The sign permanence theorem for sequences states that if a real sequence \( a_n \) converges to a limit \( L \neq 0 \), there exists an index \( N \) beyond which all terms of the sequence have the same sign as \( L \). In other words:
\[ \lim_{n\to\infty} a_n = L > 0 \, \implies \, \exists N \in \mathbb{N} \, : \, \forall n \geq N \, , \, a_n > 0 \]
If instead \( L < 0 \), then:
\[ \lim_{n\to\infty} a_n = L < 0 \, \implies \, \exists N \in \mathbb{N} \, : \, \forall n \geq N \, , \, a_n < 0 \]
By definition, the limit of \( a_n \) is \( L \) if and only if:
\[ \lim_{n\to\infty} a_n = L \, \iff \, \exists N \in \mathbb{N} \, : \, \forall n \geq N \, , \, |a_n - L| < \epsilon \]
Specifically, choosing \( \epsilon = \frac{|L|}{2} \), we get the inequality:
\[ L - \frac{|L|}{2} < a_n < L + \frac{|L|}{2} \]
Now, let's examine the following cases:
- If \( L > 0 \), then:
\[ \frac{|L|}{2} < a_n < \frac{3|L|}{2} \quad \forall n \geq N \]
- If \( L < 0 \), i.e., \( L = -|L| \), then:
\[ -\frac{3|L|}{2} < a_n < -\frac{|L|}{2} \quad \forall n \geq N \]
In both cases, for \( n \geq N \), the terms of the sequence \( a_n \) will have the same sign as \( L \).
Exercise 1: Consider the sequence \( \displaystyle a_n = \frac{1}{n} \). Let's calculate the limit of \( a_n \) as \( n \to \infty \):
\[ \lim_{n \to \infty} \frac{1}{n} = 0 \]
Although the sequence tends to 0, we cannot apply the sign permanence theorem since the limit is zero.
Exercise 2: Consider the sequence \( \displaystyle a_n = \frac{3}{n} - 2 \). Let's calculate the limit of \( a_n \) as \( n \to \infty \):
\[ \lim_{n \to \infty} \left( \frac{3}{n} - 2 \right) = -2 \]
Choose \( \epsilon = 1 \). We need to find an index \( N \) such that for every \( n \geq N \), \( |a_n + 2| < 1 \). In this case, \( \displaystyle |a_n + 2| = \left| \frac{3}{n} \right| \).
We want \( \displaystyle \frac{3}{n} < 1 \), which is satisfied for \( n > 3 \). Therefore, for every \( n \geq 3 \), \( a_n \) is negative and tends to \( -2 \), maintaining the negative sign for all \( n \geq 3 \).
Exercise 3: Consider the sequence \( \displaystyle a_n = \frac{5}{n} + 1 \). Let's calculate the limit of \( a_n \) as \( n \to \infty \):
\[ \lim_{n \to \infty} \left( \frac{5}{n} + 1 \right) = 1 \]
Choose \( \displaystyle \epsilon = \frac{1}{2} \). We need to find an index \( N \) such that for every \( n \geq N \), we have:
\[ |a_n - 1| < \frac{1}{2} \]
In this case, \( |a_n - 1| = \displaystyle \left| \frac{5}{n} \right| \).
We want \( \displaystyle \frac{5}{n} < \frac{1}{2} \), which is satisfied for \( n > 10 \). Therefore, for every \( n \geq 10 \), \( a_n \) is positive and tends to \( 1 \), maintaining the positive sign for all \( n \geq 10 \).