On this page we present 20 practice problems, solved step by step, on the sign-preservation theorem for sequences. The aim is to learn how to recognize when a sequence is eventually positive or eventually negative from the behavior of its limit.
The problems are designed to clarify the precise meaning of the word eventually, the crucial role of the hypothesis \(L\neq0\), and the cases in which the theorem cannot be applied, such as sequences whose limit is zero or sequences that have no limit at all.
In some problems we shall apply the sign-preservation theorem directly; in others we shall also determine a possible index \(N\) from which on the sign of the sequence remains stable. In this way the theoretical result is concretely linked to the study of the sign of the terms of a sequence.
Recall that if \(\lim_{n\to+\infty}a_n=L\) with \(L>0\), then there exists \(N\in\mathbb{N}\) such that \(a_n>0\) for every \(n\geq N\). If, on the other hand, \(\lim_{n\to+\infty}a_n=L\) with \(L<0\), then there exists \(N\in\mathbb{N}\) such that \(a_n<0\) for every \(n\geq N\).
In symbols:
\[ \lim_{n\to+\infty}a_n=L>0 \quad\Longrightarrow\quad \exists N\in\mathbb{N}:\forall n\geq N,\ a_n>0, \]
whereas
\[ \lim_{n\to+\infty}a_n=L<0 \quad\Longrightarrow\quad \exists N\in\mathbb{N}:\forall n\geq N,\ a_n<0. \]
Exercise 1 — level ★☆☆☆☆
Determine whether the sequence
\[ a_n=2+\frac{1}{n} \]
is eventually positive, justifying your answer by means of the sign-preservation theorem.
Answer
The sequence is eventually positive.
Solution
We compute the limit of the sequence:
\[ \lim_{n\to+\infty}\left(2+\frac{1}{n}\right)=2. \]
The limit exists, is real, and is non-zero. Moreover it is positive, since
\[ 2>0. \]
By the sign-preservation theorem, if a sequence converges to a positive limit, then its terms are eventually positive.
Hence there exists an index \(N\in\mathbb{N}\) such that, for every \(n\geq N\),
\[ a_n>0. \]
In this particular case, one sees directly that
\[ 2+\frac{1}{n}>0 \]
for every \(n\geq1\). Thus the sequence is not merely eventually positive: it is positive at every index of its domain.
Exercise 2 — level ★☆☆☆☆
Determine whether the sequence
\[ a_n=\frac{3}{n}-5 \]
is eventually negative.
Answer
The sequence is eventually negative.
Solution
We compute the limit:
\[ \lim_{n\to+\infty}\left(\frac{3}{n}-5\right)=-5. \]
The limit is a negative real number, and it is non-zero.
By the sign-preservation theorem, a sequence converging to a negative limit is eventually negative.
Hence there exists \(N\in\mathbb{N}\) such that, for every \(n\geq N\),
\[ a_n<0. \]
We may also check directly from which index this occurs. We solve:
\[ \frac{3}{n}-5<0. \]
Moving \(5\) to the right-hand side, we obtain
\[ \frac{3}{n}<5. \]
Since \(n>0\), we may multiply by \(n\) without reversing the inequality:
\[ 3<5n. \]
Hence
\[ n>\frac{3}{5}. \]
This inequality holds for every \(n\geq1\). Hence the sequence is negative for every \(n\geq1\), and therefore certainly eventually negative.
Exercise 3 — level ★☆☆☆☆
Determine whether the sequence
\[ a_n=-1+\frac{4}{n} \]
is eventually negative, and find a possible index \(N\).
Answer
The sequence is eventually negative. A possible index is \(N=5\).
Solution
We compute the limit:
\[ \lim_{n\to+\infty}\left(-1+\frac{4}{n}\right)=-1. \]
The limit is negative and non-zero. Hence, by the sign-preservation theorem, the sequence is eventually negative.
To find an index \(N\) explicitly, we solve the inequality
\[ -1+\frac{4}{n}<0. \]
Moving \(-1\) to the right-hand side:
\[ \frac{4}{n}<1. \]
Since \(n>0\), we multiply by \(n\):
\[ 4<n. \]
Thus the inequality holds for every \(n>4\), that is, for every \(n\geq5\).
