11.7 Sequence
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Compute the following limits
- (a) \(\lim_{n \to \infty} \left(1 + \frac{2}{n}\right)^n\)
Use sandwich theorem:
\(\left(1+\frac{1}{n+1}\right)^2\leq1+\frac{2}{n}\leq\left(1+\frac{1}{n}\right)^2=1+\frac{2}{n}+\frac{1}{n^2}\) Easily to check
Then \(\left(1+\frac{1}{n+1}\right)^{2n}\leq\left(1+\frac{2}{n}\right)^{n}\leq\left(1+\frac{1}{n}\right)^{2n}\)
Since \(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{2n}=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}\cdot\left(1+\frac{1}{n}\right)^{n}=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}\cdot\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}=e^2\)
Since \(\lim_{n\to\infty}\left(1+\frac{1}{n+1}\right)^{2n}=\left\lbrack\lim_{n\to\infty}\left(1+\frac{1}{n+1}\right)^{n+1}\cdot\left(1+\frac{1}{n+1}\right)^{-1}\right\rbrack^2=\left\lbrack e\cdot\lim_{n\to\infty}\left(1+\frac{1}{n+1}\right)\right\rbrack^2=e^2\)
Thus \(\lim_{n \to \infty} \left(1 + \frac{2}{n}\right)^n=e^2\) * (b) \(\lim_{n\to\infty}\left(\frac{n+3}{n+1}\right)^{n}\)
\(\lim_{n\to\infty}\left(\frac{n+3}{n+1}\right)^{n}=\lim_{n\to\infty}\left(1+\frac{2}{n+1}\right)^{n}\)
\(1+\frac{2}{n+2}\leq\left(1+\frac{2}{n+1}\right)\leq1+\frac{2}{n}\)
\(\lim_{n\to\infty}\left(1+\frac{2}{n+2}\right)^{n}=\lim_{n\to\infty}\left(1+\frac{2}{n+2}\right)^{n+2}\div\lim_{n\to\infty}\left(1+\frac{2}{n+2}\right)^2=e^2\)
Or we can
\(\lim_{n\to\infty}\left(\frac{n+3}{n+1}\right)^{n}=\lim_{n\to\infty}\left[\frac{n\left(1+\frac{3}{n}\right)}{n\left(1+\frac{1}{n}\right)}\right]^{n}=\lim_{n\to\infty}\left(\frac{1+\frac{3}{n}}{1+\frac{1}{n}}\right)^{n}\)
Thus \(\lim_{n\to\infty}\frac{\left(1+\frac{3}{n}\right)^{n}}{\left(1+\frac{1}{n}\right)^{n}}=e^2\)
Exercise: \(\lim_{n \to \infty} \left(1 + \frac{3}{n}\right)^n = e^3\)
\((1+\frac{1}{n+1})^3\leq1+\frac{3}{n}\leq(1+\frac{1}{n})^3\Rightarrow(1+\frac{1}{n+1})^{3n}\leq\left(1+\frac{3}{n}\right)^{n}\leq\left(1+\frac{1}{n}\right)^{3n}\) * (c) \(\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^n\)
\(\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^{n}=\lim_{n\to\infty}\left(\frac{n-1}{n}\right)^{n}=\lim_{n\to\infty}\left(\frac{1}{\frac{n}{n-1}}\right)^{n}\)
\(= \lim_{n \to \infty} \left(\frac{1}{1 + \frac{1}{n-1}}\right)^n = \lim_{n \to \infty} \frac{1}{\left(1 + \frac{1}{n-1}\right)^n}\)\(= \frac{1}{\lim_{n \to \infty} \left(1 + \frac{1}{n-1}\right)^n} = \frac{1}{e} = e^{-1}\) * (d) \(\lim_{n \to \infty} \frac{n!}{n^n}\)
\(\frac{a_{n+1}}{a_{n}}=\frac{(n+1)!\,n^{n}}{(n+1)^{n+1}\,n!}=\frac{n^{n}}{(n+1)^{n}}=\left(\frac{1}{1+\frac{1}{n}}\right)^{n}<1\) \(\Rightarrow a_{n+1} < a_n \Rightarrow a_n\) is monotonic decreasing.
Moreover, \(a_n = \frac{n!}{n^n} > 0\), so \(a_n\) has a lower bound.
