Theorem 5.16. Monotonic Convergence Theorem.
Every increasing (decreasing) sequence that is bounded above (below) converges.
Section 5.2 Convergence Tests for Sequences
The following statement is intuitively clear. However, its validity relies on the completeness of real numbers. It allows ones to argue that a sequence is convergent without knowing its limit.
As an application of the monotonic convergence theorem, we show that:
Example 5.17.
\(\displaystyle \lim_{n\to \infty} \left(1 +
\frac{1}{n}\right)^n\) exists.
It follows from the Binomial Theorem that
\begin{equation*}
a_n = \sum_{k=0}^n \binom{n}{k}\frac{1}{n^k}.
\end{equation*}
Since for all \(k \ge 1\text{,}\)
\begin{equation*}
\ds \binom{n}{k}\frac{1}{n^k}
= \frac{n(n-1)\cdots (n-k+1)}{k!}\frac{1}{n^k} \lt \frac{1}{k!}
\le \frac{1}{2^{k-1}},
\end{equation*}
and so \(\ds a_n \le 1 +
\sum_{k=1}^n \frac{1}{2^{k-1}} = 1 + 2 = 3.\) It remains to show that \((a_n)\) is increasing.
For each \(0 \le k \le n\text{,}\)
\begin{equation*}
\binom{n+1}{k}\frac{1}{(n+1)^k} = \frac{1}{k!}\prod_{i=0}^k
\left(1 - \frac{i}{n+1}\right) \ge \frac{1}{k!}\prod_{i=0}^k
\left(1 - \frac{i}{n}\right) = \binom{n}{k}\frac{1}{n^k}.
\end{equation*}
So,
\begin{equation*}
a_n = \sum_{k=0}^n \binom{n}{k}\frac{1}{n^k} \le \sum_{k=0}^n
\binom{n+1}{k}\frac{1}{(n+1)^k} \lt \sum_{k=0}^{n+1}
\binom{n+1}{k}\frac{1}{(n+1)^k} = a_{n+1}.
\end{equation*}
This completes the proof.
The limit of \((1+1/n)^n\) can be taken as the definition of \(e\text{,}\) the base of the natural logarithm.
The following proposition is another application of the monotonic convergence theorem.
Proposition 5.18.
The sequence \((c^n)\) converges to \(0\) if \(0 \lt c \lt
1\) or diverges to \(+\infty\) if \(c \gt 1\text{.}\)
Proof.
Suppose \(0 \lt c \lt 1\text{.}\) Then
\begin{equation*}
c \gt c^2 \gt c^3 \gt
\cdots
\end{equation*}
is decreasing and is clearly bounded below (by \(0\)). Thus, \(c^n \to L\) for some number \(L\) by the monotonic convergence theorem 5.16. Consequently the sequence
\begin{equation*}
c^2 , c^3 , c^4,
\ldots
\end{equation*}
converges to \(cL\text{.}\) However, this sequence is just the original sequence with the first term dropped, so by the uniqueness of limits, \(cL = L\text{.}\) But \(0 \lt c \lt 1\text{,}\) so \(L\) must be \(0\text{.}\)
If \(c \gt 1\text{,}\) then \(0 \lt 1/c \lt 1\text{,}\) so \(1/c^n \to
0^+\) and hence \(c^n \to +\infty\text{.}\)
Proposition 5.19. Ratio Test for sequence convergence.
Let \((x_n)\) be a sequence of non-zero real numbers and the limit \(L=\lim_n
|x_{n+1}|/|x_n|\) exists. Then
\begin{equation*}
\begin{cases}
(x_n)\ \text{is null} & \text{if}\ L \lt 1 \\
(x_n)\ \text{is unbounded} & \text{if}\ L \gt 1.
\end{cases}
\end{equation*}
Let \((x_n)\) and \((y_n)\) be two sequences of positive real numbers. We write \((x_n) \ll (y_n)\) if the sequence of ratios \((x_n/y_n)\) is null. Intuitively, \((x_n) \ll (y_n)\) means \(y_n\) is way bigger than \(x_n\) eventually.
Proposition 5.20.
For any \(a \ge 0\) and \(b \gt 1\text{,}\)
\begin{equation*}
n^a \ll b^n \ll n! \ll n^n
\end{equation*}
Proof.
Since \(b \gt 1\text{,}\) \(\ds \frac{(n+1)^a}{b^{n+1}}\frac{b^n}{n^a} =
\left(1+\frac{1}{n}\right)^a \frac{1}{b} \to \frac{1}{b}
\lt 1\text{,}\) so \(\ds \frac{n^a}{b^n} \to 0\) by the ratio test 5.19.
Also, \(\ds
\frac{b^{n+1}}{(n+1)!}\frac{n!}{b^n} = \frac{b}{n+1} \to 0\text{,}\) thus \(\ds \frac{b^n}{n!} \to 0\) by the ratio test.
