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homework 3.pdf

1. (a) (10 points) Prove that if \(0 < a < 2\), then \(a < \sqrt{2a} < 2\).

By the density of \(\R\), thus there exists a element \(x\) such that \(a<x<2\)

Then we need to check \(x=\sqrt{2a}\)

cSince \(0<a<2\), then \(2-a>0\)

Then \(0<a(2-a)<(a+2)(2-a)\Rightarrow0<2a-a^2<4-a^2\Rightarrow a^2<2a<4\Rightarrow a < \sqrt{2a} < 2\)​

(b) (25 points) Show that the following sequence converges ​\(\sqrt{2}, \sqrt{2 \sqrt{2}}, \sqrt{2 \sqrt{2 \sqrt{2}}}, \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2}}}}, \dots\)

We know that \(\sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \cdots}}}} = 2 \times 2^{\frac{1}{2}} \times 2^{\frac{1}{4}} \times 2^{\frac{1}{8}} \times \cdots\)

Thus we need to prove \(\prod_{i=1}^{\infty}2^{2^{-i}}\) is convergent.

Then \(\begin{aligned}\prod_{i=1}^{\infty}2^{2^{-i}}=2^{\sum_{i=1}^{\infty}2^{-i}}=2^{\frac{1}{1-\frac12}}=2=\lim_{i\to\infty}\prod_{i=1}^{\infty}2^{2^{-i}}\end{aligned}\) is convergent.

Or if the above proof is not a perfect way, we can use that if \(0 < a < 2\), then \(a < \sqrt{2a} < 2\).

Since we know \(1<\sqrt2<2\), then \(\sqrt2<\sqrt{2\sqrt2}<2\)

Use induction: the basic step holds for \(n=1\)

Suppose \(\sqrt{2\sqrt{2\sqrt{2\sqrt2}}}\ldots<\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt2}}}}\ldots\left(n~square~root\right)<2\)

Then we always have \(\sqrt{2\sqrt{2\sqrt{2\sqrt2}}}\ldots<\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt2}}}}\ldots\left(n+1~square~root\right)<2\)

Thus it is true for all \(n\)​

(c) (5 points) Find the limit. Hint: Note that \((x_{n+1})^2 = 2x_n\), where \(x_n\) denotes the \(n\)-th term of the sequence.

We know that the limit of a subsequence is equal to the limit of sequence.

Then we have \(\lim_{n\to\infty}x_{n}=\lim_{n\to\infty}x_{n+1}\)​

Thus \(\lim_{n\to\infty}\frac{(x_{n+1})^2}{2}=\lim_{n\to\infty}x_{n+1}\Rightarrow\frac{(x_{n+1})^2}{2}=x_{n+1}\Rightarrow x_{n+1}=0,2\)

Since we know \(x_n>0\), then \(\lim_{n\to\infty}x_{n}=2\)​

1. Give an example of each of the following, or state that such a request is impossible by referencing the proper theorem(s).

(a) (8 points) Sequences \((x_n)\) and \((y_n)\) which both diverge but \((x_n + y_n)\) converges.

True

\(x_{n}=\begin{cases}1\text{ if n is even}\\-1\text{ if n is odd}\end{cases}\), \(y_{n}=\begin{cases}1\text{ if n is odd}\\ -1\text{ if n is even}\end{cases}\)

Since \(x_n,y_n\) are not convergent, then they are divergent.

But \(x_n+y_n=0\) which is convergent

(b) (8 points) Sequences \((x_n)\) and \((y_n)\) where \((x_n)\) converges and \((y_n)\) diverges but \((x_n + y_n)\) converges.

False

Since \(x_n\) and \(x_n + y_n\) converge, then \(\lim_{n \to \infty} x_n = L\) and \(\lim_{n \to \infty} (x_n + y_n) = M\).

By property, \(\lim_{n \to \infty} x_n + \lim_{n \to \infty} y_n = M \implies \lim_{n \to \infty} y_n = M - L\).

But \(y_n\) diverges, contradiction.

Thus this proposition is impossible.

(c) (8 points) A convergent sequence \((x_n)\) with \(x_n \neq 0\) for all \(n\) such that \(\left( \frac{1}{x_n} \right)\) diverges.

True

Let \(x_n=\frac1n\), we know \(\frac1n\) is convergent.

Then \(\frac1{x_n}=n\) which is not bounded and don't converge.

(d) (8 points) An unbounded sequence \((x_n)\) and a convergent sequence \((y_n)\) such that \((x_n - y_n)\) is bounded.

False

Since \(y_n\) is convergent then \(y_n\) is also bounded.

Then \(|y_n|<K\) and since \(x_n-y_n\) is bounded, then \(|x_n-y_n|<M\)

Then \(-M<|x_{n}|-|y_{n}|<|x_{n}-y_{n}|<M\Rightarrow\left|x_{n}\right|<M+\left|y_{n}\right|<M+K\)

Then \(x_n\) is bounded which is a contradiction!

(e) (8 points) Sequences \((x_n)\) and \((y_n)\) where \((x_n)\), \((x_n y_n)\) converge but \((y_n)\) diverges.

True

\(x_n=\frac1n,y_n=n\) (\(x_n\) is convergent and \(y_n\) is divergent)

But \(x_n\cdot y_n=1\) is convergent

1. (20 points) Prove that for all real number \(L \in (0,1)\), exists a sequence of rational numbers \((a_n)\) such that \(a_n \in (0,1)\) and \(\lim_{n \to \infty} a_n = L\).

Since the density of \(\mathbb{Q}\) in \(\R\), then \(\forall x,y\in\mathbb{R},\exists a_{n}\in\mathbb{Q}:0<x<a_{n}<y<1\)​

Then let \(x=-\frac{1}{n}+L>0,y=\frac{1}{n}+L<1\), we need choose \(N_0>MAX\{\frac{1}{L},\frac{1}{1-L}\}\)​

Thus \(\forall n\geq N_0\), \(0<L-\frac{1}{n}<a_{n}<L+\frac{1}{n}<1\Rightarrow|a_{n}-L|<\frac{1}{n}\)

Then we take \(n>N\), \(|a_{n}-L|<\frac{1}{n}<\frac{1}{N}\)

Thus we just take \(N>\frac{1}{\varepsilon}\), then we have \(|a_{n}-L|<\frac{1}{n}<\frac{1}{N}<\varepsilon\), \(\forall n>N\)

Then \(\forall\varepsilon>0,\exists N>MAX\left\lbrace N_0,\frac{1}{\varepsilon}\right\rbrace,\forall n>N,\left|a_{n}-L\right|<\varepsilon\)

Thus \(a_n \in (0,1)\) and \(\lim_{n \to \infty} a_n = L\).

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