[Chapter 1~6] Problems #1

[Problem 0.0]

Let $a_n={\displaystyle (1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots +)-2\sqrt{n+1}}$.  Prove that $\{a_n\}$ has a limit.

Solution ▶

(Proof)

$\{a_n\}$ is strictly increasing and bounded above.($<1$) (See the figure below.)

 

$a_n=\text{(area of shaded rectangles)}-{\int }_1^{n+1}\frac{1}{\sqrt{x}}dx$

Thus $\{a_n\}$ is convergent by monotone convergence theorem. 

Solution ▶

 

[Problem 0.1]

Prove that the sequence $a_n={\displaystyle \frac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdot 6\cdots (2n)}}$ has a limit.

Solution ▶

(Proof)

For $n\in \mathbb{N}$, $$\begin{array}{rl}& 1\cdot 3<2^2\\ & 3\cdot 5<4^2\\ & \cdots \\ & (2n-1)(2n+1)<(2n{)}^2\end{array}$$ Multiplying both sides, $$\begin{array}{rl}& 1^2\cdot 3^2\cdots (2n-1{)}^2\cdot (2n+1)<2^2\cdot 4^2\cdots (2n{)}^2\\ & \Rightarrow \frac{1^2\cdot 3^2\cdots (2n-1{)}^2}{2^2\cdot 4^2\cdots (2n{)}^2}<\frac{1}{2n+1}\\ & \Rightarrow \frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}<\frac{1}{\sqrt{2n+1}}\end{array}$$ Since $a_n>0$ and ${\displaystyle \frac{1}{\sqrt{2n+1}}\to 0\text{as}n\to \mathrm{\infty }}$, we have $a_n\to 0$ by squeeze theorem.

Solution ▶

 

[Problem 0.2]

Prove that $n^2{a}^n\to 0$ if $|a|<1$. 

Solution ▶

(Proof)

$n^2{a}^n\to 0$ if and only if $n^2|a{|}^n\to 0$. Let $|a|={\displaystyle \frac{1}{1+k}}$ where $k>0$. By binomial theorem, $$(1+k{)}^n={\displaystyle \sum _{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)k^i\ge 1+nk+\frac{n(n-1)}{2}{k}^2+\frac{n(n-1)(n-2)}{3!}{k}^3}$$ Hence  $$\begin{array}{rl}0<n^2|a{|}^n& =n^2/(1+k{)}^n\\ & \le {\displaystyle \frac{n^2}{1+nk+\frac{n(n-1)}{2}{k}^2+\frac{n(n-1)(n-2)}{3!}{k}^3}\to 0\text{as}n\to \mathrm{\infty }}\end{array}$$

Solution ▶

 

[Problem 0.3]

Show that ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\int }_0^1(1-x^2{)}^{n}dx=0}$, without attempting to evaluate the integral explicitly.

Solution ▶

(Proof)

Let $f_n(x)=(1-x^2{)}^n$. By MVT, $$\mathrm{\exists }c\in (0,1):{\int }_0^1(1-x^2{)}^{n}dx=f_n(c)=(1-c^2{)}^n$$ Since $0<1-c^2<1$,  $(1-c^2{)}^n\to 0$. Hence ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\int }_0^1(1-x^2{)}^{n}dx=0}$. 

Solution ▶

 

[Problem 0.4]

Let $S$ and $T$ be non-empty bounded subsets of $\mathbb{R}$, and suppose that for all $s\in S$ and $t\in T$, we have $s\le t$. Prove that $supS\le infT$.

Solution ▶

(Proof)

Suppose $supS:=\alpha >\beta :=infT$ and let $ϵ=(\alpha -\beta )/2$. By definition of supremum and infimum, $$\begin{array}{rl}& \mathrm{\exists }s\in S:\alpha -ϵ=(\alpha +\beta )/2<s\le \alpha ,\\ & \mathrm{\exists }t\in T:\beta \le t<\beta +ϵ=(\alpha +\beta )/2\\ & \Rightarrow t<(\alpha +\beta )/2<s\end{array}$$ This contrary to the assumption. ($s\le t$)

Solution ▶

 

[Problem 0.5]

Let $\{a_n\}$ and $\{b_n\}$ be sequences such that $|a_{n+1}-a_n|\le b_n$ for all $n\in \mathbb{N}$. Define $B_n={\displaystyle \sum _{k=1}^n{b}_k}$ for all $n\in \mathbb{N}$. If $\{B_n\}$ is convergent, then prove that $\{a_n\}$ is convergent. 

Solution ▶

(Proof)

Since $\{B_n\}$ is convergent, it is also a Cauchy sequence. i,e, $$\mathrm{\exists }N\in \mathbb{N}:m>n>N\Rightarrow |B_m-B_{n-1}|=\left|\sum _{k=n}^m{b}_k\right|<ϵ$$ Hence, $$\begin{array}{rl}|a_m-a_n|& \le |a_{n+1}-a_n|+|a_{n+2}-a_{n+1}|+\cdots +|a_m-a_{m-1}|\\ & \le b_n+b_{n+1}+\cdots +b_m\\ & \le \left|\sum _{k=n}^m{a}_k\right|<ϵ\end{array}$$ This shows $\{a_n\}$ is a Cauchy sequence, so that is convergent.

Solution ▶

 

Reference: <Introduction to Analysis> Arthur Mattuck, 1999.