Since
\begin{equation*}
0 \le \frac{n!}{n^n} = \frac{1}{n}\frac{2}{n} \cdots
\frac{n}{n} \lt \frac{1}{n},
\end{equation*}
So \(n!/n^n \to 0\) by the Squeeze Lemma 5.11.
There is useful a characterization continuity by convergence of sequence.
Proposition 5.21.
Let \(f \colon S \to \mathbb{R}\) be a function. Then \(f\) is continuous at \(a \in
S\) if and only if for every sequence \((a_n)\) in \(S\) that converges to \(a\text{,}\) \(f(a_n) \to f(a)\text{.}\)
Now, let us give a few applications of this result.
Example 5.22.
For any \(\ve \gt 0\text{,}\) \(e^{\ve} \gt 1\text{.}\) So, according to Proposition 5.20 \(n \ll
(e^{\ve})^n\text{.}\) Taking natural log on both sides yields \(\ln(n)/n
\lt \ve\) for all \(n\) sufficiently large. Since \(\ve \gt
0\) is arbitrary, this shows that \(\ln(n)/n \to 0\text{.}\)
Example 5.23.
\(n^{1/n} \to 1\text{.}\) This follows immediately from \(\ln(n)/n \to 0\) (Example 5.22), Proposition 5.21 and continuity of \(e^x\) at \(0\text{.}\)
Another way of computing the limit is as follows: since \(1 \le
n^{1/n}\text{,}\) it suffices to show that for any \(\varepsilon \gt 0,
n^{1/n} \lt 1+\varepsilon\) eventually. But this follows immediate from Proposition 5.20 as \(n \ll (1+\varepsilon)^n\text{.}\)
Example 5.24.
Let us compute the limit of the sequence given by \(a_n =
(e^{2n} + n)^{1/n}\text{.}\)
Note that \(a_n = e^2(1+n/e^{2n})^{1/n}\) so \(\lim
a_n = e^2\) if \(b_n:=(1+n/e^{2n})^{1/n} \to 1\) or equivalently \(\ln b_n \to 0\text{.}\) Since \(n/e^{2n} \to 0\) Proposition 5.20, so it follows from the continuity of \(\ln x\text{,}\) \(\ln(1+n/e^{2n}) \to \ln(1) = 0\text{.}\) Therefore, so does \(\ln b_n = \frac{1}{n}\ln(1+n/e^{2n})\text{.}\)
Here is another way to compute the limit. Note that
\begin{equation*}
e^{2} \lt a_n=e^2\left(1 +\frac{n}{e^{2n}}\right)^{1/n}
\lt e^2(2^{1/n})
\end{equation*}
Since \(n \lt 2^n \lt e^{2n}\text{.}\) By Exercise 5.7.2 \(2^{1/n} \to 1\) and so \(a_n \to e^2\) by the Squeeze Lemma.
Here are two useful observations for computing limit of sequences:
-
If \(\lim_{x\to \infty} f(x) = L\) then \(\lim_{n \to \infty} f(n) = L\text{.}\)
-
\(\lim_{x \to \infty} f(x) = \lim_{x \to 0^+} f(1/x)\text{.}\)
Example 5.25.
\(\ds \lim_{x \to 0} \sin(x)/x = 1\text{,}\)\(\ds \lim_{x \to \infty} x\sin(1/x) = 1\text{.}\)\(n\sin(1/n) \to 1\text{.}\)Example 5.26.
Consider the sequence whose \(n\)th term is given by \(a_n =
n^2\left(1-\cos\left(\frac{c}{n}\right)\right)\) where \(c\) is a constant. Using a double-angle formula for cosine,
\begin{equation*}
a_n = 2n^2\sin^2\left(\frac{c}{2n}\right) =
\frac{c^2}{2}\left(\frac{2n}{c}\sin\left(\frac{c}{2n}\right)\right)^2,
\end{equation*}
By Example 5.25, \(a_n \to c^2/2(1)^2 =c^2/2\text{.}\)
Example 5.27.
Let us find the limit of the sequence given by \(a_n = \sqrt{n^2+4n}
- \sqrt{n^2}\text{.}\) Note that \(\sqrt{n^2+4n} \lt \sqrt{n^2+4n+4} =
n+2\text{,}\) so \(a_n \lt n+2 - \sqrt{n^2} =2\text{.}\) Also
\begin{equation*}
a_n(\sqrt{n^2+4n}+\sqrt{n^2}) = n^2 + 4n - n^2 = 4n.
\end{equation*}
But
\begin{equation*}
a_n(\sqrt{n^2+4n}+\sqrt{n^2}) \lt a_n(n+2 + n) = 2a_n(n+1).
\end{equation*}
Thus,
\begin{equation*}
\frac{4n}{2(n+1)}= \frac{2n}{n+1} \lt a_n.
\end{equation*}
Therefore, \(a_n \to 2\) by the squeeze lemma.
