Convergence and divergence of sequences

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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 \(+\infty\). □

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