The sign preservation theorem for sequences states that if a real sequence converges to a nonzero limit, then its terms eventually have the same sign as that limit.
In other words, if a sequence \((a_n)\) tends to a positive number, then from some index onward all of its terms are positive. If, on the other hand, it tends to a negative number, then from some index onward all of its terms are negative.
The word eventually is essential: the theorem does not assert that every term of the sequence has the same sign as the limit, but only that this property holds for all sufficiently large indices.
In general, to say that a property holds eventually for a sequence means that there exists an index \(N\in\mathbb{N}\) such that the property holds for every \(n\geq N\).
The case \(L=0\) must be excluded. Indeed, if the limit is zero, the theorem allows no conclusion to be drawn about the eventual sign of the terms of the sequence.
Contents
- Permanence of sign theorem for sequences
- Proof of the theorem
- Interpretation of the theorem
- Why the case \(L=0\) is excluded
- Examples
Permanence of sign theorem for sequences
Let \((a_n)\) be a real sequence and let \(L\in\mathbb{R}\) with \(L\neq0\). Suppose that
\[ \lim_{n\to+\infty}a_n=L. \]
Then the sequence \((a_n)\) eventually has the same sign as \(L\).
More precisely:
- if \(L>0\), then there exists \(N\in\mathbb{N}\) such that, for every \(n\geq N\), we have \(a_n>0\);
- if \(L<0\), then there exists \(N\in\mathbb{N}\) such that, for every \(n\geq N\), we have \(a_n<0\).
In symbols, if \(L>0\), then
\[ \lim_{n\to+\infty}a_n=L \quad\Longrightarrow\quad \exists N\in\mathbb{N}\ :\ \forall n\geq N,\ a_n>0. \]
If, instead, \(L<0\), then
\[ \lim_{n\to+\infty}a_n=L \quad\Longrightarrow\quad \exists N\in\mathbb{N}\ :\ \forall n\geq N,\ a_n<0. \]
The theorem may be summarized by saying that, when the limit is different from zero, the sign of the limit is eventually reflected in the terms of the sequence.
The condition \(L\neq0\) is indispensable. If the limit were equal to zero, it would not be possible to choose a neighborhood of \(0\) contained entirely in the positive numbers or entirely in the negative numbers.
Proof of the theorem
Suppose that
\[ \lim_{n\to+\infty}a_n=L \]
with \(L\neq0\). By the definition of limit, for every \(\varepsilon>0\) there exists \(N\in\mathbb{N}\) such that, for every \(n\geq N\), we have
\[ |a_n-L|<\varepsilon. \]
Since \(L\neq0\), we have \(|L|>0\). We may therefore choose
\[ \varepsilon=\frac{|L|}{2}. \]
By the definition of limit, there then exists \(N\in\mathbb{N}\) such that, for every \(n\geq N\),
\[ |a_n-L|<\frac{|L|}{2}. \]
This inequality is equivalent to saying that
\[ -\frac{|L|}{2}<a_n-L<\frac{|L|}{2}. \]
Adding \(L\) to each side, we obtain
\[ L-\frac{|L|}{2}<a_n<L+\frac{|L|}{2}. \]
At this point we distinguish the two possible cases.
Case \(L>0\)
If \(L>0\), then \(|L|=L\). The preceding inequality becomes
\[ L-\frac{L}{2}<a_n<L+\frac{L}{2}. \]
Hence, for every \(n\geq N\),
\[ \frac{L}{2}<a_n<\frac{3L}{2}. \]
In particular,
\[ a_n>0 \]
for every \(n\geq N\). Thus, if the limit \(L\) is positive, then the terms of the sequence are eventually positive.
We observe, moreover, that we have obtained a stronger estimate: from some index onward, not only is \(a_n>0\), but in fact \(a_n>\frac{L}{2}\).
Case \(L<0\)
If \(L<0\), then \(|L|=-L\). The inequality
\[ L-\frac{|L|}{2}<a_n<L+\frac{|L|}{2} \]
therefore becomes
\[ L+\frac{L}{2}<a_n<L-\frac{L}{2}. \]
That is,
\[ \frac{3L}{2}<a_n<\frac{L}{2}. \]
Since \(L<0\), we have
\[ \frac{3L}{2}<\frac{L}{2}<0. \]
From the inequality
\[ \frac{3L}{2}<a_n<\frac{L}{2} \]
it follows, in particular, that
\[ a_n<0 \]
for every \(n\geq N\).
Thus, if the limit \(L\) is negative, then the terms of the sequence are eventually negative.
In both cases we have proved that if a real sequence converges to a nonzero limit, then its terms eventually have the same sign as the limit.
Interpretation of the theorem
The permanence of sign theorem does not necessarily concern all the terms of the sequence, but only the terms from a certain index onward.
