Convergence and divergence of sequences

-

Consider the sequence \(a_1, a_2, a_3, \cdots\) where each \(a_i\) is a real number, and there is a real number \(a_n\) for each natural number \(n\). We usually denote such a sequence by \(\{a_n\}\) or \(\{a_n\}_{n=1}^{\infty}\). Note that a sequence is not just a set, its order matters. \(a_n\) may ``converge'' to some real number as \(n\) becomes larger and larger. But what does that mean?



Definition (Limits)

Let \(\{a_n\}\) be a sequence. If \(a_n\) approaches arbitrarily close to a constant value \(\alpha\) as \(n\) becomes arbitrarily large, we call this \(\alpha\) the limit of the sequence \(\{a_n\}\) and write

\[\lim_{n\to\infty}a_n = \alpha\]

or
\[a_n \to \alpha \text{ as } n \to \infty,\]
and we say the sequence \(\{a_n\}\) converges to \(\alpha\). In this case, we also say ``the limit of \(\{a_n\}\) is \(\alpha\).''

If the sequence \(\{a_n\}\) does not converge, we say that the sequence diverges.

Remark. When we say ``\(a_n\) approaches to \(\alpha\) as \(n\to \infty\)'', it means \(|a_n - \alpha|\) becomes smaller as \(n\) becomes larger. But we will make this notion more precise below. □

Example. Let \(\{a_n\}\) be a sequence with each term defined by \(a_n = \frac{(-1)^n}{n}\) for each \(n\in\mathbb{N}\). Then the sequence is
\[-1, \frac{1}{2}, -\frac{1}{3}, \frac{1}{4}, -\frac{1}{5}, \cdots.\]
□

Example. Let \(\{a_n\}\) be a sequence defined by \(a_n = 3n - 1\). We have
\[2, 5, 8, 11, 14, \cdots.\]
The numbers become arbitrarily large and positive numbers. Thus, this sequence diverges. □
In cases such as this last example, we say the sequence \(\{a_n\}\) diverges to the positive infinity and write
\[\lim_{n\to\infty}a_n = \infty\]
or
\[a_n \to \infty ~ (n \to \infty).\]

Example. Let \(\{a_n\}\) be defined by \(a_n = -n^2\) so we have
\[-1, -4, -9, -16, -25, \cdots.\]
The numbers become arbitrarily large negative numbers. So this sequence diverges. □
As in this example, when a sequence diverges to arbitrarily large negative values, we say the sequence \(\{a_n\}\) diverges to the negative infinity, and write
\[\lim_{n\to\infty}a_n = -\infty\]
or
\[a_n \to -\infty ~ (n \to \infty).\]

Example. Let \(\{a_n\}\) be defined by \(a_n = (-1)^n\). We have
\[-1, 1, -1, 1, \cdots\]
and this sequence diverges, but neither to the positive nor negative infinities. □

What do we exactly mean by convergence? Consider the sequence \(a_n = \frac{(-1)^n}{n}\).
\[ \begin{array}{c|cccccccc} n &1 & 2 & \cdots & 10 & 100 & 1,000 & 10,000 & \cdots\\\hline a_n & -1 & 0.5 & \cdots& 0.1 & 0.01 & 0.001 & 0.0001 & \cdots \end{array} \]
\(\lim_{n\to\infty}a_n = 0\) means that \(|a_n| = |a_n - 0|\) becomes arbitrarily small as \(n\) becomes larger. For example,
  • For \(n > 100\), \(|a_n - 0| < 0.01\).
  • For \(n > 1,000\), \(|a_n - 0| < 0.001\).
Generalizing these observations, for any arbitrarily small positive number, say \(\varepsilon = 0.0001\), we can always find some natural number, say \(N = 100,000\), such that if \(n > N\), then \(|a_n - 0| < \varepsilon\). Hence the following definition.