Therefore, choosing \(N=5\), we have
\[ a_n<0 \]
for every \(n\geq5\).
Exercise 4 — level ★☆☆☆☆
Determine whether the sequence
\[ a_n=7-\frac{10}{n} \]
is eventually positive, and find a possible index \(N\).
Answer
The sequence is eventually positive. A possible index is \(N=2\).
Solution
We compute the limit:
\[ \lim_{n\to+\infty}\left(7-\frac{10}{n}\right)=7. \]
The limit is positive and non-zero.
By the sign-preservation theorem, the sequence is eventually positive.
We now find an explicit index. We must solve:
\[ 7-\frac{10}{n}>0. \]
Moving the fraction to the right-hand side:
\[ 7>\frac{10}{n}. \]
Since \(n>0\), we multiply by \(n\):
\[ 7n>10. \]
Hence
\[ n>\frac{10}{7}. \]
The smallest integer \(n\geq1\) satisfying this condition is \(n=2\).
Hence, choosing \(N=2\), we have
\[ a_n>0 \]
for every \(n\geq2\).
Exercise 5 — level ★★☆☆☆
Determine whether the sequence
\[ a_n=\frac{2n+1}{n+3} \]
is eventually positive.
Answer
The sequence is eventually positive.
Solution
We divide both numerator and denominator by \(n\):
\[ a_n=\frac{2+\displaystyle \frac{1}{n}}{1+\displaystyle \frac{3}{n}}. \]
Passing to the limit:
\[ \lim_{n\to+\infty}\frac{2+\displaystyle \frac{1}{n}}{1+\displaystyle \frac{3}{n}}=\frac{2+0}{1+0}=2. \]
The limit is positive and non-zero.
By the sign-preservation theorem, the sequence is eventually positive.
In fact, even without the theorem we may observe that, for every \(n\geq1\),
\[ 2n+1>0 \qquad \text{and} \qquad n+3>0. \]
Hence
\[ \frac{2n+1}{n+3}>0 \]
for every \(n\geq1\).
This confirms what the theorem predicts: the sequence is certainly eventually positive.
Exercise 6 — level ★★☆☆☆
Determine whether the sequence
\[ a_n=\frac{1-3n}{2n+5} \]
is eventually negative.
Answer
The sequence is eventually negative.
Solution
We divide both numerator and denominator by \(n\):
\[ a_n=\frac{\displaystyle \frac{1}{n}-3}{2+\displaystyle \frac{5}{n}}. \]
We compute the limit:
\[ \lim_{n\to+\infty}\frac{\displaystyle \frac{1}{n}-3}{2+\displaystyle \frac{5}{n}}=\frac{0-3}{2+0}=-\frac{3}{2}. \]
The limit is negative and non-zero.
By the sign-preservation theorem, the sequence is eventually negative.
We may also check the sign directly. For \(n\geq1\) the denominator \(2n+5\) is positive, so the sign of the fraction depends on the numerator:
\[ 1-3n<0. \]
This inequality is equivalent to
\[ 1<3n, \]
that is,
\[ n>\frac{1}{3}. \]
It holds for every \(n\geq1\). Hence the sequence is negative for every \(n\geq1\), and therefore eventually negative.
Exercise 7 — level ★★☆☆☆
Determine whether the sequence
\[ a_n=\frac{n^2-4n+1}{n^2+1} \]
is eventually positive.
Answer
The sequence is eventually positive.
Solution
We divide both numerator and denominator by \(n^2\):
\[ a_n=\frac{1-\displaystyle \frac{4}{n}+\displaystyle \frac{1}{n^2}}{1+\displaystyle \frac{1}{n^2}}. \]
We compute the limit:
\[ \lim_{n\to+\infty}\frac{1-\displaystyle \frac{4}{n}+\displaystyle \frac{1}{n^2}}{1+\displaystyle \frac{1}{n^2}}=\frac{1-0+0}{1+0}=1. \]
The limit is positive and non-zero.
By the sign-preservation theorem, the sequence is eventually positive.
