Proposition 2.2.1.
The sequence \((c^n)\) is null if \(0<c<1\text{,}\) and diverges to \(+\infty\) if \(c>1\text{.}\)
Section 2.2 Convergence Tests for Sequences
In this section we study geometric sequences, sequences of the form \(c^{1/n}\text{,}\) and the ratio test. These results will be useful for deciding convergence of other sequences.
Subsection 2.2.1 Geometric Sequences
We begin with sequences of the form \((c^n)\) for \(c>0\text{.}\) The cases \(c=0\) and \(c=1\) are constant sequences, so the interesting cases are \(0<c<1\) and \(c>1\text{.}\)
Proof.
First suppose \(0 \lt c \lt 1\text{.}\) Then
\begin{equation*}
c>c^2>c^3>\cdots>0.
\end{equation*}
Hence \((c^n)\) is decreasing and bounded below, so Theorem 2.1.5 implies that it converges to some real number \(r\text{.}\) By Proposition 2.1.10, the subsequence \((c^{n+1})\) also converges to \(r\text{.}\) Since the constant sequence \(c\) converges to \(c\text{,}\) Proposition 2.1.6 gives \(c^{n+1} = c \cdot c^n \to cr\text{.}\) By Proposition 2.1.3, we must have \(r=cr\text{.}\) Because \(0<c<1\text{,}\) this implies \(r=0\text{.}\)
Now suppose \(c>1\text{.}\) Then \(0<1/c<1\text{,}\) so the first case gives \((1/c)^n\to 0\text{.}\) For any \(B > 0\text{,}\) since \((1/c)^n\to 0\text{,}\) we have \((1/c)^n<1/B\) for all sufficiently large \(n\text{.}\) Equivalently, \(c^n \gt B\) for all sufficiently large \(n\text{.}\) Thus \(c^n\to +\infty\text{.}\)
Example 2.2.2.
If \(-1<c<0\text{,}\) then \(|c|<1\text{,}\) so \((|c|^n)\to 0\) by Proposition 2.2.1. Since \(-|c|^n\le c^n\le |c|^n\text{,}\) Proposition 2.1.7 gives \(c^n\to 0\text{.}\) For example, \(((-1/2)^n)\) converges to \(0\text{.}\)
In contrast, \(((-1)^n)\) does not converge, because its odd terms are \(-1\) and its even terms are \(1\text{.}\)
Subsection 2.2.2 \(n\)-th Root
Proposition 2.2.3.
For every \(c>0\text{,}\) the sequence \((c^{1/n})\) converges to \(1\text{.}\)
Proof.
First suppose \(c\ge 1\text{.}\) Then
\begin{equation*}
c \ge c^{1/2} \ge c^{1/3} \ge \cdots \ge 1.
\end{equation*}
Thus \((c^{1/n})\) is decreasing and bounded below, so it converges by Theorem 2.1.5. Let \(L\) be its limit.
Every term satisfies \(c^{1/n}\ge 1\text{,}\) so \(L\ge 1\) by Proposition 2.1.8 applied to the sequence \((c^{1/n}-1)\text{.}\) The subsequence \((c^{1/(2n)})\) also converges to \(L\) by Proposition 2.1.10. Since \(c^{1/n}=(c^{1/(2n)})^2\text{,}\) Proposition 2.1.6 gives \(c^{1/n}\to L^2\text{.}\) By uniqueness of limits, \(L=L^2\text{.}\) Since \(L\ge1\text{,}\) it follows that \(L=1\text{.}\)
Now suppose \(0<c\le 1\text{.}\) Then \(1/c\ge1\text{,}\) so the first part shows \((1/c)^{1/n}\to 1\text{.}\) Since \(c^{1/n}=1/(1/c)^{1/n}\text{,}\) the reciprocal limit law gives \(c^{1/n}\to 1\text{.}\)
Subsection 2.2.3 Ratio Test
The next test gives a convenient condition for comparing a sequence with a geometric sequence.
Proposition 2.2.4. Ratio Test.
