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Problem 1: Let \((x_n)\) be a sequence such that \(x_2 \neq x_1\) and satisfies the following contractive condition: \(|x_{n+2}-x_{n+1}|\leq C|x_{n+1}-x_{n}|\) for \(\forall n \geq 1\) and some \(0 < C < 1\)
(a) Prove by induction that: \(|x_{n+1}-x_{n}|\leq C^{n-1}|x_2-x_1|\) for all \(n \geq 1\).
Proof
Basic step: When \(n=1\), \(|x_2-x_1|=1\cdot|x_2-x_1|=C^0\cdot|x_2-x_1|\)
Thus \(|x_2-x_1|\leq C^{1-1}|x_2-x_1|\) \(\checkmark\)
Inductive step: Suppose it is true for \(n\), then we have \(|x_{n+1}-x_{n}|\leq C^{n-1}|x_2-x_1|\) (Hypothesis)
We need to prove it is true when \(n+1\): \(|x_{n+2}-x_{n+1}|\leq C^{n}|x_2-x_1|\)
Since \(|x_{n+2}-x_{n+1}|\leq C|x_{n+1}-x_{n}|\), then we connect hypothesis with this: \(|x_{n+2}-x_{n+1}|\leq C|x_{n+1}-x_{n}|\leq C\cdot C^{n-1}\left|x_2-x_1\right|=C^{n}\left|x_2-x_1\right|\)
Thus it is true for \(n+1\)
Thus by induction, \(|x_{n+1}-x_{n}|\leq C^{n-1}|x_2-x_1|\) for all \(n \geq 1\).
(b) Show that the sequence \((x_n)\) is a Cauchy sequence.
To prove that \((x_n)\) is a Cauchy sequence, then N.T.P. \(\forall\varepsilon>0,\exists n_0\) s.t. \(m,n>n_0\Rightarrow|x_{m}-x_{n}|<\varepsilon\)
We need to construct \(|x_m-x_n|<?\)
First, let's try how to construct \(|x_{n+2}-x_{n}|\leq|x_{n+2}-x_{n+1}|+|x_{n+1}-x_{n}|\leq C^{n}|x_2-x_1|+C^{n-1}|x_2-x_1|=C^{n-1}|x_2-x_1|\left(C+1\right)\)
Thus \(|x_{m}-x_{n}|\leq|x_{m}-x_{m-1}|+|x_{m-1}-x_{m-2}|+\cdots+|x_{n+2}-x_{n+1}|+|x_{n+1}-x_{n}|\\\leq C^{m-2}|x_2-x_1|+C^{m-3}|x_2-x_1|+\cdots+C^{n}|x_2-x_1|+C^{n-1}|x_2-x_1|\\=C^{n-1}|x_2-x_1|(C^{m-n-1}+C^{m-n-2}+\cdots+C+1)\\=C^{n-1}|x_2-x_1|(\frac{1(1-C^{m-n})}{1-C})=|x_2-x_1|(\frac{C^{n-1}-C^{m-1}}{1-C})\)
Since \(|x_2-x_1|(\frac{C^{n-1}-C^{m-1}}{1-C})\leq|x_2-x_1|\frac{C^{n-1}}{1-C}\), then we can take \(n_0\) s.t. \(|x_2-x_1|\frac{C^{n_0-1}}{1-C}=\varepsilon\Rightarrow C^{n_0-1}=\frac{\varepsilon\cdot\left(1-C\right)}{|x_2-x_1|}\)
Then \(\left(n_0-1\right)\log C=\log\frac{\varepsilon\cdot\left(1-C\right)}{|x_2-x_1|}\Rightarrow n_0-1=\frac{\log\frac{\varepsilon\cdot\left(1-C\right)}{|x_2-x_1|}}{\log C}\Rightarrow n_0=\frac{\log\frac{\varepsilon\cdot\left(1-C\right)}{|x_2-x_1|}}{\log C}+1\)
Thus we can take \(n_0=\left\lceil\frac{\log\frac{\varepsilon\cdot\left(1-C\right)}{|x_2-x_1|}}{\log C}+1\right\rceil\), let's \(m>n>n_0\)
We have \(|x_{m}-x_{n}|\leq|x_2-x_1|\left(\frac{C^{n-1}-C^{m-1}}{1-C}\right)\leq|x_2-x_1|\frac{1}{1-C}\cdot\left(C^{n-1}-C^{m-1}\right)\\\leq|x_2-x_1|\frac{1}{1-C}\cdot\left(C^{n-1}\right)\leq|x_2-x_1|\frac{1}{1-C}\cdot\left(C^{n_0-1}\right)\)
\(\leq|x_2-x_1|\frac{1}{1-C}\cdot\left(C^{\left\lceil\frac{\log\frac{\varepsilon\cdot\left(1-C\right)}{|x_2-x_1|}}{\log C}\right\rceil+1-1}\right)=|x_2-x_1|\frac{1}{1-C}\cdot\left(C^{\left\lceil\frac{\log\frac{\varepsilon\cdot\left(1-C\right)}{|x_2-x_1|}}{\log C}\right\rceil}\right)\)
\(\leq|x_2-x_1|\frac{1}{1-C}\cdot\left(C^{\frac{\log\frac{\varepsilon\cdot\left(1-C\right)}{|x_2-x_1|}}{\log C}}\right)=\frac{|x_2-x_1|}{1-C}\cdot\left(C^{\log_{c}\frac{\varepsilon\cdot\left(1-C\right)}{|x_2-x_1|}}\right)=\frac{|x_2-x_1|}{1-C}\cdot\left(\frac{\varepsilon\cdot\left(1-C\right)}{|x_2-x_1|}\right)=\varepsilon\)
Thus the sequence \((x_n)\) is a Cauchy sequence.