Observe that the theorem does not assert that the sequence is positive for every \(n\). It states only that there exists an index \(N\) from which onwards all the terms are positive.
Indeed, the numerator
\[ n^2-4n+1 \]
may take negative values for some initial indices, but this does not contradict the theorem.
Since the limit is \(1>0\), from some index onwards we certainly have
\[ a_n>0. \]
Exercise 8 — level ★★☆☆☆
Determine whether the sequence
\[ a_n=\frac{-2n^2+n+4}{n^2+3} \]
is eventually negative.
Answer
The sequence is eventually negative.
Solution
We divide both numerator and denominator by \(n^2\):
\[ a_n=\frac{-2+\displaystyle \frac{1}{n}+\displaystyle \frac{4}{n^2}}{1+\displaystyle \frac{3}{n^2}}. \]
Passing to the limit:
\[ \lim_{n\to+\infty}\frac{-2+\displaystyle \frac{1}{n}+\displaystyle \frac{4}{n^2}}{1+\displaystyle \frac{3}{n^2}}=\frac{-2+0+0}{1+0}=-2. \]
The limit is negative and non-zero.
By the sign-preservation theorem, the sequence is eventually negative.
Hence there exists \(N\in\mathbb{N}\) such that, for every \(n\geq N\),
\[ a_n<0. \]
Even if the initial terms had to be checked separately, this would be irrelevant for the purposes of the theorem, since the theorem concerns the eventual behaviour of the sequence.
Exercise 9 — level ★★☆☆☆
Determine whether the sign-preservation theorem is applicable to the sequence
\[ a_n=\frac{(-1)^n}{n}. \]
Determine, moreover, whether the sequence is eventually positive or eventually negative.
Answer
The theorem is not applicable, because the limit is \(0\). The sequence is neither eventually positive nor eventually negative.
Solution
We compute the limit of the sequence:
\[ a_n=\frac{(-1)^n}{n}. \]
Since
\[ -1\leq (-1)^n\leq 1, \]
dividing by \(n>0\) we obtain
\[ -\frac{1}{n}\leq \frac{(-1)^n}{n}\leq \frac{1}{n}. \]
Since
\[ \lim_{n\to+\infty}\left(-\frac{1}{n}\right)=0 \qquad \text{and} \qquad \lim_{n\to+\infty}\frac{1}{n}=0, \]
the squeeze theorem yields
\[ \lim_{n\to+\infty}\frac{(-1)^n}{n}=0. \]
The limit exists, but it is zero.
The sign-preservation theorem requires the limit to be real and non-zero. Here the hypothesis \(L\neq 0\) is not satisfied, so the theorem is not applicable.
We now study the sign of the sequence directly. If \(n\) is even, then \((-1)^n=1\), so
\[ a_n=\frac{1}{n}>0. \]
If, on the other hand, \(n\) is odd, then \((-1)^n=-1\), so
\[ a_n=-\frac{1}{n}<0. \]
Hence the sequence changes sign infinitely often.
Recall that saying a sequence is eventually positive means that there exists an index \(N\in\mathbb{N}\) such that
\[ a_n>0 \]
for every \(n\geq N\). But this does not happen, because beyond any index there still remain infinitely many odd indices, for which \(a_n<0\).
In the same way, the sequence is not eventually negative, because beyond any index there still remain infinitely many even indices, for which \(a_n>0\).
Hence the theorem is not applicable, and the sequence is neither eventually positive nor eventually negative.
Exercise 10 — level ★★☆☆☆
Determine whether the sign-preservation theorem is applicable to the sequence
\[ a_n=(-1)^n+2. \]
Determine, moreover, whether the sequence is eventually positive.
Answer
The theorem is not applicable, because the sequence has no limit. Nevertheless the sequence is positive for every \(n\), and is therefore eventually positive.
Solution
We examine the behaviour of the sequence:
\[ a_n=(-1)^n+2. \]
If \(n\) is even, then \((-1)^n=1\), so
\[ a_n=1+2=3. \]
If \(n\) is odd, then \((-1)^n=-1\), so
\[ a_n=-1+2=1. \]
Thus the sequence takes alternately the values \(3\) and \(1\). In particular, the subsequence of even-indexed terms tends to \(3\), whereas the subsequence of odd-indexed terms tends to \(1\).