Definition (Convergence of a sequence)

The sequence \(\{a_n\}\) is said to converge to a real number \(\alpha\) if and only if the following condition holds:
  • For any \(\varepsilon > 0\), there exists \(N\in\mathbb{N}\) such that for any \(n\in\mathbb{N}\), if \(n\geq N\), then \(|a_n - \alpha| < \varepsilon\).
(Here, we implicitly assume \(\varepsilon \in\mathbb{R}\).)
Remark. We call this type of argument using \(\varepsilon\) and \(N\) the ``\(\varepsilon-N\) argument.'' □

Remark. In a logical form, the above condition for convergence can be expressed as
\[\forall \varepsilon > 0, \exists N\in\mathbb{N}, \forall n\in\mathbb{N} ~ (n\geq N\implies |a_n - \alpha|< \varepsilon).\]
□

Example. If \(a_n = \frac{n}{n+1}\), then \(\{a_n\}\) converges to \(\alpha = 1\). For example, let \(\varepsilon = 0.01\). If \(|a_n - \alpha| < \varepsilon\), then
\[1 - \frac{n}{n+1} < 0.01.\]
By solving this, we have
\[n > 99.\]
Therefore, if we set \(N = 100\), then for any \(n \geq N\), we have \(|a_n - \alpha| < \varepsilon\).
The same procedure can be applied for any values of \(\varepsilon > 0\). □

Example. Let us prove that
\[\lim_{n\to\infty}\frac{1}{n} = 0.\]
By Archimedes' principle, for any given \(\varepsilon > 0\), we can find a natural number \(N\) such that \(\varepsilon N > 1\), and hence \(\frac{1}{N} < \varepsilon\). For \(n\in\mathbb{N}\), if \(n\geq N\), then \(\frac{1}{n} \leq \frac{1}{N}\), so that
\[\left|\frac{1}{n} - 0\right| = \frac{1}{n} \leq \frac{1}{N} < \varepsilon.\]
□

Remark. Note that the choice of ``\(N\)'' depends on the value of \(\varepsilon\). For each \(\varepsilon >0\), we choose an appropriate \(N\in\mathbb{N}\). We cannot choose one \(N\in\mathbb{N}\) for all possible values of \(\varepsilon\). If that's the case, then we should have \(\frac{1}{N} < \varepsilon\) for any \(\varepsilon > 0\), which implies \(\frac{1}{N} = 0\), which is nonsense.

Example. Let us prove that \(a_n = (-1)^n\) does not converge by using the \(\varepsilon-N\) argument. 
We prove it by contradiction. 
Suppose that \(\{a_n\}\) converges to some real number \(\alpha\). Let's pick \(\varepsilon = 1\). There should exist some \(N\in\mathbb{N}\) such that, if \(n\geq N\), then
\[|a_n -\alpha| < 1.\]
When \(n\) is even, \(a_n = 1\) so \(|1 - \alpha| < 1\), in particular,
\[1 - \alpha < 1.\]
This implies that \(\alpha > 0\).

When \(n\) is odd, \(a_n = -1\) so \(|-1 -\alpha| < 1\), in particular,
\[1 + \alpha < 1.\]
This implies \(\alpha < 0\).

Therefore \(\alpha > 0\) and \(\alpha < 0\), which is a contradiction. Hence \(\{a_n\}\) does not converge. □

Definition (Divergence to \(\pm \infty\))

  1. The sequence \(\{a_n\}\) is said to diverge to the positive infinity, denoted \(\lim_{n\to\infty}a_n = +\infty\), if the following condition is satisfied.
    • For any \(M\in\mathbb{R}\), there exists \(N\in\mathbb{N}\) such that for any \(n\in\mathbb{N}\), if \(n\geq N\), then \(a_n > M\).
    Or, in a logical form, \[\forall M\in\mathbb{R}, \exists N\in\mathbb{N}, \forall n\in\mathbb{N} ~(n \geq N \implies a_n > M).\]
  2. The sequence \(\{a_n\}\) is said to diverge to the negative infinity, denoted \(\lim_{n\to\infty}a_n = -\infty\), if the following condition is satisfied.
    • For any \(M\in\mathbb{R}\), there exists \(N\in\mathbb{N}\) such that for any \(n\in\mathbb{N}\), if \(n\geq N\), then \(a_n < M\).
    Or, in a logical form, \[\forall M\in\mathbb{R}, \exists N\in\mathbb{N}, \forall n\in\mathbb{N} ~(n \geq N \implies a_n < M).\]
Example. Consider the sequence \(\{3n^2\}\). For a given arbitrary positive real number \(M>0\), let \(N = \left[\sqrt{\frac{M}{3}}\right]+1\) where \(\left[x\right]\) indicates the integer part of \(x\). Note, in particular, that \(N > \sqrt{\frac{M}{3}}\). Now, suppose \(n \geq N\). We have \(3n^2 \geq 3N^2 > M\). Therefore, the sequence \(\{3n^2\}\) diverges to \(+\infty\). □