*Problem 2: Consider the sequence \(\left(x_{n}\right)\) given by: \(x_1 = 2\) and \(x_{n+1} = 2 + \frac{1}{x_n}\). (a) Prove that the sequence \(\left(x_{n}\right)\) is bounded below.
Claim: \((x_n)\) is bounded below by \(2\)
Use induction to prove \(\forall n\geq 2,x_n\geq 2\)
Basic step: When \(n=1\), \(x_1=2\geq 2\)
Inductive step: Suppose it is true for \(n\), then \(x_n\geq 2>0\). N.T.P. \(x_{n+1}\geq 2\)
Since \(x_{n+1} = 2 + \frac{1}{x_n}\), then \(\frac{1}{x_{n}}>0\Rightarrow2+\frac{1}{x_{n}}>2\Rightarrow x_{n+1}>2\)
Thus \(x_1=2\) and \(\forall n\geq 2,x_n\geq 2\Rightarrow\) \((x_n)\) is bounded below by \(2\)
(b) Show that \((x_n)\) satisfies the contractive condition stated in the first exercise and hence it converges.
Since \(|x_{n+2}-x_{n+1}|\leq C|x_{n+1}-x_{n}|\) for \(\forall n \geq 1\) and some \(0 < C < 1\), then use hypothesis: \(|2+\frac{1}{x_{n+1}}-2+\frac{1}{x_{n}}|\leq C|x_{n+1}-x_{n}|\Leftrightarrow|\frac{1}{x_{n+1}}+\frac{1}{x_{n}}|\leq C|x_{n+1}-x_{n}|\)
Then it is equivalent to \(|\frac{x_{n}-x_{n+1}}{x_{n}\cdot x_{n+1}}|\leq C|x_{n+1}-x_{n}|\Leftrightarrow\frac{1}{\left|x_{n}\cdot x_{n+1}\right|}\leq C\) since \(x_n\neq x_{n-1}\)
Thus we only need to prove \(\frac{1}{\left|x_{n}\cdot x_{n+1}\right|}\leq C\) for some \(0 < C < 1\)
Since \(x_n\geq 2\), then \(\left|x_{n}\cdot x_{n+1}\right|\geq4\Rightarrow\frac{1}{\left|x_{n}\cdot x_{n+1}\right|}\leq\frac14\)
Then \(0<C=\frac14<1\) \(\checkmark\)
Therefore we complete the proof.
By homework 1.b \((x_n)\) is a Cauchy sequence, and also by theorem, \((x_n)\) is convergent (c) Find \(\lim_{n \to \infty} x_n\)
Since \((x_n)\) converges, then let \(\lim_{n \to \infty} x_n=L\).
Then we take a subsequence \(x_{n+1}\) and by theorem \(\lim_{n\to\infty}x_{n+1}=L\).
Since \(x_{n+1}=2+\frac{1}{x_{n}}\Rightarrow\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}\left(2+\frac{1}{x_{n}}\right)=\lim_{n\to\infty}2+\lim_{n\to\infty}\frac{1}{x_{n}}\), then \(L=2+\frac1L\)
Hence \(L=1+\sqrt2,1-\sqrt2\), but \(1-\sqrt2<0\) and \(x_n>0\), thus it is not the limit
Hence \(\lim_{n\to\infty}x_{n}=L=1+\sqrt2\)
Problem 3: Give an example for each of the following requests, or explain why the request is impossible to fulfill, referring to the proper theorem(s): (a) Two series \(\sum x_n\) and \(\sum y_n\) that are both divergent but \(\sum x_n y_n\) converges.
Possible, example:
\(\sum x_{n}=\sum\frac{1}{n},\sum y_{n}=\sum\frac{1}{n}\) which are all divergent, but \(\sum x_{n}y_{n}=\sum\frac{1}{n^2}\) is convergent
(b) A convergent series \(\sum x_n\) and a bounded sequence \((y_n)\) such that \(\sum x_n y_n\) diverges.
Possible
\(\sum x_{n}=\sum\frac{\left(-1\right)^{n}}{\sqrt{n}},y_{n}=\left(-1\right)^{n}\leq1\) which series is convergent and \((y_n)\) is bounded, then \(\sum x_{n}y_{n}=\sum\frac{1}{\sqrt{n}}\) is divergent.