By monotonic theorem: \(\lim_{n \to \infty} \frac{n!}{n^n} = L\) and \(0 \leq L\).
\(a_{n+1}=\frac{(n+1)!}{(n+1)^{n+1}}=\frac{\left(n+1\right)n!}{\left(n+1\right)\left(n+1\right)^{n}}=\frac{n!\cdot n^{n}}{(n+1)^{n}\cdot n^{n}}=\frac{n!}{n^{n}}\cdot\left(\frac{n}{n+1}\right)^{n}=a_{n}\cdot\left(\frac{n}{n+1}\right)^{n}\)
Thus, \(L=\lim_{n\to\infty}a_{n}\cdot\lim_{n\to\infty}\frac{1}{\left(1+\frac{1}{n}\right)^{n}}=L\cdot e^{-1}\)
\(L = \frac{L}{e} \Rightarrow eL = L \Rightarrow (e - 1)L = 0 \Rightarrow L = 0\)
or use sandwich theorem
\(\frac{2^{n}}{n^{n}}\leq\frac{n!}{n^{n}}=\frac{1\cdot2\cdot3\cdot\ldots\cdot n}{n\cdot n\cdot\ldots\cdot n}\leq\frac{1\cdot n\cdot n\cdot\ldots\cdot n}{n\cdot n\cdot\ldots\cdot n}=\frac{1}{n}\)
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Prove using the definition of Cauchy sequence that if \((x_n)\) and \((y_n)\) are Cauchy sequences, then \((x_n + y_n)\), \((x_n y_n)\) are Cauchy sequences.
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\(x_ny_n\)
\(x_n\) is a Cauchy sequence, then for \(\epsilon > 0\), there exists \(N_1\) such that \(|x_m - x_n| < \epsilon\) for all \(m, n \geq N_1\).
\(y_n\) is a Cauchy sequence, then for \(\epsilon > 0\), there exists \(N_2\) such that \(|y_m - y_n| < \epsilon\) for all \(m, n \geq N_2\).
I want to prove that \(x_n y_n\) is a Cauchy sequence. This implies for \(\epsilon > 0\), find \(N\) such that \(|x_m y_m - x_n y_n| < \epsilon\) for all \(m, n \geq N\)
Let's analyze:
\(|x_m y_m - x_n y_n| = |x_m y_m - x_n y_n + x_m y_n - x_m y_n| = |x_m (y_m - y_n) + y_n (x_m - x_n)|\)
\(\leq |x_m| |y_m - y_n| + |y_n| |x_m - x_n|\)
Since \(x_n\) and \(y_n\) are convergent sequences, they are bounded. Thus, there exist \(A\) and \(B\) such that \(|x_n| \leq A\) and \(|y_n| \leq B\).
\(\leq A |y_m - y_n| + B |x_m - x_n| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\)
Take \(N = \max\{N_1, N_2\}\), so for all \(m, n \geq N\), we have \(|x_m y_m - x_n y_n| < \epsilon\).\(m, n \geq N\) 3. \(x_n+y_n\)
\(x_n\) is a Cauchy sequence, then for \(\epsilon > 0\), there exists \(N_1\) such that \(|x_{m}-x_{n}|<\frac{\epsilon}{2}\) for all \(m, n \geq N_1\).
\(y_n\) is a Cauchy sequence, then for \(\epsilon > 0\), there exists \(N_2\) such that \(|y_{m}-y_{n}|<\frac{\epsilon}{2}\) for all \(m, n \geq N_2\).
I want to prove that \(x_{n}+y_{n}\) is a Cauchy sequence. This implies for \(\epsilon > 0\), find \(N\) such that \(|x_{m}+y_{m}-x_{n}+y_{n}|<\epsilon\) for all \(m, n \geq N\)
Since \(|x_{m}+y_{m}-\left(x_{n}+y_{n}\right)|=|\left(x_{m}-x_{n}\right)+\left(y_{m}-y_{n}\right)|\leq|x_{m}-x_{n}|+|y_{m}-y_{n}|\leq\varepsilon\) for all \(m,n\geq N=\max\left\lbrace N_1,N_2\right\rbrace\), then we complete the proof
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Prove that if \(x_n\) is a convergent sequence (or a Cauchy sequence), then the sequence \(|x_{n+1} - x_n|\) converges to 0.