Comments

Post a Comment

Popular posts from this blog

Open sets and closed sets in \(\mathbb{R}^n\)

-
Image
Open sets In \(\mathbb{R}\), we have the notion of an open interval such as \((a, b) = \{x \in \mathbb{R} | a < x < b\}\). We want to extend this idea to apply to \(\mathbb{R}^n\). We also introduce the notions of bounded sets and closed sets in \(\mathbb{R}^n\). Recall that the \(\varepsilon\)-neighbor of a point \(x\in\mathbb{R}^n\) is defined as \(N_{\varepsilon}(x) = \{y \in \mathbb{R}^n | d(x, y) < \varepsilon \}\) where \(d(x,y)\) is the distance between \(x\) and \(y\). Definition (Open set) A subset \(U\) of \(\mathbb{R}^n\) is said to be an  open set  if the following holds: \[\forall x \in U ~ \exists \delta > 0 ~ (N_{\delta}(x) \subset U).\tag{Eq:OpenSet}\] That is, for every point in an open set \(U\), we can always find an open ball centered at that point, that is included in \(U\). See the following figure. Perhaps, it is instructive to see what is  not  an open set. Negating (Eq:OpenSet), we have \[\exists x \in U ~ \forall \delta > 0 ~ (N_{\delta}(x) \not
Read more

Euclidean spaces

-
Image
We would like to study multivariate functions (i.e., functions of many variables), continuous multivariate functions in particular. To define continuity, we need a measure of "closeness" between points. One measure of closeness is the Euclidean distance. The set \(\mathbb{R}^n\) (with \(n \in \mathbb{N}\)) with the Euclidean distance function is called a Euclidean space. This is the space where our functions of interest live. The real line is a geometric representation of \(\mathbb{R}\), the set of all real numbers. That is, each \(a \in \mathbb{R}\) is represented as the point \(a\) on the real line. The coordinate plane , or the \(x\)-\(y\) plane , is a geometric representation of \(\mathbb{R}^2\), the set of all pairs of real numbers. Each pair of real numbers \((a, b)\) is visualized as the point \((a, b)\) in the plane. Remark . Recall that \(\mathbb{R}^2 = \mathbb{R}\times\mathbb{R} = \{(x, y) | x, y \in \mathbb{R}\}\) is the Cartesian product of \(\mathbb{R}\) with i
Read more

Newton's method

-
Image
Newton's method (the Newton-Raphson method) is a very powerful numerical method for solving nonlinear equations.  Suppose we'd like to solve a non-linear equation \(f(x) = 0\), where \(f(x)\) is a (twice) differentiable non-linear function. Newton's method generates a sequence of numbers \(c_1, c_2, c_3, \cdots\) that converges to a solution of the equation. That is, if \(\alpha\) is a solution (i.e., \(f(\alpha) = 0\)) then,  \[\lim_{n\to\infty}c_n = \alpha,\] and this sequence \(\{c_n\}\) is generated by a series of linear approximations of the function \(f(x)\). Theorem (Newton's method) Let \(f(x)\) be a function that is twice differentiable on an open interval \(I\) that contains the closed interval \([a, b]\) (i.e., \([a,b]\subset I\)) and satisfy the following conditions: \(f(a) < 0\) and \(f(b) > 0\); For all \(x\in [a, b]\), \(f'(x) > 0\) and \(f''(x) > 0\). Let us define the sequence \(\{c_n\}\) by \[ \begin{eqnarray} c_1 &
Read more