Limits of Sequences

Numerical sequences and limits of sequences are fundamental concepts in mathematical analysis. Understanding the behavior of a sequence as \(n \to \pm \infty \) is crucial in determining whether a sequence \( a_n \) is convergent, divergent, or irregular.

Table of Contents

• Convergent Sequence
• Divergent Sequence to \(+\infty\)
• Divergent Sequence to \(-\infty\)
• Irregular Sequence

A sequence \( a_n \) is said to converge to \( L \) if the absolute difference between \( a_n \) and \( L \) can be made arbitrarily small for sufficiently large \( n \).

In other words, we will say that \( a_n \) tends to \( L \) as \( n \to \infty \) if for every \( \varepsilon > 0 \), there exists a \( n_{\varepsilon} \in \mathbb{N} \) such that for all \( n \geq n_{\varepsilon} \), the absolute value \( | a_n - L | \) is less than \( \varepsilon \). More formally:

\[ \lim_{n \to \infty} a_n = L \iff \forall \varepsilon > 0 \,\, \exists n_{\varepsilon} \in \mathbb{N} \, : \, |a_n - L| < \varepsilon \quad \forall n \geq n_{\varepsilon} \]

Example: Consider the sequence \( a_n = \displaystyle \frac{1}{n} \). We will prove that \( \displaystyle \lim_{n \to \infty} a_n = 0 \).

To prove convergence, we use the definition of the limit. We need to verify that for every \( \varepsilon > 0 \), there exists a \( n_{\varepsilon} \in \mathbb{N} \) such that for all \( n \geq n_{\varepsilon} \) we have:

\[ |a_n - 0| = \left| \frac{1}{n} \right| < \varepsilon \]

From this inequality, we obtain:

\[ \frac{1}{n} < \varepsilon \implies n > \frac{1}{\varepsilon} \]

Thus, we choose \( n_{\varepsilon} = \left\lceil \frac{1}{\varepsilon} \right\rceil \) (the ceiling of \( \frac{1}{\varepsilon} \)). For every \( n \geq n_{\varepsilon} \), we have:

\[ \left| \frac{1}{n} \right| < \varepsilon \]

This proves that:

\[ \lim_{n \to \infty} a_n = 0 \]

The proof follows directly from the formal definition of the limit and shows how to properly apply the concept of convergence. This approach will later be crucial for proving the convergence of more complex sequences.

A sequence \( a_n \) is said to diverge to \( +\infty \) if its terms can be made arbitrarily large for sufficiently large \( n \).

In other words, we will say that \( a_n \) tends to \( +\infty \) as \( n \to \infty \) if for every \( M > 0 \), there exists a \( n_{M} \in \mathbb{N} \) such that for all \( n \geq n_{M} \), we have \( a_n > M \). More formally:

\[ \lim_{n \to \infty} a_n = +\infty \iff \forall M > 0 \,\, \exists n_{M} \in \mathbb{N} \, : \, a_n > M \quad \forall n \geq n_{M} \]

Example. Consider the sequence \(a_n = 2n\). To prove that it diverges to \( +\infty \), let’s take an arbitrary number \( M > 0 \). We need to find an index \( n_M \) such that for every \( n > n_M \), we have \( a_n > M \).

Take \( n_M = \left\lceil \frac{M}{2} \right\rceil \). If \( n > n_M \), then:

\[ a_n = 2n > 2\left(\frac{M}{2}\right) = M \]

Thus, we have proved that for every \( M > 0 \), there exists \( n_M = \left\lceil \frac{M}{2} \right\rceil \) such that \( a_n > M \) for every \( n > n_M \).

A sequence \( a_n \) is said to diverge to \( -\infty \) if its terms can be made arbitrarily small (negative) for sufficiently large \( n \).

In other words, we will say that \( a_n \) tends to \( -\infty \) as \( n \to \infty \) if for every \( M > 0 \), there exists a \( n_{M} \in \mathbb{N} \) such that for all \( n \geq n_{M} \), we have \( a_n < -M \). More formally:

\[ \lim_{n \to \infty} a_n = -\infty \iff \forall M > 0 \,\, \exists n_{M} \in \mathbb{N} \, : \, a_n < -M \quad \forall n \geq n_{M} \]

Example. Consider the sequence \(b_n = -3n\).

To prove that it diverges to \( -\infty \), take an arbitrary number \( M > 0 \). We need to find an index \( n_M \) such that for every \( n > n_M \), we have \( b_n < -M \).

Now, take \( n_M = \left\lceil \frac{M}{3} \right\rceil \). If \( n > n_M \), then:

\[ b_n = -3n < -3\left(\frac{M}{3}\right) = -M \]

Thus, we have proved that for every \( M > 0 \), there exists \( n_M = \left\lceil \frac{M}{3} \right\rceil \) such that \( b_n < -M \) for every \( n > n_M \).

A sequence \( a_n \) is said to be irregular if it is neither convergent nor divergent to \( +\infty \) or \( -\infty \).

Example. Consider the sequence:

\[ a_n = (-1)^n, \quad n \in \mathbb{N} \]

This sequence is neither convergent nor divergent. Notice that it is bounded because:

\[ -1 \leq (-1)^n \leq 1, \quad \forall n \in \mathbb{N} \]

Now, let us assume by contradiction that it converges to a limit \( L \):

\[ \lim_{n \to \infty} (-1)^n = L \]

Then it should also be true that:

\[ \lim_{n \to \infty} (-1)^{2n} = L \]

But \( (-1)^{2n} = 1 \) for all \( n \), so \( L = 1 \).

By the definition of limit, there should exist \( n_{\varepsilon} \) such that:

\[ \forall \varepsilon > 0 \,\, \exists n_{\varepsilon} \in \mathbb{N}: |(-1)^n - 1| < \varepsilon, \quad \forall n > n_{\varepsilon} \]

However, for odd \( n \) greater than \( n_{\varepsilon} \):

\[ |(-1)^n - 1| = |-1 - 1| = 2 \]

This is impossible for arbitrarily small \( \varepsilon \), so the sequence cannot converge. It is not divergent either because it is bounded.