Since these two values are different, the sequence does not converge to a single real limit.
The sign-preservation theorem requires the existence of a non-zero real limit. Here the limit does not exist, so the theorem is not applicable.
Nevertheless we can study the sign directly. We have seen that, for every \(n\), the sequence takes only the values \(1\) and \(3\). Hence
\[ a_n>0 \]
for every \(n\).
It follows that the sequence is eventually positive. Indeed, we may take, for instance, \(N=1\), and for every \(n\geq1\) we have \(a_n>0\).
This exercise shows that a sequence may be eventually positive even when the sign-preservation theorem is not applicable. In that case, however, eventual positivity must be established directly, and not deduced from the theorem.
Exercise 11 — level ★★☆☆☆
Determine whether the sequence
\[ a_n=\frac{n-5}{n+2} \]
is eventually positive, and find a possible index \(N\).
Answer
The sequence is eventually positive. A possible index is \(N=6\).
Solution
We compute the limit of the sequence:
\[ \lim_{n\to+\infty}\frac{n-5}{n+2}=1. \]
Indeed, numerator and denominator have the same degree, and the ratio of the leading coefficients is
\[ \frac{1}{1}=1. \]
The limit is positive and non-zero.
By the sign-preservation theorem, the sequence is eventually positive.
We now find an index \(N\) explicitly. We must solve:
\[ \frac{n-5}{n+2}>0. \]
For \(n\geq1\) the denominator is positive, since
\[ n+2>0. \]
Hence the sign of the fraction depends on the numerator:
\[ n-5>0. \]
We obtain
\[ n>5. \]
Hence, for every \(n\geq6\), we have
\[ a_n>0. \]
We may therefore choose \(N=6\).
Exercise 12 — level ★★☆☆☆
Determine whether the sequence
\[ a_n=\frac{4-2n}{n+1} \]
is eventually negative, and find a possible index \(N\).
Answer
The sequence is eventually negative. A possible index is \(N=3\).
Solution
We compute the limit:
\[ \lim_{n\to+\infty}\frac{4-2n}{n+1}=-2. \]
The limit is negative and non-zero.
By the sign-preservation theorem, the sequence is eventually negative.
We now find an explicit index. We must solve:
\[ \frac{4-2n}{n+1}<0. \]
For \(n\geq1\) the denominator is positive:
\[ n+1>0. \]
Hence the fraction is negative when the numerator is negative:
\[ 4-2n<0. \]
Moving \(2n\) to the right-hand side:
\[ 4<2n. \]
Dividing by \(2\), we obtain
\[ 2<n. \]
Hence
\[ n>2. \]
For every \(n\geq3\) the sequence is negative. We may therefore choose \(N=3\).
Exercise 13 — level ★★☆☆☆
Prove that the sequence
\[ a_n=3-\frac{2}{n} \]
is eventually greater than \(1\).
Answer
The sequence is eventually greater than \(1\). A possible index is \(N=2\).
Solution
The sign-preservation theorem concerns the sign of a sequence. To study the inequality
\[ a_n>1, \]
we move everything to the left-hand side and consider the auxiliary sequence
\[ b_n=a_n-1. \]
Since
\[ a_n=3-\frac{2}{n}, \]
we have
\[ b_n=3-\frac{2}{n}-1=2-\frac{2}{n}. \]
We compute the limit of \(b_n\):
\[ \lim_{n\to+\infty}\left(2-\frac{2}{n}\right)=2. \]
The limit is positive and non-zero. Hence, by the sign-preservation theorem,
\[ b_n>0 \]
eventually.
But \(b_n>0\) means exactly that
\[ a_n-1>0, \]
that is,
\[ a_n>1. \]
Hence \(a_n\) is eventually greater than \(1\).
