Skip to main content

Vector Calculus An Introduction

Daniel Daners

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.
🔗

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.
🔗
Limits of vectors have the same properties as limits of sequences of real numbers.
🔗

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 \(\lim_{n\to\infty}\vect x_n=(0,0,0,0)\text{.}\)
🔗
🔗
🔗
🔗
PrevTopNext