Proof
Let \(b_n = x_{n+1} - x_n\). Since \(x_n\) is convergent \(\left(\lim_{n \to \infty} x_n = L\right)\),
\(\lim_{n \to \infty} b_n = \lim_{n \to \infty} (x_{n+1} - x_n) = L - L = 0\)
Thus, \(\lim_{n \to \infty} |b_n| = \lim_{n \to \infty} |x_{n+1} - x_n| = 0\)
By exercise homework 2, \(\lim_{n \to \infty} a_n = 0 \Rightarrow \lim_{n \to \infty} |a_n| = 0\) The converse is not true: find an example of a non-convergent sequence with this property.
Counter example: \(a_n=\sqrt{n}\), \(\lim_{n\to\infty}\left|\sqrt{n+1}-\sqrt{n}\right|=0\) and \(\lim_{n\to\infty}\sqrt{n}\) diverges
Since \(\left|\sqrt{n+1} - \sqrt{n}\right| = \left|\frac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{\sqrt{n+1} + \sqrt{n}}\right|\)\(= \left|\frac{(n+1) - n}{\sqrt{n+1} + \sqrt{n}}\right| = \frac{1}{\sqrt{n+1} + \sqrt{n}}\)
Suppose \(\lim_{n\to\infty}\sqrt{n}=L\Rightarrow\lim_{n\to\infty}n=L^2\) contradiction!
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Let \((x_n)\) be a sequence, prove that if \(0 \leq r < 1\), \(C > 0\) and \(|x_{n+1} - x_n| < Cr^n\), then \((x_n)\) is a Cauchy sequence.
Proof:
We wish to prove that for \(\epsilon > 0\), there exists \(N\) such that \(|x_m - x_n| < \epsilon\) for all \(m, n \geq N\).
For \(m > n\), \(|x_m - x_n| = |x_m - x_{m-1} + x_{m-1} - x_{m-2} + \dots + x_{n+1} - x_n|\)
\(\leq |x_m - x_{m-1}| + |x_{m-1} - x_{m-2}| + \dots + |x_{n+1} - x_n|\)
\(< C \cdot r^{m-1} + C \cdot r^{m-2} + \dots + C \cdot r^n\)
\(= C(r^{m-1} + r^{m-2} + \dots + r^n) < \frac{C \cdot r^n}{1 - r}\) (since \(r < 1\) and using the sum of a geometric series)
We have: \(Cr^{n}(1+r+r^2+\dots+r^{m-1-n})\leq\frac{Cr^{n}\left(1-r^{n}\right)}{1-r}<\) \(\sum_{i=0}^{\infty}r^{i}=\frac{Cr^{n}}{1-r}\) for \(0 < r < 1\).
We want \(\frac{C r^n}{1 - r} < \epsilon\)
Applying the properties of logarithms: \(\ln \left(\frac{C r^n}{1 - r}\right) < \epsilon\)
This implies: \(n > \frac{\ln \left(\frac{\epsilon (1 - r)}{C}\right)}{\ln(r)}\)
We will accept this solution.
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Show that the sequence \((x_n)\) defined below is a Cauchy sequence.
\(x_1 = 1\) and \(x_{n+1} = 1 + \frac{1}{x_n}\) for all \(n \geq 1\).
\(x_2 = 2\), \(x_3 = \frac{3}{2}\), \(x_4 = \frac{5}{3}\), \(x_5 = \frac{8}{5}\), \(x_6 = \frac{13}{8}\)
\(x_1 < x_2\), \(x_2 > x_3\), \(x_3 < x_4\), \(x_4 > x_5\), \(x_5 < x_6\)
Not monotonic! Thus we cannot use traditional definition of convergence.
However, Cauchy sequence is a good try. From above questions, we can try to construct \(|x_{n+1} - x_n| < Cr^n\)
Claim 1: \(1 \leq x_n \leq 2\) for all \(n\). By induction
Claim 2: \(|x_{n+1} - x_n| < 2 \left(\frac{1}{2}\right)^n\) for all \(n \geq 3\). Also induction
Then by exercise 4, it is a Cauchy Sequence, thus it is convergent.
Then \(\exists\lim_{n\to\infty}x_{n}=L,1\leq L=\lim x_{n}\leq2\)
Thus \(\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}\left(1+\frac{1}{x_{n}}\right)\Rightarrow L=1+\frac{1}{L}\Rightarrow L=\frac{1+\sqrt5}{2}\)