n\to\infty}(\vect x_n+\vect y_n)
=\lim_{n\to\infty}\vect x_n+\lim_{n\to\infty}\vect y_n
=\vect x+\vect y\)
\(\displaystyle \lim_{n\to\infty}\alpha_n\vect x_n
=(\lim_{n\to\infty}\alpha_n)(\lim_{n\to\infty}\vect x_n)
=\alpha\vect x\)
\(\displaystyle \lim_{n\to\infty}(\vect x_n\cdot\vect y_n)
=(\lim_{n\to\infty}\vect x_n)\cdot(\lim_{n\to\infty}\vect y_n)
=\vect x\cdot\vect y\)
If \(N=3\) then \(\lim_{n\to\infty}(\vect x_n\times\vect y_n) =(\lim_{n\to\infty}\vect x_n)\times(\lim_{n\to\infty}\vect y_n) =\vect x\times\vect y\)
Proof.
The proof follows directly from the properties of sequences in \(\mathbb R\) and the definition of the multiplication by scalars, the scalar and the cross product.
Example 3.4.
Find the limit of
\begin{equation*}
\vect x_n=\Bigl(\frac{n-1}{n+1}, 2^{-n},\sqrt[n]{n}\Bigr)\text{.}
\end{equation*}
Solution.
We must find the limit of every component. We have
\begin{equation*}
\lim_{n\to\infty}\frac{n-1}{n+1}=1,\qquad
\lim_{n\to\infty}2^{-n}=0\qquad\text{and}\qquad
\lim_{n\to\infty}\sqrt[n]{n}=1\text{.}
\end{equation*}
Hence \(\lim_{n\to\infty}\vect x_n=(1,0,1)\text{.}\)
Example 3.5.
Find the limit of \(\vect x_n=((-1)^n, 1, 1/n)\text{.}\)
Solution.
We again have to check whether every component converges. This is not the case for the present sequence as \((-1)^n\) does not converge. Hence the sequence \(\vect x_n\) does not converge.
Example 3.6.
Find the limit of
\begin{equation*}
\vect x_n=\Bigl(\frac{n}{1+n^2}, e^{-n},
\frac{\ln n}{n},n^{-n}\Bigr)\in\mathbb R^4\text{.}
\end{equation*}
Solution.
We must find the limit of every component. We have
\begin{equation*}
\lim_{n\to\infty}\frac{n}{1+n^2}=0,\quad
\lim_{n\to\infty}e^{-n}=0,\quad
\lim_{n\to\infty}\frac{\ln n}{n}=0
\quad\text{and}\quad
\lim_{n\to\infty}n^{-n}=0\text{.}
\end{equation*}
Hence \(\lim_{n\to\infty}\vect x_n=(0,0,0,0)\text{.}\)