We also find an explicit index:
\[ 3-\frac{2}{n}>1. \]
Moving \(1\) to the left-hand side:
\[ 2-\frac{2}{n}>0. \]
Hence
\[ 2>\frac{2}{n}. \]
Since \(n>0\), we multiply by \(n\):
\[ 2n>2. \]
Dividing by \(2\), we obtain
\[ n>1. \]
Therefore, for every \(n\geq2\), we have
\[ a_n>1. \]
Exercise 14 — level ★★☆☆☆
Prove that the sequence
\[ a_n=-3+\frac{5}{n} \]
is eventually less than \(-1\).
Answer
The sequence is eventually less than \(-1\). A possible index is \(N=3\).
Solution
We wish to prove that
\[ a_n<-1 \]
eventually.
We move everything to the left-hand side:
\[ a_n+1<0. \]
We therefore consider the auxiliary sequence
\[ b_n=a_n+1. \]
Since
\[ a_n=-3+\frac{5}{n}, \]
we obtain
\[ b_n=-3+\frac{5}{n}+1=-2+\frac{5}{n}. \]
We compute the limit:
\[ \lim_{n\to+\infty}\left(-2+\frac{5}{n}\right)=-2. \]
The limit is negative and non-zero.
By the sign-preservation theorem,
\[ b_n<0 \]
eventually.
But \(b_n<0\) means
\[ a_n+1<0, \]
that is,
\[ a_n<-1. \]
Hence \(a_n\) is eventually less than \(-1\).
We now find an explicit index:
\[ -3+\frac{5}{n}<-1. \]
We add \(3\) to both sides:
\[ \frac{5}{n}<2. \]
Since \(n>0\), we multiply by \(n\):
\[ 5<2n. \]
Hence
\[ n>\frac{5}{2}. \]
The smallest integer \(n\geq1\) satisfying this condition is \(n=3\).
We may therefore choose \(N=3\).
Exercise 15 — level ★★★☆☆
Prove that the sequence
\[ a_n=\frac{n^2+2n}{n^2+n+1} \]
is eventually greater than \(\displaystyle \frac{1}{2}\).
Answer
The sequence is eventually greater than \(\displaystyle \frac{1}{2}\).
Solution
We wish to prove that
\[ a_n>\frac{1}{2} \]
eventually.
In order to apply the sign-preservation theorem, we consider the difference
\[ b_n=a_n-\frac{1}{2}. \]
If we prove that \(b_n>0\) eventually, then we shall have proved that
\[ a_n>\frac{1}{2} \]
eventually.
We compute the limit of \(a_n\):
\[ \lim_{n\to+\infty}\frac{n^2+2n}{n^2+n+1}=1. \]
Hence
\[ \lim_{n\to+\infty}b_n=\lim_{n\to+\infty}\left(a_n-\frac{1}{2}\right)=1-\frac{1}{2}=\frac{1}{2}. \]
The limit of \(b_n\) is positive and non-zero.
By the sign-preservation theorem, the sequence \(b_n\) is eventually positive.
Hence
\[ a_n-\frac{1}{2}>0 \]
eventually, that is,
\[ a_n>\frac{1}{2} \]
eventually.
We may also check directly:
\[ \frac{n^2+2n}{n^2+n+1}>\frac{1}{2}. \]
Since \(n^2+n+1>0\) for every \(n\), we may multiply by \(2(n^2+n+1)\), which is positive:
\[ 2(n^2+2n)>n^2+n+1. \]
Expanding:
\[ 2n^2+4n>n^2+n+1. \]
Moving everything to the left-hand side:
\[ n^2+3n-1>0. \]
For \(n\geq1\) we have
\[ n^2+3n-1\geq 1+3-1=3>0. \]
Hence, in fact,
\[ a_n>\frac{1}{2} \]
for every \(n\geq1\).
Exercise 16 — level ★★★☆☆
Prove that the sequence
\[ a_n=\frac{-3n^2+n+1}{n^2+4} \]
is eventually less than \(-2\).
Answer
The sequence is eventually less than \(-2\).
Solution
We wish to prove that
\[ a_n<-2 \]
eventually.
We move everything to the left-hand side:
\[ a_n+2<0. \]
We therefore consider the auxiliary sequence
\[ b_n=a_n+2. \]
If \(b_n<0\) eventually, then \(a_n<-2\) eventually.