Remark: Proof of the convergence of \(\sum\frac{\left(-1\right)^{n}}{\sqrt{n}}\)
Alternating Series Test: \(a_n = \frac{1}{\sqrt{n}}\).
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Since \(\forall\varepsilon>0,\exists n_0=\frac{1}{\varepsilon^2},\forall n>n_0,\left|\frac{1}{\sqrt{n}}\right|=\frac{1}{\sqrt{n}}<\frac{1}{\sqrt{n_0}}=\varepsilon,\) hence \(\begin{aligned}\lim_{n\to\infty}\frac{1}{\sqrt{n}}=0.\end{aligned}\)
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\(\begin{aligned}a_{n+1}=\frac{1}{\sqrt{n+1}}\quad\text{and}\quad a_{n}=\frac{1}{\sqrt{n}}.\end{aligned}Clearly:\begin{aligned}\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}\quad\text{for all}\quad n\geq1.\end{aligned}\)
By theorem, it is convergent
Proof of the divergent of \(\sum\frac{1}{\sqrt{n}}\)
since \(\frac{1}{\sqrt{n}}>\frac{1}{n}>0\) and \(\sum\frac1n\) is divergent, thus \(\frac{1}{\sqrt{n}}\) is divergent by theorem.
(c) Two sequences \((x_n)\) and \((y_n)\) where \(\sum x_n\) and \(\sum x_n y_n\) converge but \(\sum y_n\) diverges.
Possible
\(\sum x_{n}=\sum\frac{1}{n^2},\sum y_{n}=\sum\frac{1}{n},\sum x_{n}y_{n}=\frac{1}{n^3}\)where \(\sum x_n\) and \(\sum x_n y_n\) converge but \(\sum y_n\) diverges.
Remark:
Proof of the convergence of \(\sum\frac{1}{n^2}\)
Since \(0\leq\frac{1}{n^2}<\frac{1}{n\left(n-1\right)},\forall n\geq2\)
If I prove that \(\sum_{n=2}^{\infty}\frac{1}{n\left(n-1\right)}_{}<\infty\), then by theorem \(\sum_{n=2}^{\infty}\frac{1}{n^2}_{}<\infty\)
By telescope sums: \(\sum_{n=2}^{\infty}\frac{1}{n\left(n-1\right)}_{}=1-\frac{1}{n}<\infty\)
Thus \(\sum_{n=2}^{\infty}\frac{1}{n\left(n-1\right)}\)is convergent, then \(\sum\frac{1}{n^2}\) is convergent
Proof of the convergence of \(\sum\frac{1}{n^3}\)
Since \(0\leq\frac{1}{n^3}<\frac{1}{n^2},\forall n\geq2\) and \(\sum\frac{1}{n^2}\) is convergent, then by theorem \(\sum\frac{1}{n^3}\) is convergent
Problem 4: Study the following series. Decide if they are convergent or not. If they are convergent, compute the value of the series. (a) \(\sum_{n=1}^{\infty}-\frac{15}{(n+3)(n+4)}\)
Yes, it is convergent
Since \(\frac{-15}{(n+3)(n+4)} = (-15) \cdot \left( \frac{1}{n+3} - \frac{1}{n+4} \right)\)
then \(\sum_{n=1}^{\infty}-\frac{15}{(n+3)(n+4)}=\lim_{n\to\infty}\sum(-15)\cdot\left(\frac{1}{n+3}-\frac{1}{n+4}\right)\)
\(= \lim_{n \to \infty} \left( -15 \cdot \left( \frac{1}{4} - \frac{1}{n+4} \right) \right)\)\(= -\frac{15}{4} + \lim_{n \to \infty} \frac{15}{n+4} = -\frac{15}{4}\)
(b) \(\sum_{n=1}^{\infty}\frac{(-1)^{n-1}2^{n}}{9^{n-1}}\)
Yes, it is convergent
Since \(\frac{(-1)^{n-1}2^{n}}{9^{n-1}}=2\cdot\frac{(-2)^{n-1}}{9^{n-1}}=2\cdot(\frac{-2}{9})^{n-1}\), thus \(S_{n}=2\cdot\left(\frac{1\left(1-\left(-\frac29\right)^{n}\right)}{1+\frac29}\right)=\frac{18\left(1-\left(-\frac29\right)^{n}\right)}{11}=\frac{18}{11}-\frac{\left.18\left(-\frac29\right)^{n}\right)}{11}\)
Since \(-\frac29<1\), then \(\lim_{n\to\infty}\left(-\frac29\right)^{n}=0\)
Then \(\sum_{n=1}^{\infty}\frac{(-1)^{n-1}2^{n}}{9^{n-1}}=\lim_{n\to\infty}S_{n}=\lim_{n\to\infty}\left(\frac{18}{11}-\frac{\left.18\left(-\frac29\right)^{n}\right)}{11}\right)=\frac{18}{11}-\lim_{n\to\infty}\left(\frac{\left.18\left(-\frac29\right)^{n}\right)}{11}\right)=\frac{18}{11}\)