Numerical sequences and limits of sequences are fundamental concepts in mathematical analysis. Understanding the behavior of a sequence when \(n \to \pm \infty\) is crucial for determining whether a sequence \( a_n \) is convergent, divergent, or irregular.
Table of Contents
- Definition of Convergent Sequence
- Definition of Sequence Divergent to \(+\infty\)
- Definition of Sequence Divergent to \(-\infty\)
- Definition of Irregular Sequence
Definition of Convergent Sequence
A sequence \( a_n \) is said to be convergent to \( L \) if the absolute value of the difference between \( a_n \) and \( L \) can be made arbitrarily small for \( n \) sufficiently large. Intuitively, the terms of the sequence approach the limit value \( L \) indefinitely. More formally:
\[ \lim_{n \to \infty} a_n = L \iff \forall \varepsilon > 0 \quad \exists n_{\varepsilon} \in \mathbb{N} \,\,:\,\, \forall n \geq n_{\varepsilon} \quad |a_n - L| < \varepsilon \]
Geometrically, this means that for every interval \((L - \varepsilon, L + \varepsilon)\) centered at \(L\), there exists a point beyond which all terms of the sequence fall within this interval.
Exercise 1: Prove that \( a_n = \displaystyle \frac{1}{n} \) converges to zero.
\[ \lim_{n \to \infty} \frac{1}{n} = 0 \]
For every \( \varepsilon > 0 \), it suffices to choose \( n_{\varepsilon} > \displaystyle \frac{1}{\varepsilon} \). Then for every \( n \geq n_{\varepsilon} \), we have:
\[ \left| \frac{1}{n} \right| < \varepsilon \]
Therefore, the sequence converges to zero.
Exercise 2: Prove that \( a_n = \displaystyle \frac{n}{n+1} \) converges to 1.
\[ \left| \frac{n}{n+1} - 1 \right| = \frac{1}{n+1} < \varepsilon \]
This is satisfied if \( n > \displaystyle \frac{1}{\varepsilon} - 1 \). Therefore, the sequence converges to 1.
Definition of Sequence Divergent to \(+\infty\)
A sequence \( a_n \) diverges to \(+\infty\) if its terms become arbitrarily large. Formally:
\[ \lim_{n \to \infty} a_n = +\infty \iff \forall M > 0 \quad \exists n_M \in \mathbb{N} \,\,:\,\, \forall n \geq n_M \quad a_n > M \]
Exercise 3: Prove that \( a_n = 2n \) diverges to \(+\infty\).
\[ 2n > M \Rightarrow n > \frac{M}{2} \]
Choosing \( n_M > \displaystyle \frac{M}{2} \), for every \( n \geq n_M \) we have \( a_n = 2n > M \). Therefore, the sequence diverges to \(+\infty\).
Definition of Sequence Divergent to \(-\infty\)
A sequence \( a_n \) diverges to \(-\infty\) if its terms become arbitrarily negative. Formally:
\[ \lim_{n \to \infty} a_n = -\infty \iff \forall M > 0 \quad \exists n_M \in \mathbb{N} \,\,:\,\, \forall n \geq n_M \quad a_n < -M \]
Exercise 4: Prove that \( b_n = -3n \) diverges to \(-\infty\).
\[ -3n < -M \Rightarrow n > \frac{M}{3} \]
Choosing \( n_M > \displaystyle \frac{M}{3} \), we have \( b_n < -M \). Hence, \( \lim_{n \to \infty} b_n = -\infty \).
Exercise 5: Prove that \( a_n = -\ln(n) \) diverges to \(-\infty\).
\[ -\ln(n) < -M \Rightarrow n > e^M \]
Choosing \( n_M > e^M \), we have \( a_n = -\ln(n) < -M \). Therefore, the sequence diverges to \(-\infty\).
Definition of Irregular Sequence
A sequence \( a_n \) is said to be irregular if it is neither convergent nor divergent. The terms oscillate without tending to a limit value.
Exercise 6: Study the sequence \( a_n = (-1)^n \).
The sequence is bounded, but does not converge. Indeed:
\[ \lim_{n \to \infty} (-1)^{2n} = 1, \quad \lim_{n \to \infty} (-1)^{2n+1} = -1 \]
Two subsequences have different limits, so the sequence is irregular.
Exercise 7: Study the sequence \( a_n = \sin\left(\frac{n\pi}{2}\right) \).
The values are: \( 1, 0, -1, 0, 1, 0, -1, 0, \ldots \). The sequence is bounded, but does not converge. Indeed:
\[ \lim_{n \to \infty} a_{4n-3} = 1, \quad \lim_{n \to \infty} a_{4n-2} = 0 \]
Here too, two subsequences have different limits. The sequence is therefore irregular.
These examples show that being bounded does not imply convergence: a sequence can oscillate indefinitely.