We compute the limit of \(a_n\):
\[ \lim_{n\to+\infty}\frac{-3n^2+n+1}{n^2+4}=-3. \]
Hence
\[ \lim_{n\to+\infty}b_n=\lim_{n\to+\infty}(a_n+2)=-3+2=-1. \]
The limit of \(b_n\) is negative and non-zero.
By the sign-preservation theorem, \(b_n\) is eventually negative.
Hence
\[ a_n+2<0 \]
eventually, that is,
\[ a_n<-2 \]
eventually.
Let us also check directly:
\[ \frac{-3n^2+n+1}{n^2+4}<-2. \]
Since \(n^2+4>0\), we may multiply without reversing the inequality:
\[ -3n^2+n+1<-2(n^2+4). \]
Expanding the right-hand side:
\[ -3n^2+n+1<-2n^2-8. \]
We move everything to the left-hand side:
\[ -n^2+n+9<0. \]
Multiplying by \(-1\), and reversing the inequality:
\[ n^2-n-9>0. \]
This inequality is certainly true for \(n\) sufficiently large. For instance, for \(n\geq4\) we have
\[ n^2-n-9\geq 16-4-9=3>0. \]
Hence a possible index is \(N=4\).
Exercise 17 — level ★★★☆☆
Determine whether the sequence
\[ a_n=\frac{(-1)^n}{n}+4 \]
is eventually positive.
Answer
The sequence is eventually positive.
Solution
The sequence contains the oscillating term \((-1)^n\), but this does not necessarily prevent the limit from existing.
Indeed,
\[ -1\leq (-1)^n\leq 1. \]
Dividing by \(n>0\), we obtain
\[ -\frac{1}{n}\leq \frac{(-1)^n}{n}\leq \frac{1}{n}. \]
Since
\[ \lim_{n\to+\infty}\left(-\frac{1}{n}\right)=0 \qquad \text{and} \qquad \lim_{n\to+\infty}\frac{1}{n}=0, \]
the squeeze theorem gives
\[ \lim_{n\to+\infty}\frac{(-1)^n}{n}=0. \]
Hence
\[ \lim_{n\to+\infty}\left(\frac{(-1)^n}{n}+4\right)=4. \]
The limit is positive and non-zero.
By the sign-preservation theorem, the sequence is eventually positive.
In fact, in this case we may also observe directly that
\[ \frac{(-1)^n}{n}\geq -\frac{1}{n}\geq -1 \]
for every \(n\geq1\). Hence
\[ a_n=\frac{(-1)^n}{n}+4\geq -1+4=3>0. \]
Hence the sequence is positive for every \(n\geq1\), and therefore eventually positive as well.
Exercise 18 — level ★★★☆☆
Determine whether the sequence
\[ a_n=\frac{(-1)^n}{n}-2 \]
is eventually negative.
Answer
The sequence is eventually negative.
Solution
We first study the limit. Since
\[ -1\leq (-1)^n\leq 1, \]
dividing by \(n>0\) we obtain
\[ -\frac{1}{n}\leq \frac{(-1)^n}{n}\leq \frac{1}{n}. \]
Both bounds tend to \(0\), so
\[ \lim_{n\to+\infty}\frac{(-1)^n}{n}=0. \]
Therefore
\[ \lim_{n\to+\infty}\left(\frac{(-1)^n}{n}-2\right)=-2. \]
The limit is negative and non-zero.
By the sign-preservation theorem, the sequence is eventually negative.
Directly as well, since
\[ \frac{(-1)^n}{n}\leq \frac{1}{n}\leq 1 \]
for every \(n\geq1\), it follows that
\[ a_n=\frac{(-1)^n}{n}-2\leq 1-2=-1<0. \]
Hence the sequence is negative for every \(n\geq1\), and therefore certainly eventually negative.
Exercise 19 — level ★★★☆☆
Determine whether the sequence
\[ a_n=\frac{n+(-1)^n}{n} \]
is eventually positive.
Answer
The sequence is eventually positive. A possible index is \(N=2\).