Let \((x_n)\) be a sequence of nonzero real numbers, and suppose the limit
\begin{equation*}
L=\lim_n \frac{|x_{n+1}|}{|x_n|}
\end{equation*}
exists. If \(L<1\text{,}\) then \((x_n)\) is null. If \(L>1\text{,}\) then \((x_n)\) is unbounded.
Proof.
First suppose \(L<1\text{.}\) Since the terms \(|x_{n+1}|/|x_n|\) are nonnegative, we also have \(L\ge0\) by Proposition 2.1.8. We have \(L<c:=(1+L)/2<1\text{,}\) so \(|x_{n+1}|/|x_n|<c\) for all but finitely many \(n\text{.}\) Since the convergence behavior and limit of a sequence do not change when finitely many terms are modified, we may assume that \(|x_{n+1}|<c|x_n|\) for all \(n\text{.}\) Thus, for any \(n\text{,}\)
\begin{equation*}
0 \le |x_{n+1}| \lt c|x_n| \lt c^2|x_{n-1}| \lt \cdots \lt
c^n|x_1|.
\end{equation*}
By Proposition 2.2.1, the sequence \((c^n)\) is null, and then Proposition 2.1.7 shows that \((x_n)\) is also null.
Now suppose \(L>1\text{.}\) Then \(1<c:=(1+L)/2<L\) and \(|x_{n+1}|/|x_n|>c\) for all but finitely many \(n\text{.}\) Again, we may assume that the inequality, or equivalently \(|x_{n+1}|>c|x_n|\text{,}\) holds for all \(n\text{.}\) Therefore,
\begin{equation*}
|x_{n+1}| \gt c|x_n| \gt c^2|x_{n-1}| \cdots \gt c^n|x_1|.
\end{equation*}
Because \(c >1\text{,}\) Proposition 2.2.1 shows that \(c^n \to +\infty\text{.}\) Thus, \((x_n)\) is unbounded.
Example 2.2.5.
If \(L=1\text{,}\) the ratio test gives no conclusion. For example, \(x_n=1/n\) satisfies \(|x_{n+1}|/|x_n|=n/(n+1)\to 1\) and is null, while the constant sequence \(y_n=1\) satisfies \(|y_{n+1}|/|y_n|=1\to 1\) and is not null.
For positive sequences \((a_n)\) and \((b_n)\text{,}\) we write \(a_n \ll b_n\) if \((a_n/b_n)\) is null.
Proposition 2.2.6. Growth Comparison.
Let \(a\ge 0\) and \(b>1\text{.}\) Then
\begin{equation*}
n^a \ll b^n \ll n! \ll n^n.
\end{equation*}
Proof.
First,
\begin{equation*}
\frac{(n+1)^a}{b^{n+1}}\frac{b^n}{n^a}
=
\left(1+\frac{1}{n}\right)^a \frac{1}{b}
\to \frac{1}{b} < 1.
\end{equation*}
Therefore \(n^a/b^n\to 0\) by Proposition 2.2.4.
Next,
\begin{equation*}
\frac{b^{n+1}}{(n+1)!}\frac{n!}{b^n}
=
\frac{b}{n+1}
\to 0.
\end{equation*}
Hence \(b^n/n!\to 0\) by Proposition 2.2.4.
Finally,
\begin{equation*}
0 \le \frac{n!}{n^n}
=
1\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots
\frac{1}{n}
< \frac{1}{n}.
\end{equation*}
Example 2.2.7.
The sequence \((n^{1/n})\) converges to \(1\text{.}\) To see this, let \(\varepsilon>0\) and set \(b=1+\varepsilon\text{.}\) The growth comparison proposition Proposition 2.2.6 with \(a=1\) shows that \(n/b^n\to 0\text{.}\) Hence \(n<b^n\) for all sufficiently large \(n\text{.}\) Taking \(n\)-th roots, we obtain \(1\le n^{1/n}\lt b = 1+\varepsilon\) for all sufficiently large \(n\text{.}\)
Solution.
Thus \(|n^{1/n}-1|<\varepsilon\) for all sufficiently large \(n\text{.}\) Therefore \(n^{1/n}\to 1\text{.}\)