To say that \(a_n>0\) eventually means that there exists an index \(N\in\mathbb{N}\) such that
\[ a_n>0 \]
for every \(n\geq N\). The terms with index less than \(N\), on the other hand, may have any sign whatsoever.
For example, a sequence may have a few negative initial terms and then become eventually positive. If its limit is positive, the theorem guarantees that from a certain point onward the terms can no longer be negative or zero.
Likewise, if the limit is negative, then from some index onward all the terms must be negative.
The geometric idea is simple: if \(L>0\), one can choose a neighborhood of \(L\) entirely contained in the positive numbers. Since \(a_n\to L\), from some index onward all the terms \(a_n\) lie in that neighborhood, and are therefore positive.
If, on the other hand, \(L<0\), one can choose a neighborhood of \(L\) entirely contained in the negative numbers. From some index onward all the terms of the sequence lie in that neighborhood, and are therefore negative.
Why the case \(L=0\) is excluded
The condition \(L\neq0\) is essential. If a sequence converges to \(0\), the permanence of sign theorem does not allow the eventual sign of its terms to be determined.
Indeed, there is no open interval centered at \(0\) that is contained entirely in the positive numbers or entirely in the negative numbers. Every open interval centered at \(0\) contains both positive and negative numbers.
For this reason, if
\[ \lim_{n\to+\infty}a_n=0, \]
one cannot conclude, in general, that \(a_n\) is eventually positive or eventually negative.
For example, the sequence
\[ a_n=\frac{1}{n}, \qquad n\geq 1. \]
converges to \(0\) and is positive for every \(n\geq 1\). By contrast, the sequence
\[ b_n=-\frac{1}{n}, \qquad n\geq 1. \]
also converges to \(0\) and is negative for every \(n\geq 1\).
Moreover, the sequence
\[ c_n=\frac{(-1)^n}{n}, \qquad n\geq 1. \]
converges to \(0\), but changes sign infinitely many times. Thus, in the case of a zero limit, a variety of behaviors are possible.
Examples
Example 1. Consider the sequence
\[ a_n=\frac{3}{n}-2. \]
Since
\[ \lim_{n\to+\infty}\left(\frac{3}{n}-2\right)=-2, \]
and the limit is negative, the permanence of sign theorem guarantees that \(a_n\) is eventually negative.
Let us verify this directly. We seek an index \(N\in\mathbb{N}\) such that, for every \(n\geq N\), we have
\[ \frac{3}{n}-2<0. \]
Solving the inequality,
\[ \frac{3}{n}<2. \]
Since \(n>0\), we may multiply through by \(n\) without reversing the direction of the inequality:
\[ 3<2n. \]
Hence
\[ n>\frac{3}{2}. \]
Therefore, for every \(n\geq2\), we have
\[ a_n<0. \]
The sequence is thus eventually negative, in agreement with the theorem.
Example 2. Consider the sequence
\[ a_n=\frac{5}{n}+1. \]
Since
\[ \lim_{n\to+\infty}\left(\frac{5}{n}+1\right)=1, \]
and the limit is positive, the permanence of sign theorem guarantees that \(a_n\) is eventually positive.
In fact, in this case the sequence is positive for every \(n\geq 1\), because
\[ \frac{5}{n}>0 \]
for every \(n\geq 1\), and hence
\[ \frac{5}{n}+1>0. \]
This is consistent with the theorem: if a property holds for every index, then it certainly holds eventually as well.
Example 3. Consider the sequence
\[ a_n=(-1)^n+\frac{1}{n}. \]
This sequence does not converge. Indeed, the term \((-1)^n\) oscillates between \(-1\) and \(1\), while \(\displaystyle \frac{1}{n}\to0\). Hence the even-indexed and odd-indexed subsequences tend to different limits.
Consequently we cannot apply the permanence of sign theorem.
This example shows that the theorem genuinely requires the existence of a nonzero real limit. If the limit does not exist, the theorem yields no information about the eventual sign of the sequence.
Example 4. Consider the sequence
\[ a_n=\frac{(-1)^n}{n}. \]
We have
\[ \lim_{n\to+\infty}\frac{(-1)^n}{n}=0. \]
However, the sequence changes sign infinitely many times: it is positive for even indices and negative for odd indices.
The permanence of sign theorem is not applicable, because the limit equals \(0\). This example shows why the hypothesis \(L\neq0\) is indispensable.
An analogous version holds for infinite limits as well: if \(a_n\to+\infty\), then \(a_n>0\) eventually; if \(a_n\to-\infty\), then \(a_n<0\) eventually.
In conclusion, the permanence of sign theorem for sequences states that the sign of a nonzero real limit is eventually preserved in the terms of the sequence. The initial terms may behave differently, but from some index onward the sign must coincide with that of the limit.