Solution
We rewrite the sequence by separating the two terms:
\[ a_n=\frac{n}{n}+\frac{(-1)^n}{n}. \]
Hence
\[ a_n=1+\frac{(-1)^n}{n}. \]
Since
\[ \lim_{n\to+\infty}\frac{(-1)^n}{n}=0, \]
we obtain
\[ \lim_{n\to+\infty}a_n=1+0=1. \]
The limit is positive and non-zero.
By the sign-preservation theorem, the sequence is eventually positive.
This means that there exists an index \(N\in\mathbb{N}\) such that, for every \(n\geq N\),
\[ a_n>0. \]
Note, however, that the theorem guarantees the existence of such an index \(N\), but it does not assert that one may necessarily take \(N=1\).
Indeed, for \(n=1\) we have
\[ a_1=\frac{1+(-1)^1}{1}=\frac{1-1}{1}=0. \]
Hence the sequence is not positive at every index of its domain, since the first term is zero.
Nevertheless this does not contradict the theorem, since the theorem concerns the eventual behaviour of the sequence, that is, what happens from some index onwards.
To find an explicit index, we observe that, for every \(n\),
\[ -1\leq (-1)^n\leq 1. \]
Hence
\[ n+(-1)^n\geq n-1. \]
If \(n\geq2\), then
\[ n-1>0. \]
Moreover \(n>0\). Therefore, for every \(n\geq2\),
\[ \frac{n+(-1)^n}{n}>0. \]
We may therefore choose \(N=2\).
This example illustrates clearly that “eventually positive” does not mean “positive from the very first index”, but rather “positive from some index onwards”.
Exercise 20 — level ★★★★☆
Determine whether the sign-preservation theorem is applicable to the sequence
\[ a_n=\frac{\sin\left(\frac{n\pi}{2}\right)}{n}. \]
Determine, moreover, whether the sequence is eventually positive or eventually negative.
Answer
The theorem is not applicable, because the limit is \(0\). The sequence is neither eventually positive nor eventually negative.
Solution
We first study the limit of the sequence:
\[ a_n=\frac{\sin\left(\frac{n\pi}{2}\right)}{n}. \]
For every \(n\) we know that
\[ -1\leq \sin\left(\frac{n\pi}{2}\right)\leq 1. \]
Since \(n>0\), dividing by \(n\) we obtain
\[ -\frac{1}{n}\leq \frac{\sin\left(\frac{n\pi}{2}\right)}{n}\leq \frac{1}{n}. \]
Since
\[ \lim_{n\to+\infty}\left(-\frac{1}{n}\right)=0 \qquad \text{and} \qquad \lim_{n\to+\infty}\frac{1}{n}=0, \]
the squeeze theorem yields
\[ \lim_{n\to+\infty}\frac{\sin\left(\frac{n\pi}{2}\right)}{n}=0. \]
The limit exists, but it equals \(0\). Hence the sign-preservation theorem is not applicable, since it requires a non-zero real limit.
We now study the sign of the sequence. The values of
\[ \sin\left(\frac{n\pi}{2}\right) \]
repeat cyclically:
\[ 1,\ 0,\ -1,\ 0,\ 1,\ 0,\ -1,\ 0,\dots \]
Indeed:
- if \(n=4k+1\), then \(\sin\left(\frac{n\pi}{2}\right)=1\);
- if \(n=4k+2\), then \(\sin\left(\frac{n\pi}{2}\right)=0\);
- if \(n=4k+3\), then \(\sin\left(\frac{n\pi}{2}\right)=-1\);
- if \(n=4k\), then \(\sin\left(\frac{n\pi}{2}\right)=0\).
Since the denominator \(n\) is always positive, the sign of \(a_n\) depends on the numerator.
Hence the sequence takes positive values infinitely often, negative values infinitely often, and the value \(0\) infinitely often.
It cannot be eventually positive, because beyond any index there still occur both zero terms and negative terms.
It cannot be eventually negative, because beyond any index there still occur both zero terms and positive terms.
Therefore the sequence is neither eventually positive nor eventually negative.
This exercise neatly summarises the role of the hypothesis \(L\neq 0\): when the limit is \(0\), the theorem allows no conclusion about the eventual sign, and the sign must be